Greatest subsequential sum

From Rosetta Code
(Redirected from Maximum subarray)
Jump to: navigation, search
Task
Greatest subsequential sum
You are encouraged to solve this task according to the task description, using any language you may know.

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 0; thus if all elements are negative, the result must be the empty sequence.

Contents

[edit] 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;

[edit] ALGOL 68

Translation of: C
Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d
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

[edit] 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) "]"

[edit] 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

[edit] AWK

Translation of: Common Lisp
# 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

[edit] 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] -> []

[edit] 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

[edit] 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 

[edit] 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;
}

[edit] 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();
}
}
}

[edit] Clojure

Translation of: Haskell

Naive algorithm:

(defn max-subseq-sum [coll]
(->> (take-while seq (iterate rest coll)) ; tails
(mapcat #(reductions conj [] %)) ; inits
(apply max-key #(reduce + %)))) ; max sum

Sample output:

user> (max-subseq-sum [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1])
[3 5 6 -2 -1 4]

[edit] 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]
 

[edit] Common Lisp

[edit] 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

[edit] Brute Force

Translation of: PicoLisp
(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))))

[edit] D

Translation of: Python
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: []

[edit] Alternative Version

Translation of: Haskell

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;
}

[edit] 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 1x0
 

Here 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 1x0
 

To 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
>test
 

[edit] 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 O(nm2) where n is the size of the sequence and m is the number of sign changes in the sequence. I think it could be improved to O(n + m2) 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())}
`
)

[edit] Euphoria

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}
{}
{}


[edit] 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]

[edit] 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

[edit] 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 - 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 ,
 
test 11 max-sub .array \ [3 5 6 -2 -1 4 ] = 15

[edit] Fortran

Works with: Fortran version 95 and later
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

[edit] 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 

[edit] 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 = snd . foldl f ((0,[]),(0,[])) where
f ((h1,h2),sofar) x = (a,b) where
a = max (0,[]) (h1 + x, h2 ++ [x])
b = max sofar a
 
main = print $ maxsubseq [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]

[edit] Icon and Unicon

procedure main()
L1 := [-1,-2,3,5,6,-2,-1,4,-4,2,-1] # sample list
L := [-1,1,2,3,4,-11]|||L1 # prepend a local maximum into the mix
write(ximage(maxsubseq(L)))
end
 
link ximage # to show lists
 
procedure maxsubseq(L) #: return the subsequence of L with maximum positive sum
local i,maxglobal,maxglobalI,maxlocal,maxlocalI
 
maxglobal := maxlocal := 0 # global and local maxima
 
every i := 1 to *L do {
if (maxlocal := max(maxlocal +L[i],0)) > 0 then
if /maxlocalI then maxlocalI := [i,i] else maxlocalI[2] := i # local maxima subscripts
else maxlocalI := &null # reset subsequence
if maxglobal <:= maxlocal then # global maxima
maxglobalI := copy(maxlocalI)
}
return L[(\maxglobalI)[1]:maxglobalI[2]] | [] # return sub-sequence or empty list
end

[edit] J

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.

[edit] Java

Works with: Java version 1.5+

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;
}

[edit] JavaScript

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;
}

[edit] 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 sub
 

[edit] Lua

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

[edit] 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) ')

[edit] Mathematica

[edit] 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]]&]]]]]
]

[edit] Method 2

MaximumSubsequence[x_List]:=Last@SortBy[Flatten[Table[x[[a;;b]], {b,Length[x]}, {a,b}],1],Total]

[edit] 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}
{}

[edit] 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
 
 

[edit] 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

[edit] 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

[edit] Nimrod

proc maxsum(s): 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])

[edit] 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).

[edit] 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]}}

[edit] 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
};

[edit] 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

[edit] 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";

[edit] Perl 6

Translation of: Python
Works with: Rakudo version #21 "Seattle"
sub maxsubseq (*@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 fufills the "all negative" requirement of the problem.

Note that once the triangular comma bug is resolved, the inner-loop subset calculation line can be shortened to "my @subsets = [\,] @a;".

sub max_sub-seq ( *@a ) {
 
my $max_subset = [];
while @a {
my @subsets = @a.keys.map: { [ @a[0..$_] ] };
@subsets.push($max_subset);
$max_subset = @subsets.max: { [+] .list };
@a.shift;
}
 
return $max_subset;
}
 
max_sub-seq( -1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1 ).perl.say;
max_sub-seq( -2, -2, -1, 3, 5, 6, -1, 4, -4, 2, -1 ).perl.say;
max_sub-seq( -2, -2, -1, -3, -5, -6, -1, -4, -4, -2, -1 ).perl.say;
Output:
[3, 5, 6, -2, -1, 4]
[3, 5, 6, -1, 4]
[]

[edit] 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
 

[edit] 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)

[edit] 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 

[edit] Prolog

[edit] 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.

[edit] 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

[edit] Python

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]

[edit] 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()]
}

Sample output:

> max.subseq(c(-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1))
[1] 3 5 6 -2 -1 4

[edit] 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
 

[edit] 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

[edit] REXX

[edit] shortest greatest subsequential sum

This REXX version will find the sum of the shortest greatest continous subsequence.

/*REXX program finds the  shortest  greatest continous subsequence  sum.*/
parse arg @; w=words(@) /*get arg list; # words in list.*/
say 'words='w " list="@ /*show #words & LIST to console.*/
sum=0; L=0; at=w+1 /*default sum, length, starts at.*/
/* [↓] process the list. */
do j=1 for w; f=word(@,j)
do k=j to w; s=f /* [↓] process the sub-list. */
do m=j+1 to k; s=s+word(@,m)
end /*m*/
if s>sum then do; sum=s; at=j; L=k-j+1; end
end /*k*/
end /*j*/
 
seq=subword(@,at,L); if seq=='' then seq="[NULL]"
say; say 'sum='word(sum 0,1)/1 " sequence="seq
/*stick a fork in it, we're 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

[edit] longest greatest subsequential sum

This REXX version will find the sum of the longest greatest continous subsequence.

/*REXX program finds the  longest  greatest continous subsequence  sum. */
parse arg @; w=words(@) /*get arg list; # words in list.*/
say 'words='w " list="@ /*show #words & LIST to console.*/
sum=0; L=0; at=w+1 /*default sum, length, starts at.*/
/* [↓] process the list. */
do j=1 for w; f=word(@,j)
do k=j to w; s=f /* [↓] process the sub-list. */
do m=j+1 to k; s=s+word(@,m)
end /*m*/
_=k-j+1
if (s==sum & _>L) | s>sum then do; sum=s; at=j; L=_; end
end /*k*/
end /*j*/
 
seq=subword(@,at,L); if seq=='' then seq="[NULL]"
say; say 'sum='word(sum 0,1)/1 " sequence="seq
/*stick a fork in it, we're 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

[edit] 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   

[edit] Ruby

Answer is stored in "slice":

Infinity = 1.0/0
def subarray_sum(arr)
max, slice = -Infinity, []
arr.each_with_index do |n, i|
(i...arr.length).each do |j|
sum = arr[i..j].inject(0) { |x, sum| sum += x }
if sum > max
max = sum
slice = arr[i..j]
end
end
end
[max, slice]
end

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
Infinity = 1.0/0
def subarray_sum(arr)
curr,max = -Infinity,-Infinity
first,last,curr_first= 0,0,0
arr.each_with_index do |e,i|
curr = e + curr
if(e>curr)
curr = e
curr_first=i
end
if(curr > max)
max = curr
last=i
first=curr_first
end
end
return max,arr[first...last+1]
end
 
input=[1,2,3,4,5,-8,-9,-20,40,25,-5]
p subarray_sum(input)
=>[65, [40, 25]]
 
input=[-3, -1]
p subarray_sum(input)
=>[-1, [-1]]

[edit] 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

[edit] Scala

Works with: Scala version 2.8

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
}

[edit] 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).

[edit] 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:

[edit] 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).

[edit] 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]

outputs

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

[edit] 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_subsequence function 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:-0 which 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 predicate a, inductive case ayK33PfatPRT and base case a, meaning that in the base case of an empty list argument, the argument itself is returned.
  • The inductive case, ayK33PfatPRT is a concatenation (T) of two functions ayK33 and fatPR
  • The latter function, fatPR is a recursive call (R) of the enclosing recursive conditional (f) with the tail of the argument (at).
  • The remaining function, ayK33 uses the triangle-squared combinator K33 of the list-lead operator y applied to the argument a.
  • The list lead operator y by itself takes a non-empty list as an argument and returns a copy with the last item deleted.
  • The triangle-squared combinator K33 constructs a function that takes an input list of a length n, constructs a list of n 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 of y will generate the list of all prefixes of a list.

output:

<3,5,6,-2,-1,4>

[edit] 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

[edit] zkl

Translation of: F#
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()
Personal tools
Namespaces

Variants
Actions
Community
Explore
Misc
Toolbox