Greatest subsequential sum
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Given a sequence of integers, find a continuous subsequence which maximizes the sum of its elements, that is, the elements of no other single subsequence add up to a value larger than this one.
An empty subsequence is considered to have the sum of 0; thus if all elements are negative, the result must be the empty sequence.
11l
F maxsumseq(sequence)
V (start, end, sum_start) = (-1, -1, -1)
V (maxsum_, sum_) = (0, 0)
L(x) sequence
sum_ += x
I maxsum_ < sum_
maxsum_ = sum_
(start, end) = (sum_start, L.index)
E I sum_ < 0
sum_ = 0
sum_start = L.index
assert(maxsum_ == sum(sequence[start + 1 .. end]))
R sequence[start + 1 .. end]
print(maxsumseq([-1, 2, -1]))
print(maxsumseq([-1, 2, -1, 3, -1]))
print(maxsumseq([-1, 1, 2, -5, -6]))
print(maxsumseq([-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]))- Output:
[2] [2, -1, 3] [1, 2] [3, 5, 6, -2, -1, 4]
Action!
PROC PrintArray(INT ARRAY a INT first,last)
INT i
Put('[)
FOR i=first TO last
DO
IF i>first THEN Put(' ) FI
PrintI(a(i))
OD
Put(']) PutE()
RETURN
PROC Process(INT ARRAY a INT size)
INT i,j,beg,end
INT sum,best
beg=0 end=-1 best=0
FOR i=0 TO size-1
DO
sum=0
FOR j=i TO size-1
DO
sum==+a(j)
IF sum>best THEN
best=sum
beg=i
end=j
FI
OD
OD
Print("Seq=") PrintArray(a,0,size-1)
PrintF("Max sum=%i %ESubseq=",best)
PrintArray(a,beg,end) PutE()
RETURN
PROC Main()
INT ARRAY
a(11)=[1 2 3 4 5 65528 65527 65516 40 25 65531],
b(11)=[65535 65534 3 5 6 65534 65535 4 65532 2 65535],
c(5)=[65535 65534 65533 65532 65531],
d(0)=[]
Process(a,11)
Process(b,11)
Process(c,5)
Process(d,0)
RETURN- Output:
Screenshot from Atari 8-bit computer
Seq=[1 2 3 4 5 -8 -9 -20 40 25 -5] Max sum=65 Subseq=[40 25] Seq=[-1 -2 3 5 6 -2 -1 4 -4 2 -1] Max sum=15 Subseq=[3 5 6 -2 -1 4] Seq=[-1 -2 -3 -4 -5] Max sum=0 Subseq=[] Seq=[] Max sum=0 Subseq=[]
Ada
with Ada.Text_Io; use Ada.Text_Io;
procedure Max_Subarray is
type Int_Array is array (Positive range <>) of Integer;
Empty_Error : Exception;
function Max(Item : Int_Array) return Int_Array is
Start : Positive;
Finis : Positive;
Max_Sum : Integer := Integer'First;
Sum : Integer;
begin
if Item'Length = 0 then
raise Empty_Error;
end if;
for I in Item'range loop
Sum := 0;
for J in I..Item'Last loop
Sum := Sum + Item(J);
if Sum > Max_Sum then
Max_Sum := Sum;
Start := I;
Finis := J;
end if;
end loop;
end loop;
return Item(Start..Finis);
end Max;
A : Int_Array := (-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1);
B : Int_Array := Max(A);
begin
for I in B'range loop
Put_Line(Integer'Image(B(I)));
end loop;
exception
when Empty_Error =>
Put_Line("Array being analyzed has no elements.");
end Max_Subarray;
Aime
gsss(list l, integer &start, &end, &maxsum)
{
integer e, f, i, sum;
end = f = maxsum = start = sum = 0;
for (i, e in l) {
sum += e;
if (sum < 0) {
sum = 0;
f = i + 1;
} elif (maxsum < sum) {
maxsum = sum;
end = i + 1;
start = f;
}
}
}
main(void)
{
integer start, end, sum;
list l;
l = list(-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1);
gsss(l, start, end, sum);
o_("Max sum ", sum, "\n");
if (start < end) {
l.ocall(o_, 1, start, end - 1, " ");
o_newline();
}
0;
}- Output:
Max sum 15 3 5 6 -2 -1 4
ALGOL 68
main:
(
[]INT a = (-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1);
INT begin max, end max, max sum, sum;
sum := 0;
begin max := 0;
end max := -1;
max sum := 0;
FOR begin FROM LWB a TO UPB a DO
sum := 0;
FOR end FROM begin TO UPB a DO
sum +:= a[end];
IF sum > max sum THEN
max sum := sum;
begin max := begin;
end max := end
FI
OD
OD;
FOR i FROM begin max TO end max DO
print(a[i])
OD
)- Output:
+3 +5 +6 -2 -1 +4
AppleScript
Linear derivation of both sum and list, in a single fold:
-- maxSubseq :: [Int] -> [Int] -> (Int, [Int])
on maxSubseq(xs)
script go
on |λ|(ab, x)
set a to fst(ab)
set {m1, m2} to {fst(a), snd(a)}
set high to max(Tuple(0, {}), Tuple(m1 + x, m2 & {x}))
Tuple(high, max(snd(ab), high))
end |λ|
end script
snd(foldl(go, Tuple(Tuple(0, {}), Tuple(0, {})), xs))
end maxSubseq
-- TEST ---------------------------------------------------
on run
set mx to maxSubseq({-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1})
{fst(mx), snd(mx)}
end run
-- GENERIC ABSTRACTIONS -----------------------------------
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- gt :: Ord a => a -> a -> Bool
on gt(x, y)
set c to class of x
if record is c or list is c then
fst(x) > fst(y)
else
x > y
end if
end gt
-- fst :: (a, b) -> a
on fst(tpl)
if class of tpl is record then
|1| of tpl
else
item 1 of tpl
end if
end fst
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- max :: Ord a => a -> a -> a
on max(x, y)
if gt(x, y) then
x
else
y
end if
end max
-- snd :: (a, b) -> b
on snd(tpl)
if class of tpl is record then
|2| of tpl
else
item 2 of tpl
end if
end snd
-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple
- Output:
{15, {3, 5, 6, -2, -1, 4}}
Arturo
subarraySum: function [arr][
curr: 0
mx: 0
fst: size arr
lst: 0
currFst: 0
loop.with: 'i arr [e][
curr: curr + e
if e > curr [
curr: e
currFst: i
]
if curr > mx [
mx: curr
fst: currFst
lst: i
]
]
if? lst > fst -> return @[mx, slice arr fst lst]
else -> return [0, []]
]
sequences: @[
@[1, 2, 3, 4, 5, neg 8, neg 9, neg 20, 40, 25, neg 5]
@[neg 1, neg 2, 3, 5, 6, neg 2, neg 1, 4, neg 4, 2, neg 1]
@[neg 1, neg 2, neg 3, neg 4, neg 5]
@[]
]
loop sequences 'seq [
print [pad "sequence:" 15 seq]
processed: subarraySum seq
print [pad "max sum:" 15 first processed]
print [pad "subsequence:" 15 last processed "\n"]
]
- Output:
sequence: [1 2 3 4 5 -8 -9 -20 40 25 -5]
max sum: 65
subsequence: [40 25]
sequence: [-1 -2 3 5 6 -2 -1 4 -4 2 -1]
max sum: 15
subsequence: [3 5 6 -2 -1 4]
sequence: [-1 -2 -3 -4 -5]
max sum: 0
subsequence: []
sequence: []
max sum: 0
subsequence: []
ATS
(*
** This one is
** translated into ATS from the Ocaml entry
*)
(* ****** ****** *)
//
// How to compile:
// patscc -DATS_MEMALLOC_LIBC -o maxsubseq maxsubseq.dats
//
(* ****** ****** *)
//
#include
"share/atspre_staload.hats"
//
(* ****** ****** *)
typedef ints = List0(int)
(* ****** ****** *)
fun
maxsubseq
(xs: ints): (int, ints) = let
//
fun
loop
(
sum: int, seq: ints
, maxsum: int, maxseq: ints, xs: ints
) : (int, ints) =
(
case+ xs of
| nil () =>
(
maxsum
, list_vt2t(list_reverse(maxseq))
) (* end of [nil] *)
| cons (x, xs) => let
val sum = sum + x
and seq = cons (x, seq)
in
if sum < 0
then loop (0, nil, maxsum, maxseq, xs)
else (
if sum > maxsum
then loop (sum, seq, sum, seq, xs)
else loop (sum, seq, maxsum, maxseq, xs)
) (* end of [else] *)
end // end of [cons]
)
//
in
loop (0, nil, 0, nil, xs)
end // end of [maxsubseq]
implement
main0 () = () where
{
val
(maxsum
,maxseq) =
maxsubseq
(
$list{int}(~1,~2,3,5,6,~2,~1,4,~4,2,~1)
)
//
val () = println! ("maxsum = ", maxsum)
val () = println! ("maxseq = ", maxseq)
//
} (* end of [main0] *)- Output:
maxsum = 15 maxseq = 3, 5, 6, -2, -1, 4
AutoHotkey
classic algorithm:
seq = -1,-2,3,5,6,-2,-1,4,-4,2,-1
max := sum := start := 0
Loop Parse, seq, `,
If (max < sum+=A_LoopField)
max := sum, a := start, b := A_Index
Else If sum <= 0
sum := 0, start := A_Index
; read out the best subsequence
Loop Parse, seq, `,
s .= A_Index > a && A_Index <= b ? A_LoopField "," : ""
MsgBox % "Max = " max "`n[" SubStr(s,1,-1) "]"
AutoIt
Local $iArray[11] = [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
GREAT_SUB($iArray)
Local $iArray[5] = [-1, -2, -3, -4, -5]
GREAT_SUB($iArray)
Local $iArray[15] = [7, -6, -8, 5, -2, -6, 7, 4, 8, -9, -3, 2, 6, -4, -6]
GREAT_SUB($iArray)
Func GREAT_SUB($iArray)
Local $iSUM = 0, $iBEGIN_MAX = 0, $iEND_MAX = -1, $iMAX_SUM = 0
For $i = 0 To UBound($iArray) - 1
$iSUM = 0
For $k = $i To UBound($iArray) - 1
$iSUM += $iArray[$k]
If $iSUM > $iMAX_SUM Then
$iMAX_SUM = $iSUM
$iEND_MAX = $k
$iBEGIN_MAX = $i
EndIf
Next
Next
ConsoleWrite("> Array: [")
For $i = 0 To UBound($iArray) - 1
If $iArray[$i] > 0 Then ConsoleWrite("+")
ConsoleWrite($iArray[$i])
If $i <> UBound($iArray) - 1 Then ConsoleWrite(",")
Next
ConsoleWrite("]" & @CRLF & "+>Maximal subsequence: [")
$iSUM = 0
For $i = $iBEGIN_MAX To $iEND_MAX
$iSUM += $iArray[$i]
If $iArray[$i] > 0 Then ConsoleWrite("+")
ConsoleWrite($iArray[$i])
If $i <> $iEND_MAX Then ConsoleWrite(",")
Next
ConsoleWrite("]" & @CRLF & "!>SUM of subsequence: " & $iSUM & @CRLF)
EndFunc ;==>GREAT_SUB
- Output:
> Array: [-1,-2,+3,+5,+6,-2,-1,+4,-4,+2,-1] +>Maximal subsequence: [+3,+5,+6,-2,-1,+4] !>SUM of subsequence: 15 > Array: [-1,-2,-3,-4,-5] +>Maximal subsequence: [] !>SUM of subsequence: 0 > Array: [+7,-6,-8,+5,-2,-6,+7,+4,+8,-9,-3,+2,+6,-4,-6] +>Maximal subsequence: [+7,+4,+8] !>SUM of subsequence: 19
AWK
# Finds the subsequence of ary[1] to ary[len] with the greatest sum.
# Sets subseq[1] to subseq[n] and returns n. Also sets subseq["sum"].
# An empty subsequence has sum 0.
function maxsubseq(subseq, ary, len, b, bp, bs, c, cp, i) {
b = 0 # best sum
c = 0 # current sum
bp = 0 # position of best subsequence
bn = 0 # length of best subsequence
cp = 1 # position of current subsequence
for (i = 1; i <= len; i++) {
c += ary[i]
if (c < 0) {
c = 0
cp = i + 1
}
if (c > b) {
b = c
bp = cp
bn = i + 1 - cp
}
}
for (i = 1; i <= bn; i++)
subseq[i] = ary[bp + i - 1]
subseq["sum"] = b
return bn
}
Demonstration:
# Joins the elements ary[1] to ary[len] in a string.
function join(ary, len, i, s) {
s = "["
for (i = 1; i <= len; i++) {
s = s ary[i]
if (i < len)
s = s ", "
}
s = s "]"
return s
}
# Demonstrates maxsubseq().
function try(str, ary, len, max, maxlen) {
len = split(str, ary)
print "Array: " join(ary, len)
maxlen = maxsubseq(max, ary, len)
print " Maximal subsequence: " \
join(max, maxlen) ", sum " max["sum"]
}
BEGIN {
try("-1 -2 -3 -4 -5")
try("0 1 2 -3 3 -1 0 -4 0 -1 -4 2")
try("-1 -2 3 5 6 -2 -1 4 -4 2 -1")
}
- Output:
Array: [-1, -2, -3, -4, -5] Maximal subsequence: [], sum 0 Array: [0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2] Maximal subsequence: [0, 1, 2], sum 3 Array: [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] Maximal subsequence: [3, 5, 6, -2, -1, 4], sum 15
BBC BASIC
DIM A%(11) : A%() = 0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2
PRINT FNshowarray(A%()) " -> " FNmaxsubsequence(A%())
DIM B%(10) : B%() = -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1
PRINT FNshowarray(B%()) " -> " FNmaxsubsequence(B%())
DIM C%(4) : C%() = -1, -2, -3, -4, -5
PRINT FNshowarray(C%()) " -> " FNmaxsubsequence(C%())
END
DEF FNmaxsubsequence(a%())
LOCAL a%, b%, i%, j%, m%, s%, a$
a% = 1
FOR i% = 0 TO DIM(a%(),1)
s% = 0
FOR j% = i% TO DIM(a%(),1)
s% += a%(j%)
IF s% > m% THEN
m% = s%
a% = i%
b% = j%
ENDIF
NEXT
NEXT i%
IF a% > b% THEN = "[]"
a$ = "["
FOR i% = a% TO b%
a$ += STR$(a%(i%)) + ", "
NEXT
= LEFT$(LEFT$(a$)) + "]"
DEF FNshowarray(a%())
LOCAL i%, a$
a$ = "["
FOR i% = 0 TO DIM(a%(),1)
a$ += STR$(a%(i%)) + ", "
NEXT
= LEFT$(LEFT$(a$)) + "]"
- Output:
[0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2] -> [0, 1, 2] [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] -> [3, 5, 6, -2, -1, 4] [-1, -2, -3, -4, -5] -> []
Bracmat
This program iterates over all subsequences by forced backtracking, caused by the failing node ~ at the end of the middle part of the pattern. The combination of flags [% on an expression creates a pattern that succeeds if and only if the expression is successfully evaluated. sjt is an extra argument to any function that is part of a pattern and it contains the current subexpression candidate. Inside the pattern the function sum sums over all elements in sjt. The currently longest maximal subsequence is kept in seq.
( 0:?max
& :?seq
& -1 -2 3 5 6 -2 -1 4 -4 2 -1
: ?
[%( (
= s sum
. ( sum
= A
. !arg:%?A ?arg&!A+sum$!arg
| 0
)
& ( sum$!sjt:>!max:?max
& !sjt:?seq
|
)
)
$
& ~
)
?
| !seq
)3 5 6 -2 -1 4
C
#include "stdio.h"
typedef struct Range {
int start, end, sum;
} Range;
Range maxSubseq(const int sequence[], const int len) {
int maxSum = 0, thisSum = 0, i = 0;
int start = 0, end = -1, j;
for (j = 0; j < len; j++) {
thisSum += sequence[j];
if (thisSum < 0) {
i = j + 1;
thisSum = 0;
} else if (thisSum > maxSum) {
maxSum = thisSum;
start = i;
end = j;
}
}
Range r;
if (start <= end && start >= 0 && end >= 0) {
r.start = start;
r.end = end + 1;
r.sum = maxSum;
} else {
r.start = 0;
r.end = 0;
r.sum = 0;
}
return r;
}
int main(int argc, char **argv) {
int a[] = {-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1};
int alength = sizeof(a)/sizeof(a[0]);
Range r = maxSubseq(a, alength);
printf("Max sum = %d\n", r.sum);
int i;
for (i = r.start; i < r.end; i++)
printf("%d ", a[i]);
printf("\n");
return 0;
}
- Output:
Max sum = 15 3 5 6 -2 -1 4
C#
The challange
using System;
namespace Tests_With_Framework_4
{
class Program
{
static void Main(string[] args)
{
int[] integers = { -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 }; int length = integers.Length;
int maxsum, beginmax, endmax, sum; maxsum = beginmax = sum = 0; endmax = -1;
for (int i = 0; i < length; i++)
{
sum = 0;
for (int k = i; k < length; k++)
{
sum += integers[k];
if (sum > maxsum)
{
maxsum = sum;
beginmax = i;
endmax = k;
}
}
}
for (int i = beginmax; i <= endmax; i++)
Console.WriteLine(integers[i]);
Console.ReadKey();
}
}
}
C++
#include <utility> // for std::pair
#include <iterator> // for std::iterator_traits
#include <iostream> // for std::cout
#include <ostream> // for output operator and std::endl
#include <algorithm> // for std::copy
#include <iterator> // for std::output_iterator
// Function template max_subseq
//
// Given a sequence of integers, find a subsequence which maximizes
// the sum of its elements, that is, the elements of no other single
// subsequence add up to a value larger than this one.
//
// Requirements:
// * ForwardIterator is a forward iterator
// * ForwardIterator's value_type is less-than comparable and addable
// * default-construction of value_type gives the neutral element
// (zero)
// * operator+ and operator< are compatible (i.e. if a>zero and
// b>zero, then a+b>zero, and if a<zero and b<zero, then a+b<zero)
// * [begin,end) is a valid range
//
// Returns:
// a pair of iterators describing the begin and end of the
// subsequence
template<typename ForwardIterator>
std::pair<ForwardIterator, ForwardIterator>
max_subseq(ForwardIterator begin, ForwardIterator end)
{
typedef typename std::iterator_traits<ForwardIterator>::value_type
value_type;
ForwardIterator seq_begin = begin, seq_end = seq_begin;
value_type seq_sum = value_type();
ForwardIterator current_begin = begin;
value_type current_sum = value_type();
value_type zero = value_type();
for (ForwardIterator iter = begin; iter != end; ++iter)
{
value_type value = *iter;
if (zero < value)
{
if (current_sum < zero)
{
current_sum = zero;
current_begin = iter;
}
}
else
{
if (seq_sum < current_sum)
{
seq_begin = current_begin;
seq_end = iter;
seq_sum = current_sum;
}
}
current_sum += value;
}
if (seq_sum < current_sum)
{
seq_begin = current_begin;
seq_end = end;
seq_sum = current_sum;
}
return std::make_pair(seq_begin, seq_end);
}
// the test array
int array[] = { -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 };
// function template to find the one-past-end pointer to the array
template<typename T, int N> int* end(T (&arr)[N]) { return arr+N; }
int main()
{
// find the subsequence
std::pair<int*, int*> seq = max_subseq(array, end(array));
// output it
std::copy(seq.first, seq.second, std::ostream_iterator<int>(std::cout, " "));
std::cout << std::endl;
return 0;
}
Clojure
Naive algorithm:
(defn max-subseq-sum [coll]
(->> (take-while seq (iterate rest coll)) ; tails
(mapcat #(reductions conj [] %)) ; inits
(apply max-key #(reduce + %)))) ; max sum
- Output:
user> (max-subseq-sum [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1])
[3 5 6 -2 -1 4]
CoffeeScript
max_sum_seq = (sequence) ->
# This runs in linear time.
[sum_start, sum, max_sum, max_start, max_end] = [0, 0, 0, 0, 0]
for n, i in sequence
sum += n
if sum > max_sum
max_sum = sum
max_start = sum_start
max_end = i + 1
if sum < 0 # start new sequence
sum = 0
sum_start = i + 1
sequence[max_start...max_end]
# tests
console.log max_sum_seq [-1, 0, 15, 3, -9, 12, -4]
console.log max_sum_seq [-1]
console.log max_sum_seq [4, -10, 3]
Common Lisp
Linear Time
Returns the maximum subsequence sum, the subsequence with the maximum sum, and start and end indices for the subsequence within the original sequence. Based on the implementation at Word Aligned. Leading zeroes aren't trimmed from the subsequence.
(defun max-subseq (list)
(let ((best-sum 0) (current-sum 0) (end 0))
;; determine the best sum, and the end of the max subsequence
(do ((list list (rest list))
(i 0 (1+ i)))
((endp list))
(setf current-sum (max 0 (+ current-sum (first list))))
(when (> current-sum best-sum)
(setf end i
best-sum current-sum)))
;; take the subsequence of list ending at end, and remove elements
;; from the beginning until the subsequence sums to best-sum.
(let* ((sublist (subseq list 0 (1+ end)))
(sum (reduce #'+ sublist)))
(do ((start 0 (1+ start))
(sublist sublist (rest sublist))
(sum sum (- sum (first sublist))))
((or (endp sublist) (eql sum best-sum))
(values best-sum sublist start (1+ end)))))))
For example,
> (max-subseq '(-1 -2 -3 -4 -5)) 0 NIL 1 1
> (max-subseq '(0 1 2 -3 3 -1 0 -4 0 -1 -4 2)) 3 (0 1 2) 0 3
Brute Force
(defun max-subseq (seq)
(loop for subsequence in (mapcon (lambda (x) (maplist #'reverse (reverse x))) seq)
for sum = (reduce #'+ subsequence :initial-value 0)
with max-subsequence
maximizing sum into max
if (= sum max) do (setf max-subsequence subsequence)
finally (return max-subsequence))))
Component Pascal
Works with BlackBox Component Builder
MODULE OvctGreatestSubsequentialSum;
IMPORT StdLog, Strings, Args;
PROCEDURE Gss(iseq: ARRAY OF INTEGER;OUT start, end, maxsum: INTEGER);
VAR
i,j,sum: INTEGER;
BEGIN
i := 0; maxsum := 0; start := 0; end := -1;
WHILE i < LEN(iseq) - 1 DO
sum := 0; j := i;
WHILE j < LEN(iseq) -1 DO
INC(sum ,iseq[j]);
IF sum > maxsum THEN
maxsum := sum;
start := i;
end := j
END;
INC(j);
END;
INC(i)
END
END Gss;
PROCEDURE Do*;
VAR
p: Args.Params;
iseq: POINTER TO ARRAY OF INTEGER;
i, res, start, end, sum: INTEGER;
BEGIN
Args.Get(p); (* Get Params *)
NEW(iseq,p.argc);
(* Transform params to INTEGERs *)
FOR i := 0 TO p.argc - 1 DO
Strings.StringToInt(p.args[i],iseq[i],res)
END;
Gss(iseq,start,end,sum);
StdLog.String("[");
FOR i := start TO end DO
StdLog.Int(iseq[i]);
IF i < end THEN StdLog.String(",") END
END;
StdLog.String("]=");StdLog.Int(sum);StdLog.Ln;
END Do;
END OvctGreatestSubsequentialSum.
Execute:
[Ctrl-Q]OvctGreatestSubsequentialSum.Do -1 -2 3 5 6 -2 -1 4 -4 2 -2 ~ [Ctrl-Q]OvctGreatestSubsequentialSum.Do -1 -5 -3 ~
- Output:
[ 3, 5, 6, -2, -1, 4]= 15 []= 0
Crystal
Brute Force:
Answer is stored in "slice". It is very slow O(n**3)
def subarray_sum(arr)
max, slice = 0, [] of Int32
arr.each_index do |i|
(i...arr.size).each do |j|
sum = arr[i..j].sum
max, slice = sum, arr[i..j] if sum > max
end
end
[max, slice]
end
Linear Time Version:
A better answer would run in O(n) instead of O(n**2) using numerical properties to remove the need for the inner loop.
# the trick is that at any point
# in the iteration if starting a new chain is
# better than your current score with this element
# added to it, then do so.
# the interesting part is proving the math behind it
def subarray_sum(arr)
curr = max = 0
first, last, curr_first = arr.size, 0, 0
arr.each_with_index do |e, i|
curr += e
e > curr && (curr = e; curr_first = i)
curr > max && (max = curr; first = curr_first; last = i)
end
return max, arr[first..last]
end
Test:
[ [1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5],
[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1],
[-1, -2, -3, -4, -5],
[] of Int32
].each do |input|
puts "\nInput seq: #{input}"
puts " Max sum: %d\n Subseq: %s" % subarray_sum(input)
end
- Output:
Input seq: [1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5] Max sum: 65 Subseq: [40, 25] Input seq: [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] Max sum: 15 Subseq: [3, 5, 6, -2, -1, 4] Input seq: [-1, -2, -3, -4, -5] Max sum: 0 Subseq: [] Input seq: [] Max sum: 0 Subseq: []
D
import std.stdio;
inout(T[]) maxSubseq(T)(inout T[] sequence) pure nothrow @nogc {
int maxSum, thisSum, i, start, end = -1;
foreach (immutable j, immutable x; sequence) {
thisSum += x;
if (thisSum < 0) {
i = j + 1;
thisSum = 0;
} else if (thisSum > maxSum) {
maxSum = thisSum;
start = i;
end = j;
}
}
if (start <= end && start >= 0 && end >= 0)
return sequence[start .. end + 1];
else
return [];
}
void main() {
const a1 = [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1];
writeln("Maximal subsequence: ", a1.maxSubseq);
const a2 = [-1, -2, -3, -5, -6, -2, -1, -4, -4, -2, -1];
writeln("Maximal subsequence: ", a2.maxSubseq);
}
- Output:
Maximal subsequence: [3, 5, 6, -2, -1, 4] Maximal subsequence: []
Alternative Version
This version is much less efficient. The output is similar. Currently the D standard library lacks the sum, inits, tails functions, and max can't be used as the maximumBy functions (for the concatMap a map.join is enough).
import std.stdio, std.algorithm, std.range, std.typecons;
mixin template InitsTails(T) {
T[] data;
size_t pos;
@property bool empty() pure nothrow @nogc {
return pos > data.length;
}
void popFront() pure nothrow @nogc { pos++; }
}
struct Inits(T) {
mixin InitsTails!T;
@property T[] front() pure nothrow @nogc { return data[0 .. pos]; }
}
auto inits(T)(T[] seq) pure nothrow @nogc { return seq.Inits!T; }
struct Tails(T) {
mixin InitsTails!T;
@property T[] front() pure nothrow @nogc { return data[pos .. $]; }
}
auto tails(T)(T[] seq) pure nothrow @nogc { return seq.Tails!T; }
T[] maxSubseq(T)(T[] seq) pure nothrow /*@nogc*/ {
//return seq.tails.map!inits.joiner.reduce!(max!sum);
return seq.tails.map!inits.join.minPos!q{ a.sum > b.sum }[0];
}
void main() {
[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1].maxSubseq.writeln;
[-1, -2, -3, -5, -6, -2, -1, -4, -4, -2, -1].maxSubseq.writeln;
}
Delphi
See Pascal.
E
This implementation tests every combination, but it first examines the sequence to reduce the number of combinations tried: We need not consider beginning the subsequence at any point which is not the beginning, or a change from negative to positive. We need not consider ending the subsequence at any point which is not the end, or a change from positive to negative. (Zero is moot and treated as negative.)
This algorithm is therefore where is the size of the sequence and is the number of sign changes in the sequence. I think it could be improved to by recording the positive and negative intervals' sums during the initial pass and accumulating the sum of those sums in the inner for loop.
maxSubseq returns both the maximum sum found, and the interval of indexes which produces it.
pragma.enable("accumulator")
def maxSubseq(seq) {
def size := seq.size()
# Collect all intervals of indexes whose values are positive
def intervals := {
var intervals := []
var first := 0
while (first < size) {
var next := first
def seeing := seq[first] > 0
while (next < size && (seq[next] > 0) == seeing) {
next += 1
}
if (seeing) { # record every positive interval
intervals with= first..!next
}
first := next
}
intervals
}
# For recording the best result found
var maxValue := 0
var maxInterval := 0..!0
# Try all subsequences beginning and ending with such intervals.
for firstIntervalIx => firstInterval in intervals {
for lastInterval in intervals(firstIntervalIx) {
def interval :=
(firstInterval.getOptStart())..!(lastInterval.getOptBound())
def value :=
accum 0 for i in interval { _ + seq[i] }
if (value > maxValue) {
maxValue := value
maxInterval := interval
}
}
}
return ["value" => maxValue,
"indexes" => maxInterval]
}def seq := [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
def [=> value, => indexes] := maxSubseq(seq)
println(`$\
Sequence: $seq
Maximum subsequence sum: $value
Indexes: ${indexes.getOptStart()}..${indexes.getOptBound().previous()}
Subsequence: ${seq(indexes.getOptStart(), indexes.getOptBound())}
`)EchoLisp
(lib 'struct)
(struct result (score starter))
;; the score of i in sequence ( .. i j ...) is max (i , i + score (j))
;; to compute score of (a b .. x y z) :
;; start with score(z) and compute scores of y , z , ..c, b , a.
;; this is O(n)
;; return value of sub-sequence
(define (max-max L into: result)
(define value
(if
(empty? L) -Infinity
(max (first L) (+ (first L) (max-max (cdr L) result )))))
(when (> value (result-score result))
(set-result-score! result value) ;; remember best score
(set-result-starter! result L)) ;; and its location
value)
;; return (best-score (best sequence))
(define (max-seq L)
(define best (result -Infinity null))
(max-max L into: best)
(define score (result-score best))
(list score
(for/list (( n (result-starter best)))
#:break (zero? score)
(set! score (- score n))
n)))
(define L '(-1 -2 3 5 6 -2 -1 4 -4 2 -1))
(max-seq L)
→ (15 (3 5 6 -2 -1 4))
Elixir
Linear Time Version:
defmodule Greatest do
def subseq_sum(list) do
list_i = Enum.with_index(list)
acc = {0, 0, length(list), 0, 0}
{_,max,first,last,_} = Enum.reduce(list_i, acc, fn {elm,i},{curr,max,first,last,curr_first} ->
if curr < 0 do
if elm > max, do: {elm, elm, i, i, curr_first},
else: {elm, max, first, last, curr_first}
else
cur2 = curr + elm
if cur2 > max, do: {cur2, cur2, curr_first, i, curr_first},
else: {cur2, max, first, last, curr_first}
end
end)
{max, Enum.slice(list, first..last)}
end
end
Output is the same above.
Brute Force:
defmodule Greatest do
def subseq_sum(list) do
limit = length(list) - 1
ij = for i <- 0..limit, j <- i..limit, do: {i,j}
Enum.reduce(ij, {0, []}, fn {i,j},{max, subseq} ->
slice = Enum.slice(list, i..j)
sum = Enum.sum(slice)
if sum > max, do: {sum, slice}, else: {max, subseq}
end)
end
end
Test:
data = [ [1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5],
[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1],
[-1, -2, -3, -4, -5],
[] ]
Enum.each(data, fn input ->
IO.puts "\nInput seq: #{inspect input}"
{max, subseq} = Greatest.subseq_sum(input)
IO.puts " Max sum: #{max}\n Subseq: #{inspect subseq}"
end)
- Output:
Input seq: [1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5] Max sum: 65 Subseq: [40, 25] Input seq: [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] Max sum: 15 Subseq: [3, 5, 6, -2, -1, 4] Input seq: [-1, -2, -3, -4, -5] Max sum: 0 Subseq: [] Input seq: [] Max sum: 0 Subseq: []
ERRE
PROGRAM MAX_SUM
DIM A%[11],B%[10],C%[4]
!$DYNAMIC
DIM P%[0]
PROCEDURE MAX_SUBSEQUENCE(P%[],N%->A$)
LOCAL A%,B%,I%,J%,M%,S%
A%=1
FOR I%=0 TO N% DO
S%=0
FOR J%=I% TO N% DO
S%+=P%[J%]
IF S%>M% THEN
M%=S%
A%=I%
B%=J%
END IF
END FOR
END FOR
IF A%>B% THEN A$="[]" EXIT PROCEDURE END IF
A$="["
FOR I%=A% TO B% DO
A$+=STR$(P%[I%])+","
END FOR
A$=LEFT$(A$,LEN(A$)-1)+"]"
END PROCEDURE
PROCEDURE SHOW_ARRAY(P%[],N%->A$)
LOCAL I%
A$="["
FOR I%=0 TO N% DO
A$+=STR$(P%[I%])+","
END FOR
A$=LEFT$(A$,LEN(A$)-1)+"]"
END PROCEDURE
BEGIN
A%[]=(0,1,2,-3,3,-1,0,-4,0,-1,-4,2)
N%=UBOUND(A%,1)
!$DIM P%[N%]
SHOW_ARRAY(A%[],N%->A$)
PRINT(A$;" -> ";)
MAX_SUBSEQUENCE(A%[],N%->A$)
PRINT(A$)
!$ERASE P%
B%[]=(-1,-2,3,5,6,-2,-1,4,-4,2,-1)
N%=UBOUND(B%,1)
!$DIM P%[N%]
SHOW_ARRAY(B%[],N%->A$)
PRINT(A$;" -> ";)
MAX_SUBSEQUENCE(B%[],N%->A$)
PRINT(A$)
!$ERASE P%
C%[]=(-1,-2,-3,-4,-5)
N%=UBOUND(C%,1)
!$DIM P%[N%]
SHOW_ARRAY(C%[],N%->A$)
PRINT(A$;" -> ";)
MAX_SUBSEQUENCE(C%[],N%->A$)
PRINT(A$)
!$ERASE P%
END PROGRAM
Euler Math Toolbox
The following recursive system seems to have a run time of O(n), but it needs some copying, so the run time is really O(n^2).
>function %maxsubs (v,n) ...
$if n==1 then
$ if (v[1]<0) then return {zeros(1,0),zeros(1,0)}
$ else return {v,v};
$ endif;
$endif;
${v1,v2}=%maxsubs(v[1:n-1],n-1);
$m1=sum(v1); m2=sum(v2); m3=m2+v[n];
$if m3>0 then v3=v2|v[n]; else v3=zeros(1,0); endif;
$if m3>m1 then return {v2|v[n],v3};
$else return {v1,v3};
$endif;
$endfunction
>function maxsubs (v) ...
${v1,v2}=%maxsubs(v,cols(v));
$return v1
$endfunction
>maxsubs([0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4])
[ 0 1 2 ]
>maxsubs([-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1])
[ 3 5 6 -2 -1 4 ]
>maxsubs([-1, -2, -3, -4, -5])
Empty matrix of size 1x0Here is a brute force method producing and testing all sums. The runtime is O(n^3).
>function maxsubsbrute (v) ...
$ n=cols(v);
$ A=zeros(n*(n-1),n);
$ k=1;
$ for i=1 to n-1;
$ for j=i to n;
$ A[k,i:j]=1;
$ k=k+1;
$ end;
$ end;
$ k1=extrema((A.v')')[4];
$ return v[nonzeros(A[k1])];
$ endfunction
>maxsubsbrute([0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4])
[ 0 1 2 ]
>maxsubsbrute([-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1])
[ 3 5 6 -2 -1 4 ]
>maxsubsbrute([-1, -2, -3, -4, -5])
Empty matrix of size 1x0To see, if everything works, the following tests on 10 million random sequences.
>function test ...
$ loop 1 to 10000000
$ v=intrandom(1,intrandom(6)+6,20)-10;
$ if sum(maxsubs(v))!=sum(maxsubsbrute(v)) then
$ v, error("Found a wrong test example");
$ endif;
$ endfunction
>testEuphoria
function maxSubseq(sequence s)
integer sum, maxsum, first, last
maxsum = 0
first = 1
last = 0
for i = 1 to length(s) do
sum = 0
for j = i to length(s) do
sum += s[j]
if sum > maxsum then
maxsum = sum
first = i
last = j
end if
end for
end for
return s[first..last]
end function
? maxSubseq({-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1})
? maxSubseq({})
? maxSubseq({-1, -5, -3})- Output:
{3,5,6,-2,-1,4}
{}
{}
F#
let maxsubseq s =
let (_, _, maxsum, maxseq) =
List.fold (fun (sum, seq, maxsum, maxseq) x ->
let (sum, seq) = (sum + x, x :: seq)
if sum < 0 then (0, [], maxsum, maxseq)
else if sum > maxsum then (sum, seq, sum, seq)
else (sum, seq, maxsum, maxseq))
(0, [], 0, []) s
List.rev maxseq
printfn "%A" (maxsubseq [-1 ; -2 ; 3 ; 5 ; 6 ; -2 ; -1 ; 4; -4 ; 2 ; -1])
- Output:
[3; 5; 6; -2; -1; 4]
Factor
USING: kernel locals math math.order sequences ;
:: max-with-index ( elt0 ind0 elt1 ind1 -- elt ind )
elt0 elt1 < [ elt1 ind1 ] [ elt0 ind0 ] if ;
: last-of-max ( accseq -- ind ) -1 swap -1 [ max-with-index ] reduce-index nip ;
: max-subseq ( seq -- subseq )
dup 0 [ + 0 max ] accumulate swap suffix last-of-max head
dup 0 [ + ] accumulate swap suffix [ neg ] map last-of-max tail ;
( scratchpad ) { -1 -2 3 5 6 -2 -1 4 -4 2 -1 } max-subseq dup sum swap . .
{ 3 5 6 -2 -1 4 }
15
Forth
2variable best
variable best-sum
: sum ( array len -- sum )
0 -rot cells over + swap do i @ + cell +loop ;
: max-sub ( array len -- sub len )
over 0 best 2! 0 best-sum !
dup 1 do \ foreach length
2dup i - 1+ cells over + swap do \ foreach start
i j sum
dup best-sum @ > if
best-sum !
i j best 2!
else drop then
cell +loop
loop
2drop best 2@ ;
: .array ." [" dup 0 ?do over i cells + @ . loop ." ] = " sum . ;
create test -1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1 ,
create test2 -1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , 99 ,
- Output:
test 11 max-sub .array [3 5 6 -2 -1 4 ] = 15 ok
test2 11 max-sub .array [3 5 6 -2 -1 4 -4 2 99 ] = 112 ok
Fortran
program MaxSubSeq
implicit none
integer, parameter :: an = 11
integer, dimension(an) :: a = (/ -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 /)
integer, dimension(an,an) :: mix
integer :: i, j
integer, dimension(2) :: m
forall(i=1:an,j=1:an) mix(i,j) = sum(a(i:j))
m = maxloc(mix)
! a(m(1):m(2)) is the wanted subsequence
print *, a(m(1):m(2))
end program MaxSubSeq
FreeBASIC
' FB 1.05.0 Win64
Dim As Integer seq(10) = {-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1}
Dim As Integer i, j, sum, maxSum, first, last
maxSum = 0
For i = LBound(seq) To UBound(seq)
sum = 0
For j = i To UBound(seq)
' only proper sub-sequences are considered
If i = LBound(seq) AndAlso j = UBound(seq) Then Exit For
sum += seq(j)
If sum > maxSum Then
maxSum = sum
first = i
last = j
End If
Next j
Next i
If maxSum > 0 Then
Print "Maximum subsequence is from indices"; first; " to"; last
Print "Elements are : ";
For i = first To last
Print seq(i); " ";
Next
Print
Print "Sum is"; maxSum
Else
Print "Maximum subsequence is the empty sequence which has a sum of 0"
End If
Print
Print "Press any key to quit"
Sleep
- Output:
Maximum subsequence is from indices 2 to 7 Elements are : 3 5 6 -2 -1 4 Sum is 15
Go
package main
import "fmt"
func gss(s []int) ([]int, int) {
var best, start, end, sum, sumStart int
for i, x := range s {
sum += x
switch {
case sum > best:
best = sum
start = sumStart
end = i + 1
case sum < 0:
sum = 0
sumStart = i + 1
}
}
return s[start:end], best
}
var testCases = [][]int{
{-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1},
{-1, 1, 2, -5, -6},
{},
{-1, -2, -1},
}
func main() {
for _, c := range testCases {
fmt.Println("Input: ", c)
subSeq, sum := gss(c)
fmt.Println("Sub seq:", subSeq)
fmt.Println("Sum: ", sum, "\n")
}
}
- Output:
Input: [-1 -2 3 5 6 -2 -1 4 -4 2 -1] Sub seq: [3 5 6 -2 -1 4] Sum: 15 Input: [-1 1 2 -5 -6] Sub seq: [1 2] Sum: 3 Input: [] Sub seq: [] Sum: 0 Input: [-1 -2 -1] Sub seq: [] Sum: 0
Haskell
Naive approach which tests all possible subsequences, as in a few of the other examples. For fun, this is in point-free style and doesn't use loops:
import Data.List (inits, tails, maximumBy)
import Data.Ord (comparing)
subseqs :: [a] -> [[a]]
subseqs = concatMap inits . tails
maxsubseq :: (Ord a, Num a) => [a] -> [a]
maxsubseq = maximumBy (comparing sum) . subseqs
main = print $ maxsubseq [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
Secondly, the linear time constant space approach:
maxSubseq :: [Int] -> (Int, [Int])
maxSubseq =
let go x ((h1, h2), sofar) =
((,) <*> max sofar) (max (0, []) (h1 + x, x : h2))
in snd . foldr go ((0, []), (0, []))
main :: IO ()
main = print $ maxSubseq [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
- Output:
(15,[3,5,6,-2,-1,4])
Icon and Unicon
IS-BASIC
100 PROGRAM "Subseq.bas"
110 RANDOMIZE
120 NUMERIC A(1 TO 15)
130 PRINT "Sequence:"
140 FOR I=LBOUND(A) TO UBOUND(A)
150 LET A(I)=RND(11)-6
160 PRINT A(I);
170 NEXT
180 LET MAXSUM,ST=0:LET EN=-1
190 FOR I=LBOUND(A) TO UBOUND(A)
200 LET SUM=0
210 FOR J=I TO UBOUND(A)
220 LET SUM=SUM+A(J)
230 IF SUM>MAXSUM THEN LET MAXSUM=SUM:LET ST=I:LET EN=J
240 NEXT
250 NEXT
260 PRINT :PRINT "SubSequence with greatest sum:"
270 IF ST>0 THEN PRINT TAB(ST*3-2);
280 FOR I=ST TO EN
290 PRINT A(I);
300 NEXT
310 PRINT :PRINT "Sum:";MAXSUMJ
maxss=: monad define
AS =. 0,; <:/~@i.&.> #\y
MX =. (= >./) AS +/ . * y
y #~ {. MX#AS
)
y is the input vector, an integer list.
AS means "all sub-sequences." It is a binary table.
Each row indicates one sub-sequence; the count of columns equals the length of the input.
MX means "maxima." It is the first location in AS where the corresponding sum is largest.
Totals for the subsequences are calculated by the phrase 'AS +/ . * y' which is the inner product of the binary table with the input vector.
All solutions are found but only one is returned, to fit the output requirement.
If zero is the maximal sum the empty array is always returned, although sub-sequences of positive length (comprised of zeros) might be more interesting.
Example use:
maxss _1 _2 3 5 6 _2 _1 4 _4 2 _1
3 5 6 _2 _1 4
Note: if we just want the sum of the maximum subsequence, and are not concerned with the subsequence itself, we can use:
maxs=: [:>./(0>.+)/\.
Example use:
maxs _1 _2 3 5 6 _2 _1 4 _4 2 _1
15
This suggests a variant:
maxSS=:monad define
sums=: (0>.+)/\. y
start=: sums i. max=: >./ sums
max (] {.~ #@] |&>: (= +/\) i. 1:) y}.~start
)
or
maxSS2=:monad define
start=. (i. >./) (0>.+)/\. y
({.~ # |&>: [: (i.>./@,&0) +/\) y}.~start
)
These variants are considerably faster than the first implementation, on long sequences.
Java
This is not a particularly efficient solution, but it gets the job done.
The method nextChoices was modified from an RIT CS lab.
import java.util.Scanner;
import java.util.ArrayList;
public class Sub{
private static int[] indices;
public static void main(String[] args){
ArrayList<Long> array= new ArrayList<Long>(); //the main set
Scanner in = new Scanner(System.in);
while(in.hasNextLong()) array.add(in.nextLong());
long highSum= Long.MIN_VALUE;//start the sum at the lowest possible value
ArrayList<Long> highSet= new ArrayList<Long>();
//loop through all possible subarray sizes including 0
for(int subSize= 0;subSize<= array.size();subSize++){
indices= new int[subSize];
for(int i= 0;i< subSize;i++) indices[i]= i;
do{
long sum= 0;//this subarray sum variable
ArrayList<Long> temp= new ArrayList<Long>();//this subarray
//sum it and save it
for(long index:indices) {sum+= array.get(index); temp.add(array.get(index));}
if(sum > highSum){//if we found a higher sum
highSet= temp; //keep track of it
highSum= sum;
}
}while(nextIndices(array));//while we haven't tested all subarrays
}
System.out.println("Sum: " + highSum + "\nSet: " +
highSet);
}
/**
* Computes the next set of choices from the previous. The
* algorithm tries to increment the index of the final choice
* first. Should that fail (index goes out of bounds), it
* tries to increment the next-to-the-last index, and resets
* the last index to one more than the next-to-the-last.
* Should this fail the algorithm keeps starting at an earlier
* choice until it runs off the start of the choice list without
* Finding a legal set of indices for all the choices.
*
* @return true unless all choice sets have been exhausted.
* @author James Heliotis
*/
private static boolean nextIndices(ArrayList<Long> a) {
for(int i= indices.length-1;i >= 0;--i){
indices[i]++;
for(int j=i+1;j < indices.length;++j){
indices[j]= indices[j - 1] + 1;//reset the last failed try
}
if(indices[indices.length - 1] < a.size()){//if this try went out of bounds
return true;
}
}
return false;
}
}
This one runs in linear time, and isn't generalized.
private static int BiggestSubsum(int[] t) {
int sum = 0;
int maxsum = 0;
for (int i : t) {
sum += i;
if (sum < 0)
sum = 0;
maxsum = sum > maxsum ? sum : maxsum;
}
return maxsum;
}
JavaScript
Imperative
Simple brute force approach.
function MaximumSubsequence(population) {
var maxValue = 0;
var subsequence = [];
for (var i = 0, len = population.length; i < len; i++) {
for (var j = i; j <= len; j++) {
var subsequence = population.slice(i, j);
var value = sumValues(subsequence);
if (value > maxValue) {
maxValue = value;
greatest = subsequence;
};
}
}
return greatest;
}
function sumValues(arr) {
var result = 0;
for (var i = 0, len = arr.length; i < len; i++) {
result += arr[i];
}
return result;
}
Functional
Linear approach, deriving both list and sum in a single accumulating fold.
(() => {
// maxSubseq :: [Int] -> (Int, [Int])
const maxSubseq = xs =>
snd(xs.reduce((tpl, x) => {
const [m1, m2] = Array.from(fst(tpl)),
high = max(
Tuple(0, []),
Tuple(m1 + x, m2.concat(x))
);
return Tuple(high, max(snd(tpl), high));
}, Tuple(Tuple(0, []), Tuple(0, []))));
// TEST -----------------------------------------------
// main :: IO ()
const main = () => {
const mx = maxSubseq([-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]);
showLog(snd(mx), fst(mx))
}
// [3,5,6,-2,-1,4] -> 15
// GENERIC FUNCTIONS ----------------------------------
// fst :: (a, b) -> a
const fst = tpl => tpl[0];
// gt :: Ord a => a -> a -> Bool
const gt = (x, y) =>
'Tuple' === x.type ? (
x[0] > y[0]
) : (x > y);
// max :: Ord a => a -> a -> a
const max = (a, b) => gt(b, a) ? b : a;
// showLog :: a -> IO ()
const showLog = (...args) =>
console.log(
args
.map(JSON.stringify)
.join(' -> ')
);
// snd :: (a, b) -> b
const snd = tpl => tpl[1];
// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// MAIN ---
return main();
})();
- Output:
[3,5,6,-2,-1,4] -> 15
jq
This is the same linear-time algorithm as used in the #Ruby subsection on this page.
def subarray_sum:
. as $arr
| reduce range(0; length) as $i
( {"first": length, "last": 0, "curr": 0, "curr_first": 0, "max": 0};
$arr[$i] as $e
| (.curr + $e) as $curr
| . + (if $e > $curr then {"curr": $e, "curr_first": $i} else {"curr": $curr} end)
| if .curr > .max then . + {"max": $curr, "first": .curr_first, "last": $i}
else .
end)
| [ .max, $arr[ .first : (1 + .last)] ];Example:
[1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5] | subarray_sum- Output:
$ jq -c -n -f Greatest_subsequential_sum.jq
[65,[40,25]]
Jsish
From Javascript entry.
/* Greatest Subsequential Sum, in Jsish */
function sumValues(arr) {
var result = 0;
for (var i = 0, len = arr.length; i < len; i++) result += arr[i];
return result;
}
function greatestSubsequentialSum(population:array):array {
var maxValue = (population[0]) ? population[0] : 0;
var subsequence = [], greatest = [];
for (var i = 0, len = population.length; i < len; i++) {
for (var j = i; j < len; j++) {
subsequence = population.slice(i, j);
var value = sumValues(subsequence);
if (value > maxValue) {
maxValue = value;
greatest = subsequence;
};
}
}
return [maxValue, greatest];
}
if (Interp.conf('unitTest')) {
var gss = [-1,-2,3,5,6,-2,-1,4,-4,2,-1];
; gss;
; greatestSubsequentialSum(gss);
}
/*
=!EXPECTSTART!=
gss ==> [ -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 ]
greatestSubsequentialSum(gss) ==> [ 15, [ 3, 5, 6, -2, -1, 4 ] ]
=!EXPECTEND!=
*/
- Output:
prompt$ jsish --U greatestSubsequentialSum.jsi gss ==> [ -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 ] greatestSubsequentialSum(gss) ==> [ 15, [ 3, 5, 6, -2, -1, 4 ] ]
Julia
function gss(arr::Vector{<:Number})
smax = hmax = tmax = 0
for head in eachindex(arr), tail in head:length(arr)
s = sum(arr[head:tail])
if s > smax
smax = s
hmax, tmax = head, tail
end
end
return arr[hmax:tmax]
end
arr = [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
subseq = gss(arr)
s = sum(subseq)
println("Greatest subsequential sum of $arr:\n → $subseq with sum $s")
- Output:
Greatest subsequential sum of [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]: → [3, 5, 6, -2, -1, 4] with sum 15
Kotlin
// version 1.1
fun gss(seq: IntArray): Triple<Int, Int, Int> {
if (seq.isEmpty()) throw IllegalArgumentException("Array cannot be empty")
var sum: Int
var maxSum = seq[0]
var first = 0
var last = 0
for (i in 1 until seq.size) {
sum = 0
for (j in i until seq.size) {
sum += seq[j]
if (sum > maxSum) {
maxSum = sum
first = i
last = j
}
}
}
return Triple(maxSum, first, last)
}
fun main(args: Array<String>) {
val seq = intArrayOf(-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1)
val(maxSum, first, last) = gss(seq)
if (maxSum > 0) {
println("Maximum subsequence is from indices $first to $last")
print("Elements are : ")
for (i in first .. last) print("${seq[i]} ")
println("\nSum is $maxSum")
}
else
println("Maximum subsequence is the empty sequence which has a sum of 0")
}
- Output:
Maximum subsequence is from indices 2 to 7 Elements are : 3 5 6 -2 -1 4 Sum is 15
Liberty BASIC
'Greatest_subsequential_sum
N= 20 'number of elements
randomize 0.52
for K = 1 to 5
a$ = using("##",int(rnd(1)*12)-5)
for i=2 to N
a$ = a$ +","+using("##",int(rnd(1)*12)-5)
next
call maxsumseq a$
next K
sub maxsumseq a$
sum=0
maxsum=0
sumStart=1
end1 =0
start1 =1
token$="*"
i=0
while 1
i=i+1
token$=word$(a$, i, ",")
if token$ ="" then exit while 'end of stream
x=val(token$)
sum=sum+x
if maxsum<sum then
maxsum = sum
start1 = sumStart
end1 = i
else
if sum <0 then
sum=0
sumStart = i+1
end if
end if
wend
print "sequence: ";a$
print " ";
for i=1 to start1-1: print " "; :next
for i= start1 to end1: print "---"; :next
print
if end1 >0 then
print "Maximum sum subsequense: ";start1 ;" to "; end1
else
print "Maximum sum subsequense: is empty"
end if
print "Maximum sum ";maxsum
print
end subLua
function sumt(t, start, last) return start <= last and t[start] + sumt(t, start+1, last) or 0 end
function maxsub(ary, idx)
local idx = idx or 1
if not ary[idx] then return {} end
local maxsum, last = 0, idx
for i = idx, #ary do
if sumt(ary, idx, i) > maxsum then maxsum, last = sumt(ary, idx, i), i end
end
local v = maxsub(ary, idx + 1)
if maxsum < sumt(v, 1, #v) then return v end
local ret = {}
for i = idx, last do ret[#ret+1] = ary[i] end
return ret
end
M4
divert(-1)
define(`setrange',`ifelse(`$3',`',$2,`define($1[$2],$3)`'setrange($1,
incr($2),shift(shift(shift($@))))')')
define(`asize',decr(setrange(`a',1,-1,-2,3,5,6,-2,-1,4,-4,2,-1)))
define(`get',`defn(`$1[$2]')')
define(`for',
`ifelse($#,0,``$0'',
`ifelse(eval($2<=$3),1,
`pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')
define(`maxsum',0)
for(`x',1,asize,
`define(`sum',0)`'for(`y',x,asize,
`define(`sum',eval(sum+get(`a',y)))`'ifelse(eval(sum>maxsum),1,
`define(`maxsum',sum)`'define(`xmax',x)`'define(`ymax',y)')')')
divert
for(`x',xmax,ymax,`get(`a',x) ')Mathematica / Wolfram Language
Method 1
First we define 2 functions, one that gives all possibles subsequences (as a list of lists of indices) for a particular length. Then another extract those indices adds them up and looks for the largest sum.
Sequences[m_]:=Prepend[Flatten[Table[Partition[Range[m],n,1],{n,m}],1],{}]
MaximumSubsequence[x_List]:=Module[{sums},
sums={x[[#]],Total[x[[#]]]}&/@Sequences[Length[x]];
First[First[sums[[Ordering[sums,-1,#1[[2]]<#2[[2]]&]]]]]
]
Method 2
MaximumSubsequence[x_List]:=Last@SortBy[Flatten[Table[x[[a;;b]], {b,Length[x]}, {a,b}],1],Total]
Examples
MaximumSubsequence[{-1,-2,3,5,6,-2,-1,4,-4,2,-1}]
MaximumSubsequence[{2,4,5}]
MaximumSubsequence[{2,-4,3}]
MaximumSubsequence[{4}]
MaximumSubsequence[{}]
gives back:
{3,5,6,-2,-1,4}
{2,4,5}
{3}
{4}
{}
Mathprog
see wp:Special_ordered_set. Lmin specifies the minimum length of the required subsequence, and Lmax the maximum.
/*Special ordered set of type N
Nigel_Galloway
January 26th, 2012
*/
param Lmax;
param Lmin;
set SOS;
param Sx{SOS};
var db{Lmin..Lmax,SOS}, binary;
maximize s : sum{q in (Lmin..Lmax),t in (0..q-1), z in SOS: z > (q-1)} Sx[z-t]*db[q,z];
sos1 : sum{t in (Lmin..Lmax),z in SOS: z > (t-1)} db[t,z] = 1;
solve;
for{t in (Lmin..Lmax),z in SOS: db[t,z] == 1} {
printf "\nA sub-sequence of length %d sums to %f:\n", t,s;
printf{q in (z-t+1)..z} " %f", Sx[q];
}
printf "\n\n";
data;
param Lmin := 1;
param Lmax := 6;
param:
SOS: Sx :=
1 7
2 4
3 -11
4 6
5 3
6 1
;
end;
produces:
GLPSOL: GLPK LP/MIP Solver, v4.47
Parameter(s) specified in the command line:
--math GSS.mod
Reading model section from GSS.mod...
Reading data section from GSS.mod...
38 lines were read
Generating s...
Generating sos1...
Model has been successfully generated
GLPK Integer Optimizer, v4.47
2 rows, 21 columns, 41 non-zeros
21 integer variables, all of which are binary
Preprocessing...
1 row, 21 columns, 21 non-zeros
21 integer variables, all of which are binary
Scaling...
A: min|aij| = 1.000e+000 max|aij| = 1.000e+000 ratio = 1.000e+000
Problem data seem to be well scaled
Constructing initial basis...
Size of triangular part = 1
Solving LP relaxation...
GLPK Simplex Optimizer, v4.47
1 row, 21 columns, 21 non-zeros
* 0: obj = 1.000000000e+001 infeas = 0.000e+000 (0)
* 1: obj = 1.100000000e+001 infeas = 0.000e+000 (0)
OPTIMAL SOLUTION FOUND
Integer optimization begins...
+ 1: mip = not found yet <= +inf (1; 0)
+ 1: >>>>> 1.100000000e+001 <= 1.100000000e+001 0.0% (1; 0)
+ 1: mip = 1.100000000e+001 <= tree is empty 0.0% (0; 1)
INTEGER OPTIMAL SOLUTION FOUND
Time used: 0.0 secs
Memory used: 0.1 Mb (135491 bytes)
A sub-sequence of length 2 sums to 11.000000:
7.000000 4.000000
Model has been successfully processed
MATLAB / Octave
function [S,GS]=gss(a)
% Greatest subsequential sum
a =[0;a(:);0]';
ix1 = find(a(2:end) >0 & a(1:end-1) <= 0);
ix2 = find(a(2:end)<=0 & a(1:end-1) > 0);
K = 0;
S = 0;
for k = 1:length(ix1)
s = sum(a(ix1(k)+1:ix2(k)));
if (s>S)
S=s; K=k;
end;
end;
GS = a(ix1(K)+1:ix2(K));
Usage:
octave:12> [S,GS]=gss([0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2]) S = 3 GS = 1 2
NetRexx
/* REXX ***************************************************************
* 10.08.2012 Walter Pachl Pascal algorithm -> Rexx -> NetRexx
**********************************************************************/
s=' -1 -2 3 5 6 -2 -1 4 -4 2 -1'
maxSum = 0
seqStart = 0
seqEnd = -1
Loop i = 1 To s.words()
seqSum = 0
Loop j = i to s.words()
seqSum = seqSum + s.word(j)
if seqSum > maxSum then Do
maxSum = seqSum
seqStart = i
seqEnd = j
end
end
end
Say 'Sequence:'
Say s
Say 'Subsequence with greatest sum: '
If seqend<seqstart Then
Say 'empty'
Else Do
ol=' '.copies(seqStart-1)
Loop i = seqStart to seqEnd
w=s.word(i)
ol=ol||w.right(3)
End
Say ol
Say 'Sum:' maxSum
End
Output: the same as for Rexx
Nim
proc maxsum(s: openArray[int]): int =
var maxendinghere = 0
for x in s:
maxendinghere = max(maxendinghere + x, 0)
result = max(result, maxendinghere)
echo maxsum(@[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1])
- Output:
15
Oberon-2
Works with oo2c Version 2
MODULE GreatestSubsequentialSum;
IMPORT
Out,
Err,
IntStr,
ProgramArgs,
TextRider;
TYPE
IntSeq= POINTER TO ARRAY OF LONGINT;
PROCEDURE ShowUsage();
BEGIN
Out.String("Usage: GreatestSubsequentialSum {int}+");Out.Ln
END ShowUsage;
PROCEDURE Gss(iseq: IntSeq; VAR start, end, maxsum: LONGINT);
VAR
i, j, sum: LONGINT;
BEGIN
i := 0; maxsum := 0; start := 0; end := -1;
WHILE (i < LEN(iseq^)) DO
sum := 0; j := i;
WHILE (j < LEN(iseq^) - 1) DO
INC(sum,iseq[j]);
IF sum > maxsum THEN
maxsum := sum;
start := i;
end := j
END;
INC(j)
END;
INC(i)
END
END Gss;
PROCEDURE GetParams():IntSeq;
VAR
reader: TextRider.Reader;
iseq: IntSeq;
param: ARRAY 32 OF CHAR;
argc,i: LONGINT;
res: SHORTINT;
BEGIN
iseq := NIL;
reader := TextRider.ConnectReader(ProgramArgs.args);
IF reader # NIL THEN
argc := ProgramArgs.args.ArgNumber();
IF argc < 1 THEN
Err.String("There is no enough arguments.");Err.Ln;
ShowUsage;
HALT(0)
END;
reader.ReadLn; (* Skips program name *)
NEW(iseq,argc);
FOR i := 0 TO argc - 1 DO
reader.ReadLine(param);
IntStr.StrToInt(param,iseq[i],res);
END
END;
RETURN iseq
END GetParams;
PROCEDURE Do;
VAR
iseq: IntSeq;
start, end, sum, i: LONGINT;
BEGIN
iseq := GetParams();
Gss(iseq, start, end, sum);
i := start;
Out.String("[");
WHILE (i <= end) DO
Out.LongInt(iseq[i],0);
IF (i < end) THEN Out.Char(',') END;
INC(i)
END;
Out.String("]: ");Out.LongInt(sum,0);Out.Ln
END Do;
BEGIN
Do
END GreatestSubsequentialSum.
Execute:
GreatestSubsequentialSum -1 -2 3 5 6 -2 -1 4 -4 2 -2 GreatestSubsequentialSum -1 -5 -3
- Output:
[3,5,6,-2,-1,4]: 15 []: 0
OCaml
let maxsubseq =
let rec loop sum seq maxsum maxseq = function
| [] -> maxsum, List.rev maxseq
| x::xs ->
let sum = sum + x
and seq = x :: seq in
if sum < 0 then
loop 0 [] maxsum maxseq xs
else if sum > maxsum then
loop sum seq sum seq xs
else
loop sum seq maxsum maxseq xs
in
loop 0 [] 0 []
let _ =
maxsubseq [-1 ; -2 ; 3 ; 5 ; 6 ; -2 ; -1 ; 4; -4 ; 2 ; -1]
This returns a pair of the maximum sum and (one of) the maximum subsequence(s).
Oz
declare
fun {MaxSubSeq Xs}
fun {Step [Sum0 Seq0 MaxSum MaxSeq] X}
Sum = Sum0 + X
Seq = X|Seq0
in
if Sum > MaxSum then
%% found new maximum
[Sum Seq Sum Seq]
elseif Sum < 0 then
%% discard negative subseqs
[0 nil MaxSum MaxSeq]
else
[Sum Seq MaxSum MaxSeq]
end
end
[_ _ _ MaxSeq] = {FoldL Xs Step [0 nil 0 nil]}
in
{Reverse MaxSeq}
end
in
{Show {MaxSubSeq [~1 ~2 3 5 6 ~2 ~1 4 ~4 2 1]}}PARI/GP
Naive quadratic solution (with end-trimming).
grsub(v)={
my(mn=1,mx=#v,r=0,at,c);
if(vecmax(v)<=0,return([1,0]));
while(v[mn]<=0,mn++);
while(v[mx]<=0,mx--);
for(a=mn,mx,
c=0;
for(b=a,mx,
c+=v[b];
if(c>r,r=c;at=[a,b])
)
);
at
};Pascal
Program GreatestSubsequentialSum(output);
var
a: array[1..11] of integer = (-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1);
i, j: integer;
seqStart, seqEnd: integer;
maxSum, seqSum: integer;
begin
maxSum := 0;
seqStart := 0;
seqEnd := -1;
for i := low(a) to high(a) do
begin
seqSum := 0;
for j := i to high(a) do
begin
seqSum := seqSum + a[j];
if seqSum > maxSum then
begin
maxSum := seqSum;
seqStart := i;
seqEnd := j;
end;
end;
end;
writeln ('Sequence: ');
for i := low(a) to high(a) do
write (a[i]:3);
writeln;
writeln ('Subsequence with greatest sum: ');
for i := low(a) to seqStart - 1 do
write (' ':3);
for i := seqStart to seqEnd do
write (a[i]:3);
writeln;
writeln ('Sum:');
writeln (maxSum);
end.
- Output:
:> ./GreatestSubsequentialSum
Sequence:
-1 -2 3 5 6 -2 -1 4 -4 2 -1
Subsequence with greatest sum:
3 5 6 -2 -1 4
Sum:
15
Perl
O(n) running-sum method:
use strict;
sub max_sub(\@) {
my ($a, $maxs, $maxe, $s, $sum, $maxsum) = shift;
foreach (0 .. $#$a) {
my $t = $sum + $a->[$_];
($s, $sum) = $t > 0 ? ($s, $t) : ($_ + 1, 0);
if ($maxsum < $sum) {
$maxsum = $sum;
($maxs, $maxe) = ($s, $_ + 1)
}
}
@$a[$maxs .. $maxe - 1]
}
my @a = map { int(rand(20) - 10) } 1 .. 10;
my @b = (-1) x 10;
print "seq: @a\nmax: [ @{[max_sub @a]} ]\n";
print "seq: @b\nmax: [ @{[max_sub @b]} ]\n";
- Output:
seq: -7 5 -3 0 5 -5 -1 -1 -5 1 max: [ 5 -3 0 5 ] seq: -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 max: [ ]
Naive and potentionally very slow method:
use strict;
my @a = (-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1);
my @maxsubarray;
my $maxsum = 0;
foreach my $begin (0..$#a) {
foreach my $end ($begin..$#a) {
my $sum = 0;
$sum += $_ foreach @a[$begin..$end];
if($sum > $maxsum) {
$maxsum = $sum;
@maxsubarray = @a[$begin..$end];
}
}
}
print "@maxsubarray\n";
Phix
with javascript_semantics function maxSubseq(sequence s) integer maxsum = 0, first = 1, last = 0 for i=1 to length(s) do integer sumsij = 0 for j=i to length(s) do sumsij += s[j] if sumsij>maxsum then {maxsum,first,last} = {sumsij,i,j} end if end for end for return s[first..last] end function ? maxSubseq({-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1}) ? maxSubseq({}) ? maxSubseq({-1, -5, -3})
- Output:
{3,5,6,-2,-1,4}
{}
{}
PHP
<?php
function max_sum_seq($sequence) {
// This runs in linear time.
$sum_start = 0;
$sum = 0;
$max_sum = 0;
$max_start = 0;
$max_len = 0;
for ($i = 0; $i < count($sequence); $i += 1) {
$n = $sequence[$i];
$sum += $n;
if ($sum > $max_sum) {
$max_sum = $sum;
$max_start = $sum_start;
$max_len = $i + 1 - $max_start;
}
if ($sum < 0) { # start new sequence
$sum = 0;
$sum_start = $i + 1;
}
}
return array_slice($sequence, $max_start, $max_len);
}
function print_array($arr) {
if (count($arr) > 0) {
echo join(" ", $arr);
} else {
echo "(empty)";
}
echo '<br>';
}
// tests
print_array(max_sum_seq(array(-1, 0, 15, 3, -9, 12, -4)));
print_array(max_sum_seq(array(-1)));
print_array(max_sum_seq(array(4, -10, 3)));
?>
- Output:
in browser
0 15 3 -9 12
(empty)
4
Picat
Here are two versions: one iterative and one using constraint modelling (constraint programming solver).
Iterative
First build a map with all the combinations then pick the one with greatest sum.
greatest_subsequential_sum_it([]) = [] => true.
greatest_subsequential_sum_it(A) = Seq =>
P = allcomb(A),
Total = max([Tot : Tot=_T in P]),
Seq1 = [],
if Total > 0 then
[B,E] = P.get(Total),
Seq1 := [A[I] : I in B..E]
else
Seq1 := []
end,
Seq = Seq1.
allcomb(A) = Comb =>
Len = A.length,
Comb = new_map([(sum([A[I]:I in B..E])=([B,E])) : B in 1..Len, E in B..Len]).Constraint modelling
(This was inspired by a MiniZinc model created by Claudio Cesar de Sá.)
greatest_subsequential_sum_cp([]) = [] => true.
greatest_subsequential_sum_cp(A) = Seq =>
N = A.length,
% decision variables: start and end indices
Begin :: 1..N,
End :: 1..N,
% 1 if the number is in the selected sequence, 0 if not.
X = new_list(N),
X :: 0..1,
% Get the total sum (to be maximized)
TotalSum #= sum([X[I]*A[I] : I in 1..N]),
SizeWindow #= sum(X),
% Calculate the windows of the greatest subsequential sum
End #>= Begin,
End - Begin #= SizeWindow -1,
foreach(I in 1..N)
(Begin #=< I #/\ End #>= I) #<=> X[I] #= 1
end,
Vars = X ++ [Begin,End],
solve($[inout,updown,max(TotalSum)], Vars),
if TotalSum > 0 then
Seq = [A[I] : I in Begin..End]
else
Seq = []
end.Test
import cp.
go =>
LL = [[-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1],
[-1,-2, 3],
[-1,-2],
[0],
[],
[144, 5, -8, 7, 15],
[144, -145, -8, 7, 15],
[-144, 5, -8, 7, 15]
],
println("Iterative version:"),
foreach(L in LL)
printf("%w: ", L),
G = greatest_subsequential_sum_it(L),
println([gss=G, sum=sum(G)])
end,
nl,
println("Constraint model"),
foreach(L in LL)
printf("%w: ", L),
G = greatest_subsequential_sum_cp(L),
println([gss=G, sum=sum(G)])
end,
nl.- Output:
Iterative version: [-1,-2,3,5,6,-2,-1,4,-4,2,-1]: [gss = [3,5,6,-2,-1,4],sum = 15] [-1,-2,3]: [gss = [3],sum = 3] [-1,-2]: [gss = [],sum = 0] [0]: [gss = [],sum = 0] []: [gss = [],sum = 0] [144,5,-8,7,15]: [gss = [144,5,-8,7,15],sum = 163] [144,-145,-8,7,15]: [gss = [144],sum = 144] [-144,5,-8,7,15]: [gss = [7,15],sum = 22] Constraint model [-1,-2,3,5,6,-2,-1,4,-4,2,-1]: [gss = [3,5,6,-2,-1,4],sum = 15] [-1,-2,3]: [gss = [3],sum = 3] [-1,-2]: [gss = [],sum = 0] [0]: [gss = [],sum = 0] []: [gss = [],sum = 0] [144,5,-8,7,15]: [gss = [144,5,-8,7,15],sum = 163] [144,-145,-8,7,15]: [gss = [144],sum = 144] [-144,5,-8,7,15]: [gss = [7,15],sum = 22]
PicoLisp
(maxi '((L) (apply + L))
(mapcon '((L) (maplist reverse (reverse L)))
(-1 -2 3 5 6 -2 -1 4 -4 2 -1) ) )- Output:
-> (3 5 6 -2 -1 4)
PL/I
*process source attributes xref;
ss: Proc Options(Main);
/* REXX ***************************************************************
* 26.08.2013 Walter Pachl translated from REXX version 3
**********************************************************************/
Dcl HBOUND builtin;
Dcl SYSPRINT Print;
Dcl (I,J,LB,MAXSUM,SEQEND,SEQSTART,SEQSUM) Bin Fixed(15);
Dcl s(11) Bin Fixed(15) Init(-1,-2,3,5,6,-2,-1,4,-4,2,-1);
maxSum = 0;
seqStart = 0;
seqEnd = -1;
do i = 1 To hbound(s);
seqSum = 0;
Do j = i to hbound(s);
seqSum = seqSum + s(j);
if seqSum > maxSum then Do;
maxSum = seqSum;
seqStart = i;
seqEnd = j;
end;
end;
end;
Put Edit('Sequence:')(Skip,a);
Put Edit('')(Skip,a);
Do i=1 To hbound(s);
Put Edit(s(i))(f(3));
End;
Put Edit('Subsequence with greatest sum:')(Skip,a);
If seqend<seqstart Then
Put Edit('empty')(Skip,a);
Else Do;
/*ol=copies(' ',seqStart-1)*/
lb=(seqStart-1)*3;
Put Edit(' ')(Skip,a(lb));
Do i = seqStart to seqEnd;
Put Edit(s(i))(f(3));
End;
Put Edit('Sum:',maxSum)(Skip,a,f(5));
End;
End;- Output:
Sequence:
-1 -2 3 5 6 -2 -1 4 -4 2 -1
Subsequence with greatest sum:
3 5 6 -2 -1 4
Sum: 15
Potion
gss = (lst) :
# Find discrete integral
integral = (0)
accum = 0
lst each (n): accum = accum + n, integral append(accum).
# Check integral[b + 1] - integral[a] for all 0 <= a <= b < N
max = -1
max_a = 0
max_b = 0
lst length times (b) :
b times (a) :
if (integral(b + 1) - integral(a) > max) :
max = integral(b + 1) - integral(a)
max_a = a
max_b = b
.
.
.
# Print the results
if (max >= 0) :
(lst slice(max_a, max_b) join(" + "), " = ", max, "\n") join print
.
else :
"No subsequence larger than 0\n" print
.
.
gss((-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1))
gss((-1, -2, -3, -4, -5))
gss((7,-6, -8, 5, -2, -6, 7, 4, 8, -9, -3, 2, 6, -4, -6))3 + 5 + 6 + -2 + -1 + 4 = 15 No subsequence larger than 0 7 + 4 + 8 = 19
Prolog
Constraint Handling Rules
CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module CHR written by Tom Schrijvers and Jan Wielemaker.
:- use_module(library(chr)).
:- chr_constraint
init_chr/2,
seq/2,
% gss(Deb, Len, TT)
gss/3,
% gsscur(Deb, Len, TT, IdCur)
gsscur/4,
memoseq/3,
clean/0,
greatest_subsequence/0.
greatest_subsequence <=>
L = [-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1],
init_chr(1, L),
find_chr_constraint(gss(Deb, Len, V)),
clean,
writeln(L),
forall(between(1, Len, I),
( J is I+Deb-1, nth1(J, L, N), format('~w ', [N]))),
format('==> ~w ~n', [V]).
% destroy last constraint gss
clean \ gss(_,_,_) <=> true.
clean <=> true.
init_chr_end @ init_chr(_, []) <=> gss(0, 0, 0), gsscur(1,0,0,1).
init_chr_loop @ init_chr(N, [H|T]) <=> seq(N, H), N1 is N+1, init_chr(N1, T).
% here, we memorize the list
gsscur_with_negative @ gsscur(Deb, Len, TT, N), seq(N, V) <=> V =< 0 |
memoseq(Deb, Len, TT),
TT1 is TT + V,
N1 is N+1,
% if TT1 becomes negative,
% we begin a new subsequence
( TT1 < 0 -> gsscur(N1,0,0,N1)
; Len1 is Len + 1, gsscur(Deb, Len1, TT1, N1)).
gsscur_with_positive @ gsscur(Deb, Len, TT, N), seq(N, V) <=> V > 0 |
TT1 is TT + V,
N1 is N+1,
Len1 is Len + 1,
gsscur(Deb, Len1, TT1, N1).
gsscur_end @ gsscur(Deb, Len, TT, _N) <=> memoseq(Deb, Len, TT).
memoseq(_DC, _LC, TTC), gss(D, L, TT) <=> TTC =< TT |
gss(D, L, TT).
memoseq(DC, LC, TTC), gss(_D, _L, TT) <=> TTC > TT |
gss(DC, LC, TTC).
- Output:
?- greatest_subsequence. [-1,-2,3,5,6,-2,-1,4,-4,2,-1] 3 5 6 -2 -1 4 ==> 15 true ; false.
Brute Force
Works with GNU Prolog.
subseq(Sub, Seq) :- suffix(X, Seq), prefix(Sub, X).
maxsubseq(List, Sub, Sum) :-
findall(X, subseq(X, List), Subs),
maplist(sum_list, Subs, Sums),
max_list(Sums, Sum),
nth(N, Sums, Sum),
nth(N, Subs, Sub).
- Output:
| ?- maxsubseq([-1,-2,3,5,6,-2,-1,4,-4,2,-1], Sub, Sum). Sub = [3,5,6,-2,-1,4] Sum = 15 ? yes
PureBasic
If OpenConsole()
Define s$, a, b, p1, p2, sum, max, dm=(?EndOfMyData-?MyData)
Dim Seq.i(dm/SizeOf(Integer))
CopyMemory(?MyData,@seq(),dm)
For a=0 To ArraySize(seq())
sum=0
For b=a To ArraySize(seq())
sum+seq(b)
If sum>max
max=sum
p1=a
p2=b
EndIf
Next
Next
For a=p1 To p2
s$+str(seq(a))
If a<p2
s$+"+"
EndIf
Next
PrintN(s$+" = "+str(max))
Print("Press ENTER to quit"): Input()
CloseConsole()
EndIf
DataSection
MyData:
Data.i -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1
EndOfMyData:
EndDataSection
Python
Imperative
Naive, inefficient but really simple solution which tests all possible subsequences, as in a few of the other examples:
def maxsubseq(seq):
return max((seq[begin:end] for begin in xrange(len(seq)+1)
for end in xrange(begin, len(seq)+1)),
key=sum)
Classic linear-time constant-space solution based on algorithm from "Programming Pearls" book.
def maxsum(sequence):
"""Return maximum sum."""
maxsofar, maxendinghere = 0, 0
for x in sequence:
# invariant: ``maxendinghere`` and ``maxsofar`` are accurate for ``x[0..i-1]``
maxendinghere = max(maxendinghere + x, 0)
maxsofar = max(maxsofar, maxendinghere)
return maxsofar
Adapt the above-mentioned solution to return maximizing subsequence. See http://www.java-tips.org/java-se-tips/java.lang/finding-maximum-contiguous-subsequence-sum-using-divide-and-conquer-app.html
def maxsumseq(sequence):
start, end, sum_start = -1, -1, -1
maxsum_, sum_ = 0, 0
for i, x in enumerate(sequence):
sum_ += x
if maxsum_ < sum_: # found maximal subsequence so far
maxsum_ = sum_
start, end = sum_start, i
elif sum_ < 0: # start new sequence
sum_ = 0
sum_start = i
assert maxsum_ == maxsum(sequence)
assert maxsum_ == sum(sequence[start + 1:end + 1])
return sequence[start + 1:end + 1]
Modify ``maxsumseq()`` to allow any iterable not just sequences.
def maxsumit(iterable):
maxseq = seq = []
start, end, sum_start = -1, -1, -1
maxsum_, sum_ = 0, 0
for i, x in enumerate(iterable):
seq.append(x); sum_ += x
if maxsum_ < sum_:
maxseq = seq; maxsum_ = sum_
start, end = sum_start, i
elif sum_ < 0:
seq = []; sum_ = 0
sum_start = i
assert maxsum_ == sum(maxseq[:end - start])
return maxseq[:end - start]
Elementary tests:
f = maxsumit
assert f([]) == []
assert f([-1]) == []
assert f([0]) == []
assert f([1]) == [1]
assert f([1, 0]) == [1]
assert f([0, 1]) == [0, 1]
assert f([0, 1, 0]) == [0, 1]
assert f([2]) == [2]
assert f([2, -1]) == [2]
assert f([-1, 2]) == [2]
assert f([-1, 2, -1]) == [2]
assert f([2, -1, 3]) == [2, -1, 3]
assert f([2, -1, 3, -1]) == [2, -1, 3]
assert f([-1, 2, -1, 3]) == [2, -1, 3]
assert f([-1, 2, -1, 3, -1]) == [2, -1, 3]
assert f([-1, 1, 2, -5, -6]) == [1,2]
Functional
We can efficiently derive sum and sequence together, without mutation, using reduce to express a linear accumulation over a fold:
'''Greatest subsequential sum'''
from functools import (reduce)
# maxSubseq :: [Int] -> [Int] -> (Int, [Int])
def maxSubseq(xs):
'''Subsequence of xs with the maximum sum'''
def go(ab, x):
(m1, m2) = ab[0]
hi = max((0, []), (m1 + x, m2 + [x]))
return (hi, max(ab[1], hi))
return reduce(go, xs, ((0, []), (0, [])))[1]
# TEST -----------------------------------------------------------
print(
maxSubseq(
[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
)
)
- Output:
(15, [3, 5, 6, -2, -1, 4])
Quackery
[ stack ] is maxseq ( --> s )
[ stack ] is maxsum ( --> s )
[ [] maxseq put
0 maxsum put
dup dup size times
[ [] 0 rot
witheach
[ rot over join
unrot +
dup maxsum share >
if
[ dup maxsum replace
over maxseq replace ] ]
2drop behead drop dup ]
2drop
maxsum take
maxseq take ] is maxsubseqsum ( [ --> n [ )
' [ [ 1 2 3 4 5 -8 -9 -20 40 25 -5 ]
[ -1 -2 3 5 6 -2 -1 4 -4 2 -1 ]
[ -1 -2 -3 -4 -5 ]
[ ] ]
witheach
[ dup
say "Sequence: " echo cr
maxsubseqsum
say "Subsequence: " echo cr
say "Sum: " echo cr
cr ]- Output:
Sequence: [ 1 2 3 4 5 -8 -9 -20 40 25 -5 ] Subsequence: [ 40 25 ] Sum: 65 Sequence: [ -1 -2 3 5 6 -2 -1 4 -4 2 -1 ] Subsequence: [ 3 5 6 -2 -1 4 ] Sum: 15 Sequence: [ -1 -2 -3 -4 -5 ] Subsequence: [ ] Sum: 0 Sequence: [ ] Subsequence: [ ] Sum: 0
R
max.subseq <- function(x) {
cumulative <- cumsum(x)
min.cumulative.so.far <- Reduce(min, cumulative, accumulate=TRUE)
end <- which.max(cumulative-min.cumulative.so.far)
begin <- which.min(c(0, cumulative[1:end]))
if (end >= begin) x[begin:end] else x[c()]
}
- Output:
> max.subseq(c(-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1))
[1] 3 5 6 -2 -1 4
Racket
Linear time version, returns the maximum subsequence and its sum.
(define (max-subseq l)
(define-values (_ result _1 max-sum)
(for/fold ([seq '()] [max-seq '()] [sum 0] [max-sum 0])
([i l])
(cond [(> (+ sum i) max-sum)
(values (cons i seq) (cons i seq) (+ sum i) (+ sum i))]
[(< (+ sum i) 0)
(values '() max-seq 0 max-sum)]
[else
(values (cons i seq) max-seq (+ sum i) max-sum)])))
(values (reverse result) max-sum))
For example:
> (max-subseq '(-1 -2 3 5 6 -2 -1 4 -4 2 -1))
'(3 5 6 -2 -1 4)
15
Raku
(formerly Perl 6)
sub max-subseq (*@a) {
my ($start, $end, $sum, $maxsum) = -1, -1, 0, 0;
for @a.kv -> $i, $x {
$sum += $x;
if $maxsum < $sum {
($maxsum, $end) = $sum, $i;
}
elsif $sum < 0 {
($sum, $start) = 0, $i;
}
}
return @a[$start ^.. $end];
}
Another solution, not translated from any other language:
For each starting position, we calculate all the subsets starting at that position. They are combined with the best subset ($max-subset) from previous loops, to form (@subsets). The best of those @subsets is saved at the new $max-subset.
Consuming the array (.shift) allows us to skip tracking the starting point; it is always 0.
The empty sequence is used to initialize $max-subset, which fulfils the "all negative" requirement of the problem.
sub max-subseq ( *@a ) {
my $max-subset = ();
while @a {
my @subsets = [\,] @a;
@subsets.push: $max-subset;
$max-subset = @subsets.max: { [+] .list };
@a.shift;
}
return $max-subset;
}
max-subseq( -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 ).say;
max-subseq( -2, -2, -1, 3, 5, 6, -1, 4, -4, 2, -1 ).say;
max-subseq( -2, -2, -1, -3, -5, -6, -1, -4, -4, -2, -1 ).say;
- Output:
(3 5 6 -2 -1 4) (3 5 6 -1 4) ()
Raven
[ -1 -2 3 5 6 -2 -1 4 -4 2 -1 ] as $seq
1 31 shl as $max
0 $seq length 1 range each as $i
0 as $sum
$i $seq length 1 range each as $j
$seq $j get $sum + as $sum
$sum $max > if
$sum as $max
$i as $i1
$j as $j1
"Sum: " print
$i1 $j1 1 range each
#dup "$seq[%d]\n" print
$seq swap get "%d," print
$max $seq $j1 get "%d = %d\n" print- Output:
Sum: 3,5,6,-2,-1,4 = 15
REXX
shortest greatest subsequential sum
This REXX version will find the sum of the shortest greatest continuous subsequence.
/*REXX program finds and displays the longest greatest continuous subsequence sum. */
parse arg @; w= words(@); p= w + 1 /*get arg list; number words in list. */
say 'words='w " list="@ /*show number words & LIST to terminal,*/
do #=1 for w; @.#= word(@, #); end /*build an array for faster processing.*/
L=0; sum= 0 /* [↓] process the list of numbers. */
do j=1 for w /*select one number at a time from list*/
do k=j to w; _= k-j+1; s= @.j /* [↓] process a sub─list of numbers. */
do m=j+1 to k; s= s + @.m; end /*m*/
if (s==sum & _>L) | s>sum then do; sum= s; p= j; L= _; end
end /*k*/ /* [↑] chose the longest greatest sum.*/
end /*j*/
say
$= subword(@,p,L); if $=='' then $= "[NULL]" /*Englishize the null (value). */
say 'sum='sum/1 " sequence="$ /*stick a fork in it, we're all done. */
- output when the following was used for the list: -1 -2 3 5 6 -2 -1 4 -4 2 -1
words=11 list=-1 -2 3 5 6 -2 -1 4 -4 2 -1 sum=15 sequence=3 5 6 -2 -1 4
- output when the following was used for the list: 1 2 3 4 -777 1 2 3 4 0 0
words=12 list=1 2 3 4 0 -777 1 2 3 4 0 0 sum=10 sequence=1 2 3 4
longest greatest subsequential sum
This REXX version will find the sum of the longest greatest continuous subsequence.
/*REXX program finds and displays the shortest greatest continuous subsequence sum.*/
parse arg @; w= words(@); p= w + 1 /*get arg list; number words in list. */
say 'words='w " list="@ /*show number words & LIST to terminal.*/
do #=1 for w; @.#= word(@, #); end /*build an array for faster processing.*/
L=0; sum= 0 /* [↓] process the list of numbers. */
do j=1 for w /*select one number at a time from list*/
do k=j to w; s= @.j /* [↓] process a sub─list of numbers. */
do m=j+1 to k; s= s + @.m; end /*m*/
if s>sum then do; sum= s; p= j; L= k - j + 1; end
end /*k*/ /* [↑] chose greatest sum of numbers. */
end /*j*/
say
$= subword(@,p,L); if $=='' then $= "[NULL]" /*Englishize the null (value). */
say 'sum='sum/1 " sequence="$ /*stick a fork in it, we're all done. */
- output when the following was used for the list: 1 2 3 4 -777 1 2 3 4 0 0
words=12 list=1 2 3 4 0 -777 1 2 3 4 0 0 sum=10 sequence=1 2 3 4 0 0
Version 3 (translated from Pascal)
/* REXX ***************************************************************
* 09.08.2012 Walter Pachl translated Pascal algorithm to Rexx
**********************************************************************/
s=' -1 -2 3 5 6 -2 -1 4 -4 2 -1'
maxSum = 0
seqStart = 0
seqEnd = -1
do i = 1 To words(s)
seqSum = 0
Do j = i to words(s)
seqSum = seqSum + word(s,j)
if seqSum > maxSum then Do
maxSum = seqSum
seqStart = i
seqEnd = j
end
end
end
Say 'Sequence:'
Say s
Say 'Subsequence with greatest sum: '
If seqend<seqstart Then
Say 'empty'
Else Do
ol=copies(' ',seqStart-1)
Do i = seqStart to seqEnd
ol=ol||right(word(s,i),3)
End
Say ol
Say 'Sum:' maxSum
End
- Output:
Sequence:
-1 -2 3 5 6 -2 -1 4 -4 2 -1
Subsequence with greatest sum:
3 5 6 -2 -1 4
Sum: 15
Ring
# Project : Greatest subsequential sum
aList1 = [0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2]
see "[0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2] -> " + sum(aList1) + nl
aList2 = [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
see "[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] -> " + sum(aList2) + nl
aList3 = [-1, -2, -3, -4, -5]
see "[-1, -2, -3, -4, -5] -> " + sum(aList3) + nl
aList4 = []
see "[] - > " + sum(aList4) + nl
func sum aList
sumold = []
sumnew = []
snew = 0
flag = 0
if len(aList) = 0
return 0
ok
for s=1 to len(aList)
if aList[s] > -1
flag = 1
ok
next
if flag = 0
return "[]"
ok
for n=1 to len(aList)
sumold = []
sold = 0
for m=n to len(aList)
add(sumold, aList[m])
sold = sold + aList[m]
if sold > snew
snew = sold
sumnew = sumold
ok
next
next
return showarray(sumnew)
func showarray(a)
conv = "["
for i = 1 to len(a)
conv = conv + string(a[i]) + ", "
next
conv = left(conv, len(conv) - 2) + "]"
return convOutput:
[0, 1, 2, -3, 3, -1, 0, -4, 0, -1, -4, 2] -> [0, 1, 2] [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] -> [3, 5, 6, -2, -1, 4] [-1, -2, -3, -4, -5] -> [] [] - > 0
RPL
≪ DUP SIZE -> input size
≪ { }
CASE
size NOT THEN END @ empty list case
size 1 == THEN @ singleton case
IF input 1 GET 0 ≥ THEN DROP input END
END
input 0 < ΠLIST NOT THEN @ for any list with at least 1 item > 0
input ≪ MAX ≫ STREAM @ initialize sum with maximum item
+ LASTARG SWAP DROP
1 size 2 - FOR len
1 size len - FOR j
input j DUP len + SUB
DUP ∑LIST 3 PICK
IF OVER < THEN 4 ROLL 4 ROLL END
DROP2
NEXT NEXT
DROP
END
END
≫ ≫ 'BIGSUB' STO
{ { -1 } { -1 2 -1 } { -1 2 -1 3 -1 } { -1 1 2 -5 -6 } { -1 -2 3 5 6 -2 -1 4 -4 2 -1 } }
1 ≪ BIGSUB ≫ DOLIST
- Output:
1: { { } { 2 } { 2 -1 3 } { 1 2 } { 3 5 6 -2 -1 4 } }
Using matrices
≪ DUP SIZE → input size
≪ { }
CASE
size 1 == THEN @ singleton case
IF input 1 GET 0 ≥ THEN DROP input END
END
size THEN
DROP
size DUP ≪ → j k ≪ IF j k ≤ THEN input j k SUB 0 + ∑LIST ELSE 0 END ≫ ≫ LCXM
@ forall(i=1:an,j=1:an) mix(i,j) = sum(a(i:j))
OBJ→ ΠLIST →LIST DUP ≪ MAX ≫ STREAM POS @ m = maxloc(mix)
1 - size IDIV2 R→C (1,1) + input SWAP C→R SUB @ print *, a(m(1):m(2))
END
END
≫ ≫ 'BIGSUB' STO
Efficient solution
Uses only basic RPL instructions for maximum compatibility.
≪ DUP SIZE → s length ≪ { } IF length THEN 0 (1 0) 1 length FOR j 0 j length FOR k s k GET + IF 3 PICK OVER < THEN ROT ROT DROP2 j k R→C OVER END NEXT DROP NEXT SWAP DROP IF DUP IM THEN s SWAP C→R SUB + ELSE DROP END END ≫ ≫ 'BIGSUB' STO
Ruby
Brute Force:
Answer is stored in "slice". It is very slow O(n**3)
def subarray_sum(arr)
max, slice = 0, []
arr.each_index do |i|
(i...arr.length).each do |j|
sum = arr[i..j].inject(0, :+)
max, slice = sum, arr[i..j] if sum > max
end
end
[max, slice]
end
Test:
[ [1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5],
[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1],
[-1, -2, -3, -4, -5],
[]
].each do |input|
puts "\nInput seq: #{input}"
puts " Max sum: %d\n Subseq: %s" % subarray_sum(input)
end
- Output:
Input seq: [1, 2, 3, 4, 5, -8, -9, -20, 40, 25, -5] Max sum: 65 Subseq: [40, 25] Input seq: [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] Max sum: 15 Subseq: [3, 5, 6, -2, -1, 4] Input seq: [-1, -2, -3, -4, -5] Max sum: 0 Subseq: [] Input seq: [] Max sum: 0 Subseq: []
Linear Time Version:
A better answer would run in O(n) instead of O(n**2) using numerical properties to remove the need for the inner loop.
# the trick is that at any point
# in the iteration if starting a new chain is
# better than your current score with this element
# added to it, then do so.
# the interesting part is proving the math behind it
def subarray_sum(arr)
curr = max = 0
first, last, curr_first = arr.size, 0, 0
arr.each_with_index do |e,i|
curr += e
if e > curr
curr = e
curr_first = i
end
if curr > max
max = curr
first = curr_first
last = i
end
end
return max, arr[first..last]
end
The test result is the same above.
Run BASIC
seq$ = "-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1"
max = -999
for i = 1 to 11
sum = 0
for j = i to 11
sum = sum + val(word$(seq$,j,","))
If sum > max then
max = sum
i1 = i
j1 = j
end if
next j
next i
print "Sum:";
for i = i1 to j1
print word$(seq$,i,",");",";
next i
print " = ";max- Output:
Sum: 3, 5, 6, -2, -1, 4, = 15
Rust
Naive brute force
fn main() {
let nums = [1,2,39,34,20, -20, -16, 35, 0];
let mut max = 0;
let mut boundaries = 0..0;
for length in 0..nums.len() {
for start in 0..nums.len()-length {
let sum = (&nums[start..start+length]).iter()
.fold(0, |sum, elem| sum+elem);
if sum > max {
max = sum;
boundaries = start..start+length;
}
}
}
println!("Max subsequence sum: {} for {:?}", max, &nums[boundaries]);;
}
- Output:
Max subsequence sum: 96 for [1, 2, 39, 34, 20]
Scala
The first solution solves the problem as specified, the second gives preference to the longest subsequence in case of ties. They are both vulnerable to integer overflow.
The third solution accepts any type N for which there's a Numeric[N], which includes all standard numeric types, and can be extended to include user defined numeric classes.
The last solution keeps to linear time by increasing complexity slightly.
def maxSubseq(l: List[Int]) = l.scanRight(Nil : List[Int]) {
case (el, acc) if acc.sum + el < 0 => Nil
case (el, acc) => el :: acc
} max Ordering.by((_: List[Int]).sum)
def biggestMaxSubseq(l: List[Int]) = l.scanRight(Nil : List[Int]) {
case (el, acc) if acc.sum + el < 0 => Nil
case (el, acc) => el :: acc
} max Ordering.by((ss: List[Int]) => (ss.sum, ss.length))
def biggestMaxSubseq[N](l: List[N])(implicit n: Numeric[N]) = {
import n._
l.scanRight(Nil : List[N]) {
case (el, acc) if acc.sum + el < zero => Nil
case (el, acc) => el :: acc
} max Ordering.by((ss: List[N]) => (ss.sum, ss.length))
}
def linearBiggestMaxSubseq[N](l: List[N])(implicit n: Numeric[N]) = {
import n._
l.scanRight((zero, Nil : List[N])) {
case (el, (acc, _)) if acc + el < zero => (zero, Nil)
case (el, (acc, ss)) => (acc + el, el :: ss)
} max Ordering.by((t: (N, List[N])) => (t._1, t._2.length)) _2
}
Scheme
(define (maxsubseq in)
(let loop
((_sum 0) (_seq (list)) (maxsum 0) (maxseq (list)) (l in))
(if (null? l)
(cons maxsum (reverse maxseq))
(let* ((x (car l)) (sum (+ _sum x)) (seq (cons x _seq)))
(if (> sum 0)
(if (> sum maxsum)
(loop sum seq sum seq (cdr l))
(loop sum seq maxsum maxseq (cdr l)))
(loop 0 (list) maxsum maxseq (cdr l)))))))
This returns a cons of the maximum sum and (one of) the maximum subsequence(s).
Seed7
$ include "seed7_05.s7i";
const func array integer: maxSubseq (in array integer: sequence) is func
result
var array integer: maxSequence is 0 times 0;
local
var integer: number is 0;
var integer: index is 0;
var integer: currentSum is 0;
var integer: currentStart is 1;
var integer: maxSum is 0;
var integer: startPos is 0;
var integer: endPos is 0;
begin
for number key index range sequence do
currentSum +:= number;
if currentSum < 0 then
currentStart := succ(index);
currentSum := 0;
elsif currentSum > maxSum then
maxSum := currentSum;
startPos := currentStart;
endPos := index;
end if;
end for;
if startPos <= endPos and startPos >= 1 and endPos >= 1 then
maxSequence := sequence[startPos .. endPos];
end if;
end func;
const proc: main is func
local
const array integer: a1 is [] (-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1);
const array integer: a2 is [] (-1, -2, -3, -5, -6, -2, -1, -4, -4, -2, -1);
var integer: number is 0;
begin
write("Maximal subsequence:");
for number range maxSubseq(a1) do
write(" " <& number);
end for;
writeln;
write("Maximal subsequence:");
for number range maxSubseq(a2) do
write(" " <& number);
end for;
writeln;
end func;- Output:
Maximal subsequence: 3 5 6 -2 -1 4 Maximal subsequence:
Sidef
func maxsubseq(*a) {
var (start, end, sum, maxsum) = (-1, -1, 0, 0)
a.each_kv { |i, x|
sum += x
if (maxsum < sum) {
maxsum = sum
end = i
}
elsif (sum < 0) {
sum = 0
start = i
}
}
a.slice(start+1).first(end-start)
}
say maxsubseq(-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1)
say maxsubseq(-2, -2, -1, 3, 5, 6, -1, 4, -4, 2, -1)
say maxsubseq(-2, -2, -1, -3, -5, -6, -1, -4, -4, -2, -1)
- Output:
[3, 5, 6, -2, -1, 4] [3, 5, 6, -1, 4] []
SparForte
As a structured script.
#!/usr/local/bin/spar
pragma annotate( summary, "gss" )
@( description, "greatest sequential sum" )
@( description, "Given a sequence of integers, find a continuous subsequence which maximizes the" )
@( description, "sum of its elements, that is, the elements of no other single subsequence add" )
@( description, "up to a value larger than this one. An empty subsequence is considered to have" )
@( description, "the sum 0; thus if all elements are negative, the result must be the empty" )
@( description, "sequence." )
@( see_also, "http://rosettacode.org/wiki/Greatest_subsequential_sum" )
@( author, "Ken O. Burtch" );
pragma license( unrestricted );
pragma restriction( no_external_commands );
procedure gss is
type int_array is array( 1..11 ) of integer;
a : constant int_array := (-1 , -2 , 3 , 5 , 6 , -2 , -1 , 4 , -4 , 2 , -1);
length : constant integer := arrays.length( a );
beginmax : integer := 0;
endmax : integer := -1;
maxsum : integer := 0;
running_sum : integer := 0;
begin
for start in arrays.first(a)..length-1 loop
running_sum := 0;
for finish in start..length-1 loop
running_sum := @ + a(finish);
if running_sum > maxsum then
maxsum := running_sum;
beginmax := start;
endmax := finish;
end if;
end loop;
end loop;
for i in beginmax..endmax loop
? a(i);
end loop;
end gss;
SQL
This is not a particularly efficient solution, but it gets the job done.
/*
This is a code implementation for finding one or more contiguous subsequences in a general sequence with the maximum sum of its elements.
p_list -- List of elements of the general sequence of integers separated by a delimiter.
p_delimiter -- proper delimiter
*/
with
function greatest_subsequential_sum(p_list in varchar2, p_delimiter in varchar2) return varchar2 is
-- Variablen
v_list varchar2(32767) := trim(both p_delimiter from p_list);
v_substr_i varchar2(32767);
v_substr_j varchar2(32767);
v_substr_out varchar2(32767);
v_res integer := 0;
v_res_out integer := 0;
--
begin
--
v_list := regexp_replace(v_list,''||chr(92)||p_delimiter||'{2,}',p_delimiter);
--
for i in 1..nvl(regexp_count(v_list,'[^'||p_delimiter||']+'),0)
loop
v_substr_i := substr(v_list,regexp_instr(v_list,'[^'||p_delimiter||']+',1,i));
--
for j in reverse 1..regexp_count(v_substr_i,'[^'||p_delimiter||']+')
loop
--
v_substr_j := trim(both p_delimiter from substr(v_substr_i,1,regexp_instr(v_substr_i,'[^'||p_delimiter||']+',1,j,1)));
execute immediate 'select sum('||replace(v_substr_j,p_delimiter,'+')||') from dual' into v_res;
--
if v_res > v_res_out then
v_res_out := v_res;
v_substr_out := '{'||v_substr_j||'}';
elsif v_res = v_res_out then
v_res_out := v_res;
v_substr_out := v_substr_out||',{'||v_substr_j||'}';
end if;
--
end loop;
--
end loop;
--
v_substr_out := trim(both ',' from nvl(v_substr_out,'{}'));
v_substr_out := case when regexp_count(v_substr_out,'},{')>0 then 'subsequences '||v_substr_out else 'a subsequence '||v_substr_out end;
return 'The maximum sum '||v_res_out||' belongs to '||v_substr_out||' of the main sequence {'||p_list||'}';
end;
--Test
select greatest_subsequential_sum('-1|-2|-3|-4|-5|', '|') as "greatest subsequential sum" from dual
union all
select greatest_subsequential_sum('', '') from dual
union all
select greatest_subsequential_sum(' ', ' ') from dual
union all
select greatest_subsequential_sum(';;;;;;+1;;;;;;;;;;;;;2;+3;4;;;;-5;;;;', ';') from dual
union all
select greatest_subsequential_sum('-1,-2,+3,,,,,,,,,,,,+5,+6,-2,-1,+4,-4,+2,-1', ',') from dual
union all
select greatest_subsequential_sum(',+7,-6,-8,+5,-2,-6,+7,+4,+8,-9,-3,+2,+6,-4,-6,,', ',') from dual
union all
select greatest_subsequential_sum('01 +2 3 +4 05 -8 -9 -20 40 25 -5', ' ') from dual
union all
select greatest_subsequential_sum('1 2 3 0 0 -99 02 03 00001 -99 3 2 1 -99 3 1 2 0', ' ') from dual
union all
select greatest_subsequential_sum('0,0,1,0', ',') from dual
union all
select greatest_subsequential_sum('0,0,0', ',') from dual
union all
select greatest_subsequential_sum('1,-1,+1', ',') from dual;
- Output:
The maximum sum 0 belongs to a subsequence {} of the main sequence {-1|-2|-3|-4|-5|}
The maximum sum 0 belongs to a subsequence {} of the main sequence {}
The maximum sum 0 belongs to a subsequence {} of the main sequence { }
The maximum sum 10 belongs to a subsequence {+1;2;+3;4} of the main sequence {;;;;;;+1;;;;;;;;;;;;;2;+3;4;;;;-5;;;;}
The maximum sum 15 belongs to a subsequence {+3,+5,+6,-2,-1,+4} of the main sequence {-1,-2,+3,,,,,,,,,,,,+5,+6,-2,-1,+4,-4,+2,-1}
The maximum sum 19 belongs to a subsequence {+7,+4,+8} of the main sequence {,+7,-6,-8,+5,-2,-6,+7,+4,+8,-9,-3,+2,+6,-4,-6,,}
The maximum sum 65 belongs to a subsequence {40 25} of the main sequence {01 +2 3 +4 05 -8 -9 -20 40 25 -5}
The maximum sum 6 belongs to subsequences {1 2 3 0 0},{1 2 3 0},{1 2 3},{02 03 00001},{3 2 1},{3 1 2 0},{3 1 2} of the main sequence {1 2 3 0 0 -99 02 03 00001 -99 3 2 1 -99 3 1 2 0}
The maximum sum 1 belongs to subsequences {0,0,1,0},{0,0,1},{0,1,0},{0,1},{1,0},{1} of the main sequence {0,0,1,0}
The maximum sum 0 belongs to subsequences {0,0,0},{0,0},{0},{0,0},{0},{0} of the main sequence {0,0,0}
The maximum sum 1 belongs to subsequences {1,-1,+1},{1},{+1} of the main sequence {1,-1,+1}
Standard ML
val maxsubseq = let
fun loop (_, _, maxsum, maxseq) [] = (maxsum, rev maxseq)
| loop (sum, seq, maxsum, maxseq) (x::xs) = let
val sum = sum + x
val seq = x :: seq
in
if sum < 0 then
loop (0, [], maxsum, maxseq) xs
else if sum > maxsum then
loop (sum, seq, sum, seq) xs
else
loop (sum, seq, maxsum, maxseq) xs
end
in
loop (0, [], 0, [])
end;
maxsubseq [~1, ~2, 3, 5, 6, ~2, ~1, 4, ~4, 2, ~1]
This returns a pair of the maximum sum and (one of) the maximum subsequence(s).
Swift
func maxSubseq(sequence: [Int]) -> (Int, Int, Int) {
var maxSum = 0, thisSum = 0, i = 0
var start = 0, end = -1
for (j, seq) in sequence.enumerated() {
thisSum += seq
if thisSum < 0 {
i = j + 1
thisSum = 0
} else if (thisSum > maxSum) {
maxSum = thisSum
start = i
end = j
}
}
return start <= end && start >= 0 && end >= 0
? (start, end + 1, maxSum) : (0, 0, 0)
}
let a = [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]
let (start, end, maxSum) = maxSubseq(sequence: a)
print("Max sum = \(maxSum)")
print(a[start..<end])
- Output:
Max sum = 15 [3, 5, 6, -2, -1, 4]
Tcl
package require Tcl 8.5
set a {-1 -2 3 5 6 -2 -1 4 -4 2 -1}
# from the Perl solution
proc maxsumseq1 {a} {
set len [llength $a]
set maxsum 0
for {set start 0} {$start < $len} {incr start} {
for {set end $start} {$end < $len} {incr end} {
set sum 0
incr sum [expr [join [lrange $a $start $end] +]]
if {$sum > $maxsum} {
set maxsum $sum
set maxsumseq [lrange $a $start $end]
}
}
}
return $maxsumseq
}
# from the Python solution
proc maxsumseq2 {sequence} {
set start -1
set end -1
set maxsum_ 0
set sum_ 0
for {set i 0} {$i < [llength $sequence]} {incr i} {
set x [lindex $sequence $i]
incr sum_ $x
if {$maxsum_ < $sum_} {
set maxsum_ $sum_
set end $i
} elseif {$sum_ < 0} {
set sum_ 0
set start $i
}
}
assert {$maxsum_ == [maxsum $sequence]}
assert {$maxsum_ == [sum [lrange $sequence [expr {$start + 1}] $end]]}
return [lrange $sequence [expr {$start + 1}] $end]
}
proc maxsum {sequence} {
set maxsofar 0
set maxendinghere 0
foreach x $sequence {
set maxendinghere [expr {max($maxendinghere + $x, 0)}]
set maxsofar [expr {max($maxsofar, $maxendinghere)}]
}
return $maxsofar
}
proc assert {condition {message "Assertion failed!"}} {
if { ! [uplevel 1 [list expr $condition]]} {
return -code error $message
}
}
proc sum list {
expr [join $list +]
}
puts "sequence: $a"
puts "maxsumseq1: [maxsumseq1 $a]"
puts [time {maxsumseq1 $a} 1000]
puts "maxsumseq2: [maxsumseq2 $a]"
puts [time {maxsumseq2 $a} 1000]
- Output:
sequence: -1 -2 3 5 6 -2 -1 4 -4 2 -1 maxsumseq1: 3 5 6 -2 -1 4 367.041 microseconds per iteration maxsumseq2: 3 5 6 -2 -1 4 74.623 microseconds per iteration
Ursala
This example solves the problem by the naive algorithm of testing all possible subsequences.
#import std
#import int
max_subsequence = zleq$^l&r/&+ *aayK33PfatPRTaq ^/~& sum:-0
#cast %zL
example = max_subsequence <-1,-2,3,5,6,-2,-1,4,-4,2,-1>The general theory of operation is as follows.
- The
max_subsequencefunction is a composition of three functions, one to generate the sequences, one to sum all of them, and one to pick out the one with the maximum sum. - The function that sums all the sequences is
* ^/~& sum:-0which applies to every member of a list (by the*operator) and forms a pair (using the^operator) of the identity function (~&) of its argument, and the reduction (:-) of the sum over a list with a vacuous case result of 0. - The function that picks out the maximum sum is
zleq$^l&r/&, which uses the maximizing operator ($^) over a list of pairs with respect to the integer ordering relation (zleq) applied to the right sides of the pairs (&r), after which the left side (l) of the maximizing pair is extracted. The/&inserts an extra pair(<>,0)at the beginning of the list before searching it in case it's empty or has only negative sums. - The function that generates all the sequences is
~&aayK33PfatPRTaq, which appears as a suffix of the*operator rather than being used explicitly. - The sequence generating function is in the form of a recursive conditional (
q) with predicatea, inductive caseayK33PfatPRTand base casea, meaning that in the base case of an empty list argument, the argument itself is returned. - The inductive case,
ayK33PfatPRTis a concatenation (T) of two functionsayK33andfatPR - The latter function,
fatPRis a recursive call (R) of the enclosing recursive conditional (f) with the tail of the argument (at). - The remaining function,
ayK33uses the triangle-squared combinatorK33of the list-lead operatoryapplied to the argumenta. - The list lead operator
yby itself takes a non-empty list as an argument and returns a copy with the last item deleted. - The triangle-squared combinator
K33constructs a function that takes an input list of a length , constructs a list of copies of it, and applies its operand 0 times to the head, once to the head of tail, twice to the head of the tail of the tail, and so on. Hence, an operand ofywill generate the list of all prefixes of a list.
- Output:
<3,5,6,-2,-1,4>
Wren
var gss = Fn.new { |s|
var best = 0
var start = 0
var end = 0
var sum = 0
var sumStart = 0
var i = 0
for (x in s) {
sum = sum + x
if (sum > best) {
best = sum
start = sumStart
end = i + 1
} else if (sum < 0) {
sum = 0
sumStart = i + 1
}
i = i + 1
}
return [s[start...end], best]
}
var tests = [
[-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1],
[-1, 1, 2, -5, -6],
[],
[-1, -2, -1]
]
for (test in tests) {
System.print("Input: %(test)")
var res = gss.call(test)
var subSeq = res[0]
var sum = res[1]
System.print("Sub seq: %(subSeq)")
System.print("Sum: %(sum)\n")
}
- Output:
Input: [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1] Sub seq: [3, 5, 6, -2, -1, 4] Sum: 15 Input: [-1, 1, 2, -5, -6] Sub seq: [1, 2] Sum: 3 Input: [] Sub seq: [] Sum: 0 Input: [-1, -2, -1] Sub seq: [] Sum: 0
XPL0
include c:\cxpl\codes;
int Array, Size, Sum, Best, I, Lo, Hi, BLo, BHi;
[Array:= [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1];
Size:= 11;
Best:= -100000;
for Lo:= 0 to Size-1 do
for Hi:= Lo to Size-1 do
[Sum:= 0;
for I:= Lo to Hi do
Sum:= Sum + Array(I);
if Sum > Best then
[Best:= Sum; BLo:= Lo; BHi:= Hi];
];
Text(0, "Sequence = ");
for I:= 0 to Size-1 do
[IntOut(0, Array(I)); Text(0, " ")];
CrLf(0);
Text(0, "Greatest = ");
for I:= BLo to BHi do
[IntOut(0, Array(I)); Text(0, " ")];
CrLf(0);
Text(0, "Sum = "); IntOut(0, Best); CrLf(0);
]- Output:
Sequence = -1 -2 3 5 6 -2 -1 4 -4 2 -1 Greatest = 3 5 6 -2 -1 4 Sum = 15
zkl
fcn maxsubseq(s){
s.reduce(fcn([(sum, seq, maxsum, maxseq)], x){
sum=sum+x; seq=T(x).extend(seq);
if(sum < 0) return(0,T,maxsum,maxseq);
if (sum>maxsum) return(sum, seq, sum, seq);
return(sum, seq, maxsum, maxseq);
},
T(0,T,0,T))[3].reverse(); // -->maxseq.reverse()
}s:=maxsubseq(T(-1,-2,3,5,6,-2,-1,4,-4,2,-1));
println(s.sum()," : ",s);
s:=maxsubseq(T(-1,-2)); println(s.sum()," : ",s);
s:=maxsubseq(T); println(s.sum()," : ",s);- Output:
15 : L(3,5,6,-2,-1,4) 0 : L() 0 : L()
ZX Spectrum Basic
10 DATA 12,0,1,2,-3,3,-1,0,-4,0,-1,-4,2
20 DATA 11,-1,-2,3,5,6,-2,-1,4,-4,2,-1
30 DATA 5,-1,-2,-3,-4,-5
40 FOR n=1 TO 3
50 READ l
60 DIM a(l)
70 FOR i=1 TO l
80 READ a(i)
90 PRINT a(i);
100 IF i<l THEN PRINT ", ";
110 NEXT i
120 PRINT
130 LET a=1: LET m=0: LET b=0
140 FOR i=1 TO l
150 LET s=0
160 FOR j=i TO l
170 LET s=s+a(j)
180 IF s>m THEN LET m=s: LET a=i: LET b=j
190 NEXT j
200 NEXT i
210 IF a>b THEN PRINT "[]": GO TO 280
220 PRINT "[";
230 FOR i=a TO b
240 PRINT a(i);
250 IF i<b THEN PRINT ", ";
260 NEXT i
270 PRINT "]"
280 NEXT nCategories:
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