Equilibrium index: Difference between revisions
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=={{header|Wren}}==
{{libheader|fmt}}
<lang ecmascript>var equilibrium = Fn.new { |a|▼
<lang ecmascript>import "/fmt" for Fmt
var len = a.count
var equi = []
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}
return equi
}
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System.print("The equilibrium indices for the following sequences are:\n")
for (test in tests) {
System.print("%(
}</lang>
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Revision as of 15:32, 9 May 2020
You are encouraged to solve this task according to the task description, using any language you may know.
An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence :
3 is an equilibrium index, because:
6 is also an equilibrium index, because:
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence .
- Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
11l
<lang 11l>F eqindex(arr)
R (0 .< arr.len).filter(i -> sum(@arr[0.<i]) == sum(@arr[i+1..]))
print(eqindex([-7, 1, 5, 2, -4, 3, 0]))</lang>
- Output:
[3, 6]
ABAP
<lang ABAP>REPORT equilibrium_index.
TYPES: y_i TYPE STANDARD TABLE OF i WITH EMPTY KEY.
cl_demo_output=>display( REDUCE y_i( LET sequences = VALUE y_i( ( -7 ) ( 1 ) ( 5 ) ( 2 ) ( -4 ) ( 3 ) ( 0 ) )
total_sum = REDUCE #( INIT sum = 0 FOR sequence IN sequences NEXT sum = sum + ( sequence ) ) IN INIT x = VALUE y_i( ) y = 0 FOR i = 1 UNTIL i > lines( sequences ) LET z = sequences[ i ] IN NEXT x = COND #( WHEN y = ( total_sum - y - z ) THEN VALUE y_i( BASE x ( i - 1 ) ) ELSE x ) y = y + z ) ).</lang>
Ada
Generic solution that returns a Vector of Indices.
equilibrium.ads: <lang Ada>with Ada.Containers.Vectors;
generic
type Index_Type is range <>; type Element_Type is private; Zero : Element_Type; with function "+" (Left, Right : Element_Type) return Element_Type is <>; with function "-" (Left, Right : Element_Type) return Element_Type is <>; with function "=" (Left, Right : Element_Type) return Boolean is <>; type Array_Type is private; with function Element (From : Array_Type; Key : Index_Type) return Element_Type is <>;
package Equilibrium is
package Index_Vectors is new Ada.Containers.Vectors (Index_Type => Positive, Element_Type => Index_Type);
function Get_Indices (From : Array_Type) return Index_Vectors.Vector;
end Equilibrium;</lang> equilibrium.adb: <lang Ada>package body Equilibrium is
function Get_Indices (From : Array_Type) return Index_Vectors.Vector is Result : Index_Vectors.Vector; Right_Sum, Left_Sum : Element_Type := Zero; begin for Index in Index_Type'Range loop Right_Sum := Right_Sum + Element (From, Index); end loop; for Index in Index_Type'Range loop Right_Sum := Right_Sum - Element (From, Index); if Left_Sum = Right_Sum then Index_Vectors.Append (Result, Index); end if; Left_Sum := Left_Sum + Element (From, Index); end loop; return Result; end Get_Indices;
end Equilibrium;</lang> Test program using two different versions, one with vectors and one with arrays: <lang Ada>with Ada.Text_IO; with Equilibrium; with Ada.Containers.Vectors;
procedure Main is
subtype Index_Type is Positive range 1 .. 7; package Vectors is new Ada.Containers.Vectors (Element_Type => Integer, Index_Type => Index_Type); type Plain_Array is array (Index_Type) of Integer; function Element (From : Plain_Array; Key : Index_Type) return Integer is begin return From (Key); end Element;
package Vector_Equilibrium is new Equilibrium (Index_Type => Index_Type, Element_Type => Integer, Zero => 0, Array_Type => Vectors.Vector, Element => Vectors.Element); package Array_Equilibrium is new Equilibrium (Index_Type => Index_Type, Element_Type => Integer, Zero => 0, Array_Type => Plain_Array);
My_Vector : Vectors.Vector; My_Array : Plain_Array := (-7, 1, 5, 2, -4, 3, 0); Vector_Result : Vector_Equilibrium.Index_Vectors.Vector; Array_Result : Array_Equilibrium.Index_Vectors.Vector := Array_Equilibrium.Get_Indices (My_Array);
begin
Vectors.Append (My_Vector, -7); Vectors.Append (My_Vector, 1); Vectors.Append (My_Vector, 5); Vectors.Append (My_Vector, 2); Vectors.Append (My_Vector, -4); Vectors.Append (My_Vector, 3); Vectors.Append (My_Vector, 0); Vector_Result := Vector_Equilibrium.Get_Indices (My_Vector); Ada.Text_IO.Put_Line ("Results:"); Ada.Text_IO.Put ("Array: "); for I in Array_Result.First_Index .. Array_Result.Last_Index loop Ada.Text_IO.Put (Integer'Image (Array_Equilibrium.Index_Vectors.Element (Array_Result, I))); end loop; Ada.Text_IO.New_Line; Ada.Text_IO.Put ("Vector: "); for I in Vector_Result.First_Index .. Vector_Result.Last_Index loop Ada.Text_IO.Put (Integer'Image (Vector_Equilibrium.Index_Vectors.Element (Vector_Result, I))); end loop; Ada.Text_IO.New_Line;
end Main;</lang>
- Output:
(Index_Type is based on 1):
Results: Array: 4 7 Vector: 4 7
Version that works with Ada 95, too:
equilibrium.adb: <lang Ada>with Ada.Text_IO;
procedure Equilibrium is
type Integer_Sequence is array (Positive range <>) of Integer; function Seq_Img (From : Integer_Sequence) return String is begin if From'First /= From'Last then return " " & Integer'Image (From (From'First)) & Seq_Img (From (From'First + 1 .. From'Last)); else return " " & Integer'Image (From (From'First)); end if; end Seq_Img;
type Boolean_Sequence is array (Positive range <>) of Boolean; function Seq_Img (From : Boolean_Sequence) return String is begin if From'First > From'Last then return ""; end if; if From (From'First) then return Integer'Image (From'First) & Seq_Img (From (From'First + 1 .. From'Last)); else return Seq_Img (From (From'First + 1 .. From'Last)); end if; end Seq_Img;
function Get_Indices (From : Integer_Sequence) return Boolean_Sequence is Result : Boolean_Sequence (From'Range) := (others => False); Left_Sum, Right_Sum : Integer := 0; begin for Index in From'Range loop Right_Sum := Right_Sum + From (Index); end loop; for Index in From'Range loop Right_Sum := Right_Sum - From (Index); Result (Index) := Left_Sum = Right_Sum; Left_Sum := Left_Sum + From (Index); end loop; return Result; end Get_Indices;
X1 : Integer_Sequence := (-7, 1, 5, 2, -4, 3, 0); X1_Result : Boolean_Sequence := Get_Indices (X1); X2 : Integer_Sequence := ( 2, 4, 6); X2_Result : Boolean_Sequence := Get_Indices (X2); X3 : Integer_Sequence := ( 2, 9, 2); X3_Result : Boolean_Sequence := Get_Indices (X3); X4 : Integer_Sequence := ( 1, -1, 1, -1, 1 ,-1, 1); X4_Result : Boolean_Sequence := Get_Indices (X4);
begin
Ada.Text_IO.Put_Line ("Results:"); Ada.Text_IO.New_Line; Ada.Text_IO.Put_Line ("X1:" & Seq_Img (X1)); Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X1_Result)); Ada.Text_IO.New_Line; Ada.Text_IO.Put_Line ("X2:" & Seq_Img (X2)); Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X2_Result)); Ada.Text_IO.New_Line; Ada.Text_IO.Put_Line ("X3:" & Seq_Img (X3)); Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X3_Result)); Ada.Text_IO.New_Line; Ada.Text_IO.Put_Line ("X4:" & Seq_Img (X4)); Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X4_Result));
end Equilibrium;</lang>
- Output:
Results: X1: -7 1 5 2 -4 3 0 Eqs: 4 7 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 2 X4: 1 -1 1 -1 1 -1 1 Eqs: 1 2 3 4 5 6 7
Aime
<lang aime>list eqindex(list l) {
integer e, i, s, sum; list x;
s = sum = 0; l.ucall(add_i, 1, sum); for (i, e in l) { if (s * 2 + e == sum) { x.append(i); } s += e; }
x;
}
integer main(void) {
list(-7, 1, 5, 2, -4, 3, 0).eqindex.ucall(o_, 0, "\n");
0;
}</lang>
ALGOL 68
<lang algol68>MODE YIELDINT = PROC(INT)VOID;
PROC gen equilibrium index = ([]INT arr, YIELDINT yield)VOID: (
INT sum := 0; FOR i FROM LWB arr TO UPB arr DO sum +:= arr[i] OD;
INT left:=0, right:=sum; FOR i FROM LWB arr TO UPB arr DO right -:= arr[i]; IF left = right THEN yield(i) FI; left +:= arr[i] OD
);
test:(
[]INT arr = []INT(-7, 1, 5, 2, -4, 3, 0)[@0];
- FOR INT index IN # gen equilibrium index(arr, # ) DO ( #
- (INT index)VOID:
print(index)
- OD # );
print(new line)
)</lang>
- Output:
+3 +6
AppleScript
(ES6 version)
<lang applescript>-- equilibriumIndices :: [Int] -> [Int] on equilibriumIndices(xs)
script balancedPair on |λ|(a, pair, i) set {x, y} to pair if x = y then {i - 1} & a else a end if end |λ| end script script plus on |λ|(a, b) a + b end |λ| end script -- Fold over zipped pairs of sums from left -- and sums from right foldr(balancedPair, {}, ¬ zip(scanl1(plus, xs), scanr1(plus, xs)))
end equilibriumIndices
-- TEST ----------------------------------------------------------------------- on run
map(equilibriumIndices, {¬ {-7, 1, 5, 2, -4, 3, 0}, ¬ {2, 4, 6}, ¬ {2, 9, 2}, ¬ {1, -1, 1, -1, 1, -1, 1}, ¬ {1}, ¬ {}}) --> {{3, 6}, {}, {1}, {0, 1, 2, 3, 4, 5, 6}, {0}, {}}
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- 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
-- foldr :: (a -> b -> a) -> a -> [b] -> a on foldr(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs repeat with i from lng to 1 by -1 set v to |λ|(v, item i of xs, i, xs) end repeat return v end tell
end foldr
-- init :: [a] -> [a] on init(xs)
set lng to length of xs if lng > 1 then items 1 thru -2 of xs else if lng > 0 then {} else missing value end if
end init
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell
end map
-- min :: Ord a => a -> a -> a on min(x, y)
if y < x then y else x end if
end min
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then f else script property |λ| : f end script end if
end mReturn
-- scanl :: (b -> a -> b) -> b -> [a] -> [b] on scanl(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs set lst to {startValue} repeat with i from 1 to lng set v to |λ|(v, item i of xs, i, xs) set end of lst to v end repeat return lst end tell
end scanl
-- scanl1 :: (a -> a -> a) -> [a] -> [a] on scanl1(f, xs)
if length of xs > 0 then scanl(f, item 1 of xs, tail(xs)) else {} end if
end scanl1
-- scanr :: (b -> a -> b) -> b -> [a] -> [b] on scanr(f, startValue, xs)
tell mReturn(f) set v to startValue set lng to length of xs set lst to {startValue} repeat with i from lng to 1 by -1 set v to |λ|(v, item i of xs, i, xs) set end of lst to v end repeat return reverse of lst end tell
end scanr
-- scanr1 :: (a -> a -> a) -> [a] -> [a] on scanr1(f, xs)
if length of xs > 0 then scanr(f, item -1 of xs, init(xs)) else {} end if
end scanr1
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then items 2 thru -1 of xs else {} end if
end tail
-- zip :: [a] -> [b] -> [(a, b)] on zip(xs, ys)
set lng to min(length of xs, length of ys) set lst to {} repeat with i from 1 to lng set end of lst to {item i of xs, item i of ys} end repeat return lst
end zip</lang>
- Output:
{{3, 6}, {}, {1}, {0, 1, 2, 3, 4, 5, 6}, {0}, {}}
AutoHotkey
<lang AutoHotkey>Equilibrium_index(list, BaseIndex=0){ StringSplit, A, list, `, Loop % A0 { i := A_Index , Pre := Post := 0 loop, % A0 if (A_Index < i) Pre += A%A_Index% else if (A_Index > i) Post += A%A_Index% if (Pre = Post) Res .= (Res?", ":"") i - (BaseIndex?0:1) } return Res }</lang> Examples:<lang AutoHotkey>list = -7, 1, 5, 2, -4, 3, 0 MsgBox % Equilibrium_index(list)</lang>
- Output:
3, 6
AWK
<lang AWK>
- syntax: GAWK -f EQUILIBRIUM_INDEX.AWK
BEGIN {
main("-7 1 5 2 -4 3 0") main("2 4 6") main("2 9 2") main("1 -1 1 -1 1 -1 1") exit(0)
} function main(numbers, x) {
x = equilibrium(numbers) printf("numbers: %s\n",numbers) printf("indices: %s\n\n",length(x)==0?"none":x)
} function equilibrium(numbers, arr,i,leftsum,leng,str,sum) {
leng = split(numbers,arr," ") for (i=1; i<=leng; i++) { sum += arr[i] } for (i=1; i<=leng; i++) { sum -= arr[i] if (leftsum == sum) { str = str i " " } leftsum += arr[i] } return(str)
} </lang>
Output:
numbers: -7 1 5 2 -4 3 0 indices: 4 7 numbers: 2 4 6 indices: none numbers: 2 9 2 indices: 2 numbers: 1 -1 1 -1 1 -1 1 indices: 1 2 3 4 5 6 7
Batch File
<lang dos>@echo off setlocal enabledelayedexpansion
call :equilibrium-index "-7 1 5 2 -4 3 0" call :equilibrium-index "2 4 6" call :equilibrium-index "2 9 2" call :equilibrium-index "1 -1 1 -1 1 -1 1" pause>nul exit /b
%== The Function ==%
- equilibrium-index <sequence with quotes>
::Set the pseudo-array sequence... set "seq=%~1" set seq.length=0 for %%S in (!seq!) do ( set seq[!seq.length!]=%%S set /a seq.length+=1 ) ::Initialization of other variables... set "equilms=" set /a last=seq.length - 1 ::The main checking... for /l %%e in (0,1,!last!) do ( set left=0 set right=0
for /l %%i in (0,1,!last!) do ( if %%i lss %%e (set /a left+=!seq[%%i]!) if %%i gtr %%e (set /a right+=!seq[%%i]!) ) if !left!==!right! ( if defined equilms ( set "equilms=!equilms! %%e" ) else ( set "equilms=%%e" ) ) ) echo [!equilms!] goto :EOF %==/The Function ==%</lang>
- Output:
[3 6] [] [1] [0 1 2 3 4 5 6]
BBC BASIC
BBC BASIC's SUM function is useful for this task. <lang bbcbasic> DIM list(6)
list() = -7, 1, 5, 2, -4, 3, 0 PRINT "Equilibrium indices are " FNequilibrium(list()) END DEF FNequilibrium(l()) LOCAL i%, r, s, e$ s = SUM(l()) FOR i% = 0 TO DIM(l(),1) IF r = s - r - l(i%) THEN e$ += STR$(i%) + "," r += l(i%) NEXT = LEFT$(e$)</lang>
Output:
Equilibrium indices are 3,6
C
<lang C>#include <stdio.h>
- include <stdlib.h>
int list[] = {-7, 1, 5, 2, -4, 3, 0};
int eq_idx(int *a, int len, int **ret) { int i, sum, s, cnt; /* alloc long enough: if we can afford the original list, * we should be able to afford to this. Beats a potential
* million realloc() calls. Even if memory is a real concern, * there's no garantee the result is shorter than the input anyway */ cnt = s = sum = 0;
*ret = malloc(sizeof(int) * len);
for (i = 0; i < len; i++)
sum += a[i];
for (i = 0; i < len; i++) { if (s * 2 + a[i] == sum) { (*ret)[cnt] = i;
cnt++; }
s += a[i]; }
/* uncouraged way to use realloc since it can leak memory, for example */
*ret = realloc(*ret, cnt * sizeof(int));
return cnt; }
int main() { int i, cnt, *idx; cnt = eq_idx(list, sizeof(list) / sizeof(int), &idx);
printf("Found:"); for (i = 0; i < cnt; i++)
printf(" %d", idx[i]);
printf("\n");
return 0; }</lang>
C#
<lang csharp>using System; using System.Collections.Generic; using System.Linq;
class Program {
static IEnumerable<int> EquilibriumIndices(IEnumerable<int> sequence) { var left = 0; var right = sequence.Sum(); var index = 0; foreach (var element in sequence) { right -= element; if (left == right) { yield return index; } left += element; index++; } }
static void Main() { foreach (var index in EquilibriumIndices(new[] { -7, 1, 5, 2, -4, 3, 0 })) { Console.WriteLine(index); } }
}</lang>
- Output:
<lang>3 6</lang>
C++
<lang cpp>#include <algorithm>
- include <iostream>
- include <numeric>
- include <vector>
template <typename T> std::vector<size_t> equilibrium(T first, T last) {
typedef typename std::iterator_traits<T>::value_type value_t;
value_t left = 0; value_t right = std::accumulate(first, last, value_t(0)); std::vector<size_t> result;
for (size_t index = 0; first != last; ++first, ++index) { right -= *first; if (left == right) { result.push_back(index); } left += *first; } return result;
}
template <typename T> void print(const T& value) {
std::cout << value << "\n";
}
int main() {
const int data[] = { -7, 1, 5, 2, -4, 3, 0 };
std::vector<size_t> indices(equilibrium(data, data + 7));
std::for_each(indices.begin(), indices.end(), print<size_t>);
}</lang>
- Output:
3 6
Clojure
<lang clojure>(defn equilibrium [lst]
(loop [acc '(), i 0, left 0, right (apply + lst), lst lst] (if (empty? lst)
(reverse acc) (let [[x & xs] lst right (- right x) acc (if (= left right) (cons i acc) acc)] (recur acc (inc i) (+ left x) right xs)))))</lang>
- Output:
> (equilibrium [-7, 1, 5, 2, -4, 3, 0]) (3 6)
Common Lisp
<lang lisp>(defun dflt-on-nil (v dflt)
(if v v dflt))
(defun eq-index (v)
(do* ((stack nil) (i 0 (+ 1 i)) (rest v (cdr rest)) (lsum 0) (rsum (apply #'+ (cdr v)))) ;; Reverse here is not strictly necessary ((null rest) (reverse stack)) (if (eql lsum rsum) (push i stack)) (setf lsum (+ lsum (car rest))) (setf rsum (- rsum (dflt-on-nil (cadr rest) 0)))))</lang>
- Output:
<lang>(eq-index '(-7 1 5 2 -4 3 0)) (3 6)</lang>
D
More Functional Style
<lang d>import std.stdio, std.algorithm, std.range, std.functional;
auto equilibrium(Range)(Range r) pure nothrow @safe /*@nogc*/ {
return r.length.iota.filter!(i => r[0 .. i].sum == r[i + 1 .. $].sum);
}
void main() {
[-7, 1, 5, 2, -4, 3, 0].equilibrium.writeln;
}</lang>
- Output:
[3, 6]
Less Functional Style
Same output. <lang d>import std.stdio, std.algorithm;
size_t[] equilibrium(T)(in T[] items) @safe pure nothrow {
size_t[] result; T left = 0, right = items.sum;
foreach (immutable i, e; items) { right -= e; if (right == left) result ~= i; left += e; } return result;
}
void main() {
[-7, 1, 5, 2, -4, 3, 0].equilibrium.writeln;
}</lang>
Elena
ELENA 5.0 : <lang elena>import extensions; import system'routines; import system'collections; import extensions'routines;
class EquilibriumEnumerator : Enumerator {
int left; int right; int index; Enumerator en; constructor new(Enumerator en) { this en := en; self.reset() } constructor new(Enumerable list) <= new(list.enumerator()); constructor new(o) <= new(cast Enumerable(o)); bool next() { index += 1; while(en.next()) { var element := en.get(); right -= element; bool found := (left == right); left += element; if (found) { ^ true }; index += 1 }; ^ false } reset() { en.reset(); left := 0; right := en.summarize(); index := -1; en.reset(); } get() = index; enumerable() => en;
}
public program() {
EquilibriumEnumerator.new(new int[]{ -7, 1, 5, 2, -4, 3, 0 }) .forEach:printingLn
}</lang>
3 6
Elixir
computes either side each time. <lang elixir>defmodule Equilibrium do
def index(list) do last = length(list) Enum.filter(0..last-1, fn i -> Enum.sum(Enum.slice(list, 0, i)) == Enum.sum(Enum.slice(list, i+1..last)) end) end
end</lang>
faster version: <lang elixir>defmodule Equilibrium do
def index(list), do: index(list,0,0,Enum.sum(list),[]) defp index([],_,_,_,acc), do: Enum.reverse(acc) defp index([h|t],i,left,right,acc) when left==right-h, do: index(t,i+1,left+h,right-h,[i|acc]) defp index([h|t],i,left,right,acc) , do: index(t,i+1,left+h,right-h,acc)
end</lang>
Test: <lang elixir>indices = [
[-7, 1, 5, 2,-4, 3, 0], [2, 4, 6], [2, 9, 2], [1,-1, 1,-1, 1,-1, 1]
] Enum.each(indices, fn list ->
IO.puts "#{inspect list} => #{inspect Equilibrium.index(list)}"
end)</lang>
- Output:
[-7, 1, 5, 2, -4, 3, 0] => [3, 6] [2, 4, 6] => [] [2, 9, 2] => [1] [1, -1, 1, -1, 1, -1, 1] => [0, 1, 2, 3, 4, 5, 6]
ERRE
<lang ERRE> PROGRAM EQUILIBRIUM
DIM LISTA[6]
PROCEDURE EQ(LISTA[]->RES$)
LOCAL I%,R,S,E$ FOR I%=0 TO UBOUND(LISTA,1) DO S+=LISTA[I%] END FOR FOR I%=0 TO UBOUND(LISTA,1) DO IF R=S-R-LISTA[I%] THEN E$+=STR$(I%)+"," END IF R+=LISTA[I%] END FOR RES$=LEFT$(E$,LEN(E$)-1)
END PROCEDURE
BEGIN
LISTA[]=(-7,1,5,2,-4,3,0) EQ(LISTA[]->RES$) PRINT("Equilibrium indices are";RES$)
END PROGRAM </lang> Output:
Equilibrium indices are 3, 6
Euphoria
<lang euphoria>function equilibrium(sequence s)
integer lower_sum, higher_sum sequence indices lower_sum = 0 higher_sum = 0 for i = 1 to length(s) do higher_sum += s[i] end for indices = {} for i = 1 to length(s) do higher_sum -= s[i] if lower_sum = higher_sum then indices &= i end if lower_sum += s[i] end for return indices
end function
? equilibrium({-7,1,5,2,-4,3,0})</lang>
- Output:
(Remember that indices are 1-based in Euphoria)
{4,7}
Factor
Executed in the listener. Note that accum-left
and accum-right
have different outputs than accumulate
as they drop the final result.
<lang factor>USE: math.vectors
- accum-left ( seq id quot -- seq ) accumulate nip ; inline
- accum-right ( seq id quot -- seq ) [ <reversed> ] 2dip accum-left <reversed> ; inline
- equilibrium-indices ( seq -- inds )
0 [ + ] [ accum-left ] [ accum-right ] 3bi [ = ] 2map V{ } swap dup length iota [ [ suffix ] curry [ ] if ] 2each ;</lang>
- Output:
<lang factor>( scratchpad ) { -7 1 5 2 -4 3 0 } equilibrium-indices . V{ 3 6 }</lang>
Fortran
Array indices are 1-based. <lang fortran>program Equilibrium
implicit none integer :: array(7) = (/ -7, 1, 5, 2, -4, 3, 0 /) call equil_index(array)
contains
subroutine equil_index(a)
integer, intent(in) :: a(:) integer :: i
do i = 1, size(a) if(sum(a(1:i-1)) == sum(a(i+1:size(a)))) write(*,*) i end do
end subroutine end program</lang>
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
Sub equilibriumIndices (a() As Integer, b() As Integer)
If UBound(a) = -1 Then Return empty array
Dim sum As Integer = 0 Dim count As Integer = 0 For i As Integer = LBound(a) To UBound(a) : sum += a(i) : Next Dim sumLeft As Integer = 0, sumRight As Integer = 0
For i As Integer = LBound(a) To UBound(a) sumRight = sum - sumLeft - a(i) If sumLeft = sumRight Then Redim Preserve b(0 To Count) b(count) = i count += 1 End If sumLeft += a(i) Next
End Sub
Dim a(0 To 6) As Integer = { -7, 1, 5, 2, -4, 3, 0 } Dim b() As Integer equilibriumIndices a(), b() If UBound(b) = -1 Then
Print "There are no equilibrium indices"
ElseIf UBound(b) = LBound(b) Then
Print "The only equilibrium index is : "; b(LBound(b))
Else
Print "The equilibrium indices are : " For i As Integer = LBound(b) To UBound(b) : Print b(i); " "; : Next
End If
Print Print "Press any key to quit" Sleep</lang>
- Output:
The equilibrium indices are : 3 6
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
<lang go>package main
import (
"fmt" "math/rand" "time"
)
func main() {
fmt.Println(ex([]int32{-7, 1, 5, 2, -4, 3, 0}))
// sequence of 1,000,000 random numbers, with values // chosen so that it will be likely to have a couple // of equalibrium indexes. rand.Seed(time.Now().UnixNano()) verylong := make([]int32, 1e6) for i := range verylong { verylong[i] = rand.Int31n(1001) - 500 } fmt.Println(ex(verylong))
}
func ex(s []int32) (eq []int) {
var r, l int64 for _, el := range s { r += int64(el) } for i, el := range s { r -= int64(el) if l == r { eq = append(eq, i) } l += int64(el) } return
}</lang>
- Output:
[3 6] [145125 947872]
Haskell
<lang haskell>import System.Random (randomRIO) import Data.List (elemIndices, takeWhile) import Control.Monad (replicateM) import Control.Arrow ((&&&))
equilibr xs =
elemIndices True . map (\(a, b) -> sum a == sum b) . takeWhile (not . null . snd) $ flip ((&&&) <$> take <*> (drop . pred)) xs <$> [1 ..]
langeSliert = replicateM 2000 (randomRIO (-15, 15) :: IO Int) >>= print . equilibr</lang> Small example <lang haskell>*Main> equilibr [-7, 1, 5, 2, -4, 3, 0] [3,6]</lang> Long random list in langeSliert (several tries for this one) <lang haskell>*Main> langeSliert [231,245,259,265,382,1480,1611,1612]</lang>
Or, using default Prelude functions:
<lang haskell>equilibriumIndices :: [Int] -> [Int] equilibriumIndices xs =
let matchIndex (x, y, i) a | x == y = i : a | otherwise = a in foldr matchIndex [] (zip3 (scanl1 (+) xs) -- Sums from the left (scanr1 (+) xs) -- Sums from the right [0 ..] -- Indices )
TEST----------------------------
main :: IO () main =
mapM_ print (equilibriumIndices <$> [ [-7, 1, 5, 2, -4, 3, 0] , [2, 4, 6] , [2, 9, 2] , [1, -1, 1, -1, 1, -1, 1] , [1] , [] ])</lang>
- Output:
[3,6] [] [1] [0,1,2,3,4,5,6] [0] []
Icon and Unicon
<lang Icon>procedure main(arglist) L := if *arglist > 0 then arglist else [-7, 1, 5, 2, -4, 3, 0] # command line args or default every writes( "equilibrium indicies of [ " | (!L ||" ") | "] = " | (eqindex(L)||" ") | "\n" ) end
procedure eqindex(L) # generate equilibrium points in a list L or fail local s,l,i
every (s := 0, i := !L) do
s +:= numeric(i) | fail # sum and validate
every (l := 0, i := 1 to *L) do {
if l = (s-L[i])/2 then suspend i l +:= L[i] # sum of left side }
end</lang>
- Output:
equilibrium indicies of [ -7 1 5 2 -4 3 0 ] = 4 7
J
<lang j>equilidx=: +/\ I.@:= +/\.</lang>
- Example use:
<lang j> equilidx _7 1 5 2 _4 3 0 3 6</lang>
Java
<lang java5> public class Equlibrium { public static void main(String[] args) { int[] sequence = {-7, 1, 5, 2, -4, 3, 0}; equlibrium_indices(sequence); }
public static void equlibrium_indices(int[] sequence){ //find total sum int totalSum = 0; for (int n : sequence) { totalSum += n; } //compare running sum to remaining sum to find equlibrium indices int runningSum = 0; for (int i = 0; i < sequence.length; i++) { int n = sequence[i]; if (totalSum - runningSum - n == runningSum) { System.out.println(i); } runningSum += n; } } } </lang>
- Output:
3 6
JavaScript
ES5
<lang javascript>function equilibrium(a) {
var N = a.length, i, l = [], r = [], e = [] for (l[0] = a[0], r[N - 1] = a[N - 1], i = 1; i<N; i++) l[i] = l[i - 1] + a[i], r[N - i - 1] = r[N - i] + a[N - i - 1] for (i = 0; i < N; i++) if (l[i] === r[i]) e.push(i) return e
}
// test & output [ [-7, 1, 5, 2, -4, 3, 0], // 3, 6
[2, 4, 6], // empty [2, 9, 2], // 1 [1, -1, 1, -1, 1, -1, 1], // 0,1,2,3,4,5,6 [1], // 0 [] // empty
].forEach(function(x) {
console.log(equilibrium(x))
});</lang>
- Output:
<lang JavaScript>[[3,6],[],[1],[0,1,2,3,4,5,6],[0],[]]</lang>
ES6 Procedural
Two pass O(n), returning only the first equilibrium index. <lang JavaScript>function equilibrium(arr) {
let sum = arr.reduce((a, b) => a + b); let leftSum = 0;
for (let i = 0; i < arr.length; ++i) { sum -= arr[i];
if (leftSum === sum) { return i; }
leftSum += arr[i]; }
return -1;
} </lang>
- Output:
<lang JavaScript>3, -1, 1, 0, 0</lang>
ES6 Functional
A composition of pure generic functions, returning all equilibrium indices.
<lang javascript>(() => {
'use strict';
// equilibriumIndices :: [Int] -> [Int] const equilibriumIndices = xs => zip(scanl1(add)(xs))( scanr1(add)(xs) ).reduceRight((a, xy, i) => xy[0] === xy[1] ? ( [i, ...a] ) : a, []);
// ------------------------TEST------------------------ const main = () => { console.log(JSON.stringify([ [-7, 1, 5, 2, -4, 3, 0], [2, 4, 6], [2, 9, 2], [1, -1, 1, -1, 1, -1, 1], [1], [] ].map(equilibriumIndices))); // -> [[3, 6], [], [1], [0, 1, 2, 3, 4, 5, 6], [0], []] };
// -----------------GENERIC FUNCTIONS------------------
// Tuple (,) :: a -> b -> (a, b) const Tuple = a => b => ({ type: 'Tuple', '0': a, '1': b, length: 2 });
// add (+) :: Num a => a -> a -> a const add = a => // Curried addition. b => a + b;
// scanl :: (b -> a -> b) -> b -> [a] -> [b] const scanl = f => // scanl is like foldl or reduce, but // returns a succession of intermediate // values, building from the left. v => xs => xs.reduce((a, x) => { const v = f(a[0])(x); return Tuple(v)(a[1].concat(v)); }, Tuple(v)([v]))[1];
// scanr :: (b -> a -> b) -> b -> [a] -> [b] const scanr = f => // scanr is like foldr or reduceRight, but // returns a succession of intermediate // values, building from the right. v => xs => xs.reduceRight((a, x) => { const v = f(x)(a[0]); return Tuple(v)([v].concat(a[1])); }, Tuple(v)([v]))[1];
// scanl1 :: (a -> a -> a) -> [a] -> [a] const scanl1 = f => // scanl1 is a variant of scanl that has no // seed-value argument, and assumes that // xs is not empty. xs => xs.length > 0 ? ( scanl(f)( xs[0] )(xs.slice(1)) ) : [];
// scanr1 :: (a -> a -> a) -> [a] -> [a] const scanr1 = f => // scanr1 is a variant of scanr that has no // seed-value argument, and assumes that // xs is not empty. xs => xs.length > 0 ? ( scanr(f)( xs.slice(-1)[0] )(xs.slice(0, -1)) ) : [];
// zip :: [a] -> [b] -> [(a, b)] const zip = xs => ys => Array.from({ length: Math.min(xs.length, ys.length) }, (_, i) => Tuple(xs[i])(ys[i]));
// MAIN --- return main();
})();</lang>
- Output:
[[3,6],[],[1],[0,1,2,3,4,5,6],[0],[]]
jq
The following implementation will work with jq 1.4 but for input arrays larger than 1e4 in length, a version of jq with tail-call optimization (TCO) should probably be used.
Since the task description indicates that the array might be very long:
- the implementation uses a 0-arity inner function to do the heavy lifting;
- the algorithm walks along the array so as to minimize both memory requirements and the number of arithmetic operations;
- the answers are emitted as a stream.
The top-level function is defined as a 0-arity filter that emits answers as a stream, as is idiomatic in jq. <lang jq># The index origin is 0 in jq. def equilibrium_indices:
def indices(a; mx): def report: # [i, headsum, tailsum] .[0] as $i | if $i == mx then empty # all done else .[1] as $h | (.[2] - a[$i]) as $t | (if $h == $t then $i else empty end), ( [ $i + 1, $h + a[$i], $t ] | report ) end; [0, 0, (a|add)] | report; . as $in | indices($in; $in|length);</lang>
Example 1: <lang jq>[-7, 1, 5, 2, -4, 3, 0] | equilibrium_indices</lang>
- Output:
$ jq -M -n -f equilibrium_indices.jq 3 6
Example 2: <lang jq>def count(g): reduce g as $i (0; .+1);
- Create an array of length n with "init" elements:
def array(n;init): reduce range(0;n) as $i ([]; . + [0]);
count( array(1e4;0) | equilibrium_indices )</lang>
- Output:
$ jq -M -n -f equilibrium_indices.jq 10000
Julia
<lang julia>function equindex2pass(data::Array)
rst = Vector{Int}(0) suml, sumr, ddelayed = 0, sum(data), 0 for (i, d) in enumerate(data) suml += ddelayed sumr -= d ddelayed = d if suml == sumr push!(rst, i) end end return rst
end
@show equindex2pass([1, -1, 1, -1, 1, -1, 1]) @show equindex2pass([1, 2, 2, 1]) @show equindex2pass([-7, 1, 5, 2, -4, 3, 0])</lang>
- Output:
equindex2pass([1, -1, 1, -1, 1, -1, 1]) = [1, 2, 3, 4, 5, 6, 7] equindex2pass([1, 2, 2, 1]) = Int64[] equindex2pass([-7, 1, 5, 2, -4, 3, 0]) = [4, 7]
K
<lang K> f:{&{(+/y# x)=+/(y+1)_x}[x]'!#x}
f -7 1 5 2 -4 3 0
3 6
f 2 4 6
!0
f 2 9 2
,1
f 1 -1 1 -1 1 -1 1
0 1 2 3 4 5 6</lang>
Kotlin
<lang scala>// version 1.1
fun equilibriumIndices(a: IntArray): MutableList<Int> {
val ei = mutableListOf<Int>() if (a.isEmpty()) return ei // empty list val sumAll = a.sumBy { it } var sumLeft = 0 var sumRight: Int for (i in 0 until a.size) { sumRight = sumAll - sumLeft - a[i] if (sumLeft == sumRight) ei.add(i) sumLeft += a[i] } return ei
}
fun main(args: Array<String>) {
val a = intArrayOf(-7, 1, 5, 2, -4, 3, 0) val ei = equilibriumIndices(a) when (ei.size) { 0 -> println("There are no equilibrium indices") 1 -> println("The only equilibrium index is : ${ei[0]}") else -> println("The equilibrium indices are : ${ei.joinToString(", ")}") }
}</lang>
- Output:
The equilibrium indices are : 3, 6
Liberty BASIC
<lang lb> a(0)=-7 a(1)=1 a(2)=5 a(3)=2 a(4)=-4 a(5)=3 a(6)=0
print "EQ Indices are ";EQindex$("a",0,6)
wait
function EQindex$(b$,mini,maxi)
if mini>=maxi then exit function sum=0 for i = mini to maxi sum=sum+eval(b$;"(";i;")") next sumA=0:sumB=sum for i = mini to maxi sumB = sumB - eval(b$;"(";i;")") if sumA=sumB then EQindex$=EQindex$+str$(i)+", " sumA = sumA + eval(b$;"(";i;")") next if len(EQindex$)>0 then EQindex$=mid$(EQindex$, 1, len(EQindex$)-2) 'remove last ", "
end function </lang>
- Output:
EQ Indices are 3, 6
Logo
<lang logo>to equilibrium.iter :i :before :after :tail :ret
if equal? :before :after [make "ret lput :i :ret] if empty? butfirst :tail [output :ret] output equilibrium.iter :i+1 (:before+first :tail) (:after-first butfirst :tail) (butfirst :tail) :ret
end to equilibrium.index :list
output equilibrium.iter 1 0 (apply "sum butfirst :list) :list []
end
show equilibrium_index [-7 1 5 2 -4 3 0] ; [4 7]</lang>
Mathematica / Wolfram Language
Mathematica indexes are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays. <lang Mathematica>equilibriumIndex[data_]:=Reap[
Do[If[Total[data;; n - 1] == Total[datan + 1 ;;],Sow[n]], {n, Length[data]}]]2, 1</lang>
- Usage:
equilibriumIndex[{-7 , 1, 5 , 2, -4 , 3, 0}] {4, 7}
MATLAB
MATLAB arrays are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays. <lang MATLAB>function indicies = equilibriumIndex(list)
indicies = [];
for i = (1:numel(list)) if ( sum(-list(1:i)) == sum(-list(i:end)) ) indicies = [indicies i]; end end
end</lang>
- Output:
<lang MATLAB>>> equilibriumIndex([-7 1 5 2 -4 3 0])
ans =
4 7</lang>
NetRexx
<lang NetRexx>/* NetRexx */ options replace format comments java crossref symbols nobinary
numeric digits 20 runSample(arg) return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -- @see http://www.geeksforgeeks.org/equilibrium-index-of-an-array/ method equilibriumIndex(sequence) private static
es = loop ix = 1 to sequence.words() sum = 0 loop jx = 1 to sequence.words() if jx < ix then sum = sum + sequence.word(jx) if jx > ix then sum = sum - sequence.word(jx) end jx if sum = 0 then es = es ix end ix return es
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ method runSample(arg) private static
-- Note: A Rexx object based list of "words" starts at index 1 sequences = [ - '-7 1 5 2 -4 3 0', - -- 4 7 ' 2 4 6' , - -- (no equilibrium point) ' 0 2 4 0 6 0' , - -- 4 ' 2 9 2' , - -- 2 ' 1 -1 1 -1 1 -1 1' - -- 1 2 3 4 5 6 7 ] loop sequence over sequences say 'For sequence "'sequence.space(1, ',')'" the equilibrium indices are: \-' say '"'equilibriumIndex(sequence).space(1, ',')'"' end sequence return
</lang>
- Output:
For sequence "-7,1,5,2,-4,3,0" the equilibrium indices are: "4,7" For sequence "2,4,6" the equilibrium indices are: "" For sequence "0,2,4,0,6,0" the equilibrium indices are: "4" For sequence "2,9,2" the equilibrium indices are: "2" For sequence "1,-1,1,-1,1,-1,1" the equilibrium indices are: "1,2,3,4,5,6,7"
Nim
<lang nim>import math, sequtils
iterator eqindex(data) =
var suml, ddelayed = 0 var sumr = sum(data) for i,d in data: suml += ddelayed sumr -= d ddelayed = d if suml == sumr: yield i
const d = @[@[-7, 1, 5, 2, -4, 3, 0],
@[2, 4, 6], @[2, 9, 2], @[1, -1, 1, -1, 1, -1, 1]]
for data in d:
echo "d = ", data echo "eqIndex(d) -> ", toSeq(eqindex(data))</lang>
Objeck
<lang objeck>class Rosetta {
function : Main(args : String[]) ~ Nil { sequence := [-7, 1, 5, 2, -4, 3, 0]; EqulibriumIndices(sequence); } function : EqulibriumIndices(sequence : Int[]) ~ Nil { # find total sum totalSum := 0; each(i : sequence) { totalSum += sequence[i]; }; # compare running sum to remaining sum to find equlibrium indices runningSum := 0; each(i : sequence) { n := sequence[i]; if (totalSum - runningSum - n = runningSum) { i->PrintLine(); }; runningSum += n; }; }
}</lang>
Output:
3 6
OCaml
<lang ocaml>let lst = [ -7; 1; 5; 2; -4; 3; 0 ] let sum = List.fold_left ( + ) 0 lst
let () =
let rec aux acc i left right = function | x::xs -> let right = right - x in let acc = if left = right then i::acc else acc in aux acc (succ i) (left + x) right xs | [] -> List.rev acc in let res = aux [] 0 0 sum lst in print_string "Results:"; List.iter (Printf.printf " %d") res; print_newline()</lang>
Oforth
Oforth collections are 1-based
<lang Oforth>: equilibrium(l) | ls rs i e |
0 ->ls l sum ->rs ListBuffer new l size loop: i [ l at(i) ->e rs e - dup ->rs ls == ifTrue: [ i over add ] ls e + ->ls ] ;</lang>
- Output:
>equilibrium([-7, 1, 5, 2, -4, 3, 0]) println [4, 7]
PARI/GP
This uses 1-based vectors instead of 0-based arrays; subtract 1 from each index if you prefer the other style. <lang parigp>equilib(v)={
my(a=sum(i=2,#v,v[i]),b=0,u=[]); for(i=1,#v-1, if(a==b, u=concat(u,i)); b+=v[i]; a-=v[i+1] ); if(b,u,concat(u,#v))
};</lang>
Pascal
<lang pascal>Program EquilibriumIndexDemo(output);
function ArraySum(list: array of integer; first, last: integer): integer;
var i: integer; begin ArraySum := 0; for i := first to last do // not taken if first > last ArraySum := ArraySum + list[i]; end;
procedure EquilibriumIndex(list: array of integer; offset: integer);
var i: integer; begin for i := low(list) to high(list) do if ArraySum(list, low(list), i-1) = ArraySum(list, i+1, high(list)) then write(offset + i:3); end;
var {** The base index of the array is fully taken care off and can be any number. **}
numbers: array [1..7] of integer = (-7, 1, 5, 2, -4, 3, 0); i: integer;
begin
write('List of numbers: '); for i := low(numbers) to high(numbers) do write(numbers[i]:3); writeln; write('Equilibirum indices: '); EquilibriumIndex(numbers, low(numbers)); writeln;
end.</lang>
- Output:
:> ./EquilibriumIndex List of numbers: -7 1 5 2 -4 3 0 Equilibirum indices: 4 7
alternative
slightly modified.Calculating the sum only once.Using a zero-based array type.Data type could be any type of signed integer or float. But beware, that during building the sum, the limits of the data type mustn't be violated. <lang pascal>Program EquilibriumIndexDemo(output); {$IFDEF FPC}{$Mode delphi}{$ENDIF} type
tEquiData = shortInt;//Int64;extended ,double tnumList = array of tEquiData; tresList = array of LongInt;
const
cNumbers: array [11..17] of tEquiData = (-7, 1, 5, 2, -4, 3, 0);
function ArraySum(const list: tnumList):tEquiData; var
i: integer;
begin
result := 0; for i := Low(list) to High(list) do result := result+list[i];
end;
procedure EquilibriumIndex(const list:tnumList;
var indices:tresList);
var
pC : ^tEquiData; LeftSum, RightSum : tEquiData; i,idx,HiList: integer;
begin
HiList := High(List); RightSum :=ArraySum(list); setlength(indices,10); idx := 0;
i := -Hilist; pC := @List[0]; LeftSum:= 0; repeat Rightsum:= RightSum-pC^; IF LeftSum = RightSum then Begin indices[idx] := Hilist+i; inc(idx); IF idx > high(indices) then setlength(indices, idx+10); end; inc(i); leftSum := leftsum+pC^; inc(pC); until i>=0; leftSum := leftsum+pC^; IF LeftSum = RightSum then Begin indices[idx] := Hilist+i; inc(idx); end; setlength(indices,idx);
end;
procedure TestRun(const numbers:tnumList); var
indices : tresList; i: integer;
Begin
write('List of numbers: '); for i := low(numbers) to high(numbers) do write(numbers[i]:3); writeln; EquilibriumIndex(numbers,indices); write('Equilibirum indices: '); EquilibriumIndex(numbers,indices); for i := low(indices) to high(indices) do write(indices[i]:3); writeln; writeln;
end;
var
numbers: tnumList; I: integer;
begin
setlength(numbers,High(cNumbers)-Low(cNumbers)+1); move(cNumbers[Low(cNumbers)],numbers[0],sizeof(cnumbers)); TestRun(numbers); for i := low(numbers) to high(numbers) do numbers[i]:= 0; TestRun(numbers);
end.</lang>
- Output:
List of numbers: -7 1 5 2 -4 3 0Equilibirum indices: 3 6
List of numbers: 0 0 0 0 0 0 0
Equilibirum indices: 0 1 2 3 4 5 6
Perl
<lang perl>sub eq_index {
my ( $i, $sum, %sums ) = ( 0, 0 );
for (@_) { push @{ $sums{ $sum * 2 + $_ } }, $i++; $sum += $_; }
return join ' ', @{ $sums{$sum} || [] }, "\n";
}
print eq_index qw( -7 1 5 2 -4 3 0 ); # 3 6 print eq_index qw( 2 4 6 ); # (no eq point) print eq_index qw( 2 9 2 ); # 1 print eq_index qw( 1 -1 1 -1 1 -1 1 ); # 0 1 2 3 4 5 6</lang>
Phix
<lang Phix>function equilibrium(sequence s) atom lower_sum = 0 atom higher_sum = sum(s) sequence res = {}
for i=1 to length(s) do higher_sum -= s[i] if lower_sum=higher_sum then res &= i end if lower_sum += s[i] end for return res
end function
? equilibrium({-7,1,5,2,-4,3,0})</lang>
- Output:
(Remember that indices are 1-based in Phix)
{4,7}
PHP
<lang php><?php $arr = array(-7, 1, 5, 2, -4, 3, 0);
function getEquilibriums($arr) {
$right = array_sum($arr); $left = 0; $equilibriums = array(); foreach($arr as $key => $value){ $right -= $value; if($left == $right) $equilibriums[] = $key; $left += $value; } return $equilibriums;
}
echo "# results:\n"; foreach (getEquilibriums($arr) as $r) echo "$r, "; ?></lang>
- Output:
# results: 3, 6,
PicoLisp
<lang PicoLisp>(de equilibria (Lst)
(make (let Sum 0 (for ((I . L) Lst L (cdr L)) (and (= Sum (sum prog (cdr L))) (link I)) (inc 'Sum (car L)) ) ) ) )</lang>
- Output:
: (equilibria (-7 1 5 2 -4 3 0)) -> (4 7) : (equilibria (make (do 10000 (link (rand -10 10))))) -> (4091 6174 6198 7104 7112 7754)
PowerShell
In real life in PowerShell, one would likely leverage pipelines, ForEach-Object, Where-Object, and Measure-Object for tasks such as this. Normally in PowerShell, speed is an important, but not primary consideration, and the advantages of pipelines tend to outweigh the overhead incurred. However, for this particular task, keeping in mind that “the sequence may be very long,” this code was optimized primarly for speed. <lang PowerShell> function Get-EquilibriumIndex ( $Sequence )
{ $Indexes = 0..($Sequence.Count - 1) $EqulibriumIndex = @() ForEach ( $TestIndex in $Indexes ) { $Left = 0 $Right = 0 ForEach ( $Index in $Indexes ) { If ( $Index -lt $TestIndex ) { $Left += $Sequence[$Index] } ElseIf ( $Index -gt $TestIndex ) { $Right += $Sequence[$Index] } } If ( $Left -eq $Right ) { $EqulibriumIndex += $TestIndex } } return $EqulibriumIndex }
</lang> <lang PowerShell> Get-EquilibriumIndex -7, 1, 5, 2, -4, 3, 0 </lang>
- Output:
3 6
Prolog
<lang prolog>equilibrium_index(List, Index) :-
append(Front, [_|Back], List), sumlist(Front, Sum), sumlist(Back, Sum), length(Front, Len), Index is Len.</lang>
Example:
<lang prolog>?- equilibrium_index([-7, 1, 5, 2, -4, 3, 0], Index). Index = 3 ; Index = 6 ; false.</lang>
PureBasic
<lang PureBasic>If OpenConsole()
Define i, c=CountProgramParameters()-1 For i=0 To c Define j, LSum=0, RSum=0 For j=0 To c If ji RSum+Val(ProgramParameter(j)) EndIf Next j If LSum=RSum: PrintN(Str(i)): EndIf Next i
EndIf</lang>
- Output:
> Equilibrium.exe -7 1 5 2 -4 3 0 3 6
Python
Two Pass
Uses an initial summation of the whole list then visits each item of the list adding it to the left-hand sum (after a delay); and subtracting the item from the right-hand sum. I think it should be quicker than algorithms that scan the list creating left and right sums for each index as it does ~2N add/subtractions rather than n*n. <lang python>def eqindex2Pass(data):
"Two pass" suml, sumr, ddelayed = 0, sum(data), 0 for i, d in enumerate(data): suml += ddelayed sumr -= d ddelayed = d if suml == sumr: yield i</lang>
Multi Pass
This is the version that does more summations, but may be faster for some sizes of input as the sum function is implemented in C internally: <lang python>def eqindexMultiPass(data):
"Multi pass" for i in range(len(data)): suml, sumr = sum(data[:i]), sum(data[i+1:]) if suml == sumr: yield i</lang>
Shorter alternative: <lang python>def eqindexMultiPass(s):
return [i for i in xrange(len(s)) if sum(s[:i]) == sum(s[i+1:])]
print eqindexMultiPass([-7, 1, 5, 2, -4, 3, 0])</lang>
One Pass
This routine would need careful evaluation against the two-pass solution above as, although it only runs through the data once, it may create a dict that is as long as the input data in its worst case of an input of say a simple 1, 2, 3, ... counting sequence. <lang python>from collections import defaultdict
def eqindex1Pass(data):
"One pass" l, h = 0, defaultdict(list) for i, c in enumerate(data): l += c h[l * 2 - c].append(i) return h[l]</lang>
Tests
<lang python>f = (eqindex2Pass, eqindexMultiPass, eqindex1Pass) d = ([-7, 1, 5, 2, -4, 3, 0],
[2, 4, 6], [2, 9, 2], [1, -1, 1, -1, 1, -1, 1])
for data in d:
print("d = %r" % data) for func in f: print(" %16s(d) -> %r" % (func.__name__, list(func(data))))</lang>
- Sample output:
d = [-7, 1, 5, 2, -4, 3, 0] eqindex2Pass(d) -> [3, 6] eqindexMultiPass(d) -> [3, 6] eqindex1Pass(d) -> [3, 6] d = [2, 4, 6] eqindex2Pass(d) -> [] eqindexMultiPass(d) -> [] eqindex1Pass(d) -> [] d = [2, 9, 2] eqindex2Pass(d) -> [1] eqindexMultiPass(d) -> [1] eqindex1Pass(d) -> [1] d = [1, -1, 1, -1, 1, -1, 1] eqindex2Pass(d) -> [0, 1, 2, 3, 4, 5, 6] eqindexMultiPass(d) -> [0, 1, 2, 3, 4, 5, 6] eqindex1Pass(d) -> [0, 1, 2, 3, 4, 5, 6]
In terms of itertools.accumulate
The left scan is efficiently derived by the accumulate function in the itertools module.
The right scan can be derived from the left as a map or equivalent list comprehension: <lang python>"""Equilibrium index"""
from itertools import (accumulate)
- equilibriumIndices :: [Num] -> [Int]
def equilibriumIndices(xs):
List indices at which the sum of values to the left equals the sum of values to the right. def go(xs): Left scan from accumulate, right scan derived from left ls = list(accumulate(xs)) n = ls[-1] return [i for (i, (x, y)) in enumerate(zip( ls, [n] + [n - x for x in ls[0:-1]] )) if x == y] return go(xs) if xs else []
- TEST -------------------------------------------------
- main :: IO ()
def main():
Tabulated test results print( tabulated('Equilibrium indices:\n')( equilibriumIndices )([ [-7, 1, 5, 2, -4, 3, 0], [2, 4, 6], [2, 9, 2], [1, -1, 1, -1, 1, -1, 1], [1], [] ]) )
- GENERIC -------------------------------------------------
- tabulated :: String -> (a -> b) -> [a] -> String
def tabulated(s):
heading -> function -> input List -> tabulated output string def go(f, xs): def width(x): return len(str(x)) w = width(max(xs, key=width)) return s + '\n' + '\n'.join([ str(x).rjust(w, ' ') + ' -> ' + str(f(x)) for x in xs ]) return lambda f: lambda xs: go(f, xs)
if __name__ == '__main__':
main()</lang>
- Output:
Equilibrium indices: [-7, 1, 5, 2, -4, 3, 0] -> [3, 6] [2, 4, 6] -> [] [2, 9, 2] -> [1] [1, -1, 1, -1, 1, -1, 1] -> [0, 1, 2, 3, 4, 5, 6] [1] -> [0] [] -> []
Racket
<lang racket>
- lang racket
(define (subsums xs)
(for/fold ([sums '()] [sum 0]) ([x xs]) (values (cons (+ x sum) sums) (+ x sum))))
(define (equivilibrium xs)
(define-values (sums total) (subsums xs)) (for/list ([sum (reverse sums)] [x xs] [i (in-naturals)] #:when (= (- sum x) (- total sum))) i))
(equivilibrium '(-7 1 5 2 -4 3 0)) </lang>
- Output:
<lang racket> '(3 6) </lang>
Raku
(formerly Perl 6) <lang perl6>sub equilibrium_index(@list) {
my ($left,$right) = 0, [+] @list;
gather for @list.kv -> $i, $x { $right -= $x; take $i if $left == $right; $left += $x; }
}
my @list = -7, 1, 5, 2, -4, 3, 0; .say for equilibrium_index(@list).grep(/\d/);</lang> And here's an FP solution that manages to remain O(n): <lang perl6>sub equilibrium_index(@list) {
my @a = [\+] @list; my @b = reverse [\+] reverse @list; ^@list Zxx (@a »==« @b);
}</lang> The [\+] is a reduction that returns a list of partial results. The »==« is a vectorized equality comparison; it returns a vector of true and false. The Zxx is a zip with the list replication operator, so we return only the elements of the left list where the right list is true (which is taken to mean 1 here). And the ^@list is just shorthand for 0 ..^ @list. We could just as easily have used @list.keys there.
Single-pass solution
The task can be restated in a way that removes the "right side" from the calculation.
C is the current element,
L is the sum of elements left of C,
R is the sum of elements right of C,
S is the sum of the entire list.
By definition, L + C + R == S for any choice of C, and L == R for any C that is an equilibrium point.
Therefore (by substituting L for R), L + C + L == S at all equilibrium points.
Restated, 2L + C == S.
<lang perl6># Original example, with expanded calculations:
0 1 2 3 4 5 6 # Index -7 1 5 2 -4 3 0 # C (Value at index) 0 -7 -6 -1 1 -3 0 # L (Sum of left) -7 -13 -7 0 -2 -3 0 # 2L+C</lang>
If we build a hash as we walk the list, with 2L+C as hash keys, and arrays of C-indexes as hash values, we get: <lang perl6>{
-7 => [ 0, 2 ], -13 => [ 1 ], 0 => [ 3, 6 ], -2 => [ 4 ], -3 => [ 5 ],
}</lang> After we have finished walking the list, we will have the sum (S), which we look up in the hash. Here S=0, so the equilibrium points are 3 and 6.
Note: In the code below, it is more convenient to calculate 2L+C *after* L has already been incremented by C; the calculation is simply 2L-C, because each L has an extra C in it. 2(L-C)+C == 2L-C. <lang perl6>sub eq_index ( *@list ) {
my $sum = 0;
my %h = @list.keys.classify: { $sum += @list[$_]; $sum * 2 - @list[$_]; };
return %h{$sum} // [];
}
say eq_index < -7 1 5 2 -4 3 0 >; # 3 6 say eq_index < 2 4 6 >; # (no eq point) say eq_index < 2 9 2 >; # 1 say eq_index < 1 -1 1 -1 1 -1 1 >; # 0 1 2 3 4 5 6</lang> The .classify method creates a hash, with its code block's return value as key. Each hash value is an Array of all the inputs that returned that key.
We could have used .pairs instead of .keys to save the cost of @list lookups, but that would change each %h value to an Array of Pairs, which would complicate the return line.
Red
<lang Rebol>Red [] eqindex: func [a [block!]] [
collect [ repeat ind length? a [ if (sum skip a ind) = sum copy/part a ind - 1 [ keep ind ] ] ]
] prin "(1 based) equ indices are: " probe eqindex [-7 1 5 2 -4 3 0] </lang>
- Output:
(1 based) equ indices are: [4 7]
REXX
version 1
This REXX version utilizes a zero-based stemmed array to mimic the illustrative example in this Rosetta Code task's
prologue, which uses a zero-based index.
<lang rexx>/*REXX program calculates and displays the equilibrium index for a numeric array (list).*/
parse arg x /*obtain the optional arguments from CL*/
if x= then x= copies(" 7 -7", 50) 7 /*Not specified? Then use the default.*/
say ' array list: ' space(x) /*echo the array list to the terminal. */
- = words(x) /*the number of elements in the X list.*/
do j=0 for #; @.j= word(x, j+1) /*zero─start is for zero─based array. */ end /*j*/ /* [↑] assign @.0 @.1 @.3 ··· */
say /* ··· and also display a blank line. */ answer= equilibriumIDX(); w= words(answer) /*calculate the equilibrium index. */ say 'equilibrium' word("(none) index: indices:", 1 + (w>0) + (w>1)) answer exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ equilibriumIDX: $=; do i=0 for #; sum= 0
do k=0 for #; sum= sum + @.k * sign(k-i); end /*k*/ if sum==0 then $= $ i end /*i*/ /* [↑] Zero? Found an equilibrium index*/ return $ /*return equilibrium list (may be null)*/</lang>
- output when using the input of: -7 1 5 2 -4 3 0
array list: -7 1 5 2 -4 3 0 equilibrium indices: 3 6
- output when using the input of: 2 9 2
array list: 2 9 2 equilibrium index: 1
- output when using the input of: 5 4 4 5
array list: 5 4 4 5 equilibrium (none)
- output when using the default input
array list: 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 equilibrium indices: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
version 2
<lang rexx>/* REXX ---------------------------------------------------------------
- 30.06.2014 Walter Pachl
- --------------------------------------------------------------------*/
parse arg l say ' array list:' strip(l) x.=0 Do i=1 To words(l)
x.i=word(l,i) End
n=i-1 ans=strip(equilibriumIndices()) n=words(ans) Select
When n=0 Then Say 'Theres no equilibrium index' When n=1 Then Say 'equilibrium index :' ans Otherwise Say 'equilibrium indices:' ans End
Say '---' exit equilibriumIndices: procedure expose x. n sum.=0 sum=0 eil= Do i=1 To n
sum=sum+x.i sum.i=sum End
Do i=1 To n
im1=i-1 If sum.im1=(sum.n-x.i)/2 Then eil=eil im1 End
Return eil</lang> output
array list: -7 1 5 2 -4 3 0 equilibrium indices: 3 6 --- array list: 2 9 2 equilibrium index : 1 --- array list: 1 -1 1 -1 1 -1 1 equilibrium indices: 0 1 2 3 4 5 6 --- array list: 1 1 There's no equilibrium index ---
Ring
<lang ring> list = [-7, 1, 5, 2, -4, 3, 0] see "equilibrium indices are : " + equilibrium(list) + nl
func equilibrium l
r = 0 s = 0 e = "" for n = 1 to len(l) s += l[n] next for i = 1 to len(l) if r = s - r - l[i] e += string(i-1) + "," ok r += l[i] next e = left(e,len(e)-1) return e
</lang> Output:
equilibrium indices are : 3,6
Ruby
- Functional Style
<lang ruby>def eq_indices(list)
list.each_index.select do |i| list[0...i].sum == list[i+1..-1].sum end
end</lang>
- Tail Recursion
- This one would be good if Ruby did tail-call optimization (TCO).
- MRI does not do TCO; so this function fails with a long list (by overflowing the call stack).
<lang ruby>def eq_indices(list)
result = [] list.empty? and return result final = list.size - 1 helper = lambda do |left, current, right, index| left == right and result << index # Push index to result? index == final and return # Terminate recursion? new = list[index + 1] helper.call(left + current, new, right - new, index + 1) end helper.call 0, list.first, list.drop(1).sum, 0 result
end</lang>
- Imperative Style (faster)
<lang ruby>def eq_indices(list)
left, right = 0, list.sum equilibrium_indices = [] list.each_with_index do |val, i| right -= val equilibrium_indices << i if right == left left += val end equilibrium_indices
end</lang>
- Test
<lang ruby>indices = [
[-7, 1, 5, 2,-4, 3, 0], [2, 4, 6], [2, 9, 2], [1,-1, 1,-1, 1,-1, 1]
] indices.each do |x|
puts "%p => %p" % [x, eq_indices(x)]
end</lang>
- Output:
[-7, 1, 5, 2, -4, 3, 0] => [3, 6] [2, 4, 6] => [] [2, 9, 2] => [1] [1, -1, 1, -1, 1, -1, 1] => [0, 1, 2, 3, 4, 5, 6]
Scala
<lang Scala> def getEquilibriumIndex(A: Array[Int]): Int = {
val bigA: Array[BigInt] = A.map(BigInt(_)) val partialSums: Array[BigInt] = bigA.scanLeft(BigInt(0))(_+_).tail def lSum(i: Int): BigInt = if (i == 0) 0 else partialSums(i - 1) def rSum(i: Int): BigInt = partialSums.last - partialSums(i) def isRandLSumEqual(i: Int): Boolean = lSum(i) == rSum(i) (0 until partialSums.length).find(isRandLSumEqual).getOrElse(-1) } </lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
const array integer: numList is [] (-7, 1, 5, 2, -4, 3, 0);
const func array integer: equilibriumIndex (in array integer: elements) is func
result var array integer: indexList is 0 times 0; local var integer: element is 0; var integer: index is 0; var integer: sum is 0; var integer: subSum is 0; var integer: count is 0; begin indexList := length(elements) times 0; for element range elements do sum +:= element; end for; for element key index range elements do if 2 * subSum + element = sum then incr(count); indexList[count] := index; end if; subSum +:= element; end for; indexList := indexList[.. count]; end func;
const proc: main is func
local var array integer: indexList is 0 times 0; var integer: element is 0; begin indexList := equilibriumIndex(numList); write("Found:"); for element range indexList do write(" " <& element); end for; writeln; end func;</lang>
- Output:
Found: 4 7
Sidef
<lang ruby>func eq_index(nums) {
var (i, sum, sums) = (0, 0, Hash.new); nums.each { |n| sums{2*sum + n} := [] -> append(i++); sum += n; } sums{sum} \\ [];
}</lang>
Test: <lang ruby>var indices = [
[-7, 1, 5, 2,-4, 3, 0], [2, 4, 6], [2, 9, 2], [1,-1, 1,-1, 1,-1, 1],
]
for x in indices {
say ("%s => %s" % @|[x, eq_index(x)].map{.dump});
}</lang>
- Output:
[-7, 1, 5, 2, -4, 3, 0] => [3, 6] [2, 4, 6] => [] [2, 9, 2] => [1] [1, -1, 1, -1, 1, -1, 1] => [0, 1, 2, 3, 4, 5, 6]
Swift
<lang swift>extension Collection where Element: Numeric {
func equilibriumIndexes() -> [Index] { guard !isEmpty else { return [] }
let sumAll = reduce(0, +) var ret = [Index]() var sumLeft: Element = 0 var sumRight: Element
for i in indices { sumRight = sumAll - sumLeft - self[i]
if sumLeft == sumRight { ret.append(i) }
sumLeft += self[i] }
return ret }
}
let arr = [-7, 1, 5, 2, -4, 3, 0]
print("Equilibrium indexes of \(arr): \(arr.equilibriumIndexes())")</lang>
- Output:
Equilibrium indexes of [-7, 1, 5, 2, -4, 3, 0]: [3, 6]
Tcl
<lang tcl>proc listEquilibria {list} {
set after 0 foreach item $list {incr after $item} set result {} set idx 0 set before 0 foreach item $list {
incr after [expr {-$item}] if {$after == $before} { lappend result $idx } incr before $item incr idx
} return $result
}</lang>
- Example of use
<lang tcl>set testData {-7 1 5 2 -4 3 0} puts Equilibria=[join [listEquilibria $testData] ", "]</lang>
- Output:
Equilibria=3, 6
Ursala
<lang Ursala>#import std
- import int
edex = num@yK33ySPtK33xtS2px; ~&nS+ *~ ==+ ~~r sum:-0
- cast %nL
example = edex <-7,1,5,2,-4,3,0></lang>
- Output:
<3,6>
VBScript
Solution adopted from http://www.geeksforgeeks.org/equilibrium-index-of-an-array/ . <lang vb>arr = Array(-7,1,5,2,-4,3,0) WScript.StdOut.Write equilibrium(arr,UBound(arr)) WScript.StdOut.WriteLine
Function equilibrium(arr,n) sum = 0 leftsum = 0 'find the sum of the whole array For i = 0 To UBound(arr) sum = sum + arr(i) Next For i = 0 To UBound(arr) sum = sum - arr(i) If leftsum = sum Then equilibrium = equilibrium & i & ", " End If leftsum = leftsum + arr(i) Next End Function</lang>
- Output:
Indices: 3, 6,
Wren
<lang ecmascript>import "/fmt" for Fmt
var equilibrium = Fn.new { |a|
var len = a.count var equi = [] if (len == 0) return equi // sequence has no indices at all var rsum = a.reduce { |acc, x| acc + x } var lsum = 0 for (i in 0...len) { rsum = rsum - a[i] if (rsum == lsum) equi.add(i) lsum = lsum + a[i] } return equi
}
var tests = [
[-7, 1, 5, 2, -4, 3, 0], [2, 4, 6], [2, 9, 2], [1, -1, 1, -1, 1, -1, 1], [1], []
]
System.print("The equilibrium indices for the following sequences are:\n") for (test in tests) {
System.print("%(Fmt.s(24, test)) -> %(equilibrium.call(test))")
}</lang>
- Output:
The equilibrium indices for the following sequences are: [-7, 1, 5, 2, -4, 3, 0] -> [3, 6] [2, 4, 6] -> [] [2, 9, 2] -> [1] [1, -1, 1, -1, 1, -1, 1] -> [0, 1, 2, 3, 4, 5, 6] [1] -> [0] [] -> []
XPL0
<lang XPL0>code Ran=1, ChOut=8, IntOut=11; def Size = 1_000_000; int I, S, A(Size), Hi(Size), Lo(Size); [for I:= 0 to Size-1 do A(I):= Ran(100) - 50; S:= 0; for I:= 0 to Size-1 do [S:= S+A(I); Lo(I):= S]; S:= 0; for I:= Size-1 downto 0 do [S:= S+A(I); Hi(I):= S]; for I:= 0 to Size-1 do
if Lo(I) = Hi(I) then [IntOut(0, I); ChOut(0, ^ )];
]</lang>
- Output:
502910 504929 508168
Yorick
Yorick arrays are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays. <lang yorick>func equilibrium_indices(A) {
return where(A(psum) == A(::-1)(psum)(::-1));
}</lang>
- Example interactive usage:
> equilibrium_indices([-7, 1, 5, 2, -4, 3, 0]) [4,7]
zkl
<lang zkl>fcn equilibrium(lst){ // two pass
reg acc=List(), left=0,right=lst.sum(0),i=0; foreach x in (lst){ right-=x; if(left==right) acc.write(i); i+=1; left+=x; } acc
}</lang>
<lang zkl>fcn equilibrium(lst){ // lst should immutable, n^2
(0).filter(lst.len(),'wrap(n){ lst[0,n].sum(0) == lst[n+1,*].sum(0) })
}</lang> If the input list is immutable, no new lists are generated (other than accumulating the result). <lang zkl>equilibrium(T(-7, 1, 5, 2, -4, 3, 0)).println();</lang>
- Output:
L(3,6)
ZX Spectrum Basic
<lang zxbasic>10 DATA 7,-7,1,5,2,-4,3,0 20 READ n 30 DIM a(n): LET sum=0: LET leftsum=0: LET s$="" 40 FOR i=1 TO n: READ a(i): LET sum=sum+a(i): NEXT i 50 FOR i=1 TO n 60 LET sum=sum-a(i) 70 IF leftsum=sum THEN LET s$=s$+STR$ i+" " 80 LET leftsum=leftsum+a(i) 90 NEXT i 100 PRINT "Numbers: "; 110 FOR i=1 TO n: PRINT a(i);" ";: NEXT i 120 PRINT '"Indices: ";s$</lang>
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