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{{task}}
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
{{trans|Python}}
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]))
{{out}}
[3, 6]
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 ) ).
Ada
{{works with|Ada 2005}} Generic solution that returns a Vector of Indices.
equilibrium.ads:
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;
equilibrium.adb:
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;
Test program using two different versions, one with vectors and one with arrays:
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;
{{out}} (Index_Type is based on 1):
Results:
Array: 4 7
Vector: 4 7
Version that works with Ada 95, too: {{works with|Ada 95}} {{works with|Ada 2005}} equilibrium.adb:
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;
{{out}}
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
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;
}
ALGOL 68
{{trans|C}} {{works with|ALGOL 68|Revision 1 - no extensions to language used}} {{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}} {{works with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d]}}
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)
)
{{out}}
+3 +6
AppleScript
{{Trans|JavaScript}}(ES6 version)
-- 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
{{Out}}
{{3, 6}, {}, {1}, {0, 1, 2, 3, 4, 5, 6}, {0}, {}}
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
}
Examples:
list = -7, 1, 5, 2, -4, 3, 0
MsgBox % Equilibrium_index(list)
{{out}}
3, 6
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)
}
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
@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 ==%
{{Out}}
[3 6]
[]
[1]
[0 1 2 3 4 5 6]
BBC BASIC
BBC BASIC's '''SUM''' function is useful for this task.
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$)
'''Output:'''
Equilibrium indices are 3,6
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; }
C++
#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>); }
{{out}}
3
6
C#
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); } } }
{{out}}
## Clojure
{{trans|Ocaml}}
```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)))))
{{out}}
> (equilibrium [-7, 1, 5, 2, -4, 3, 0])
(3 6)
Common 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)))))
{{out}}
## D
### More Functional Style
```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;
}
{{out}}
[3, 6]
Less Functional Style
{{trans|PHP}} Same output.
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; }
Elena
ELENA 4.1 :
import extensions;
import system'routines;
import system'collections;
import extensions'routines;
class EquilibriumEnumerator : Enumerator
{
int left;
int right;
int index;
Enumerator en;
constructor(Enumerator en)
{
this en := en;
self.reset()
}
constructor(Enumerable list)
<= (list.enumerator());
constructor(o)
<= (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()
{
new EquilibriumEnumerator(new int[]::( -7, 1, 5, 2, -4, 3, 0 ))
.forEach:printingLn
}
3
6
Elixir
{{trans|Ruby}} computes either side each time.
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
faster version:
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
'''Test:'''
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)
{{out}}
[-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
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
'''Output:'''
Equilibrium indices are 3, 6
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})
{{out}} ''(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.
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 ;
{{out}}
( scratchpad ) { -7 1 5 2 -4 3 0 } equilibrium-indices .
V{ 3 6 }
=={{header|Fōrmulæ}}==
In [http://wiki.formulae.org/Equilibrium_index 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 ([http://wiki.formulae.org/Editing_F%C5%8Drmul%C3%A6_expressions 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.
Fortran
{{works with|Fortran|90 and later}} Array indices are 1-based.
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
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
{{out}}
The equilibrium indices are :
3 6
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 }
{{out}}
[3 6]
[145125 947872]
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
Small example
*Main> equilibr [-7, 1, 5, 2, -4, 3, 0] [3,6]
Long random list in langeSliert (several tries for this one)
*Main> langeSliert [231,245,259,265,382,1480,1611,1612]
Or, using default Prelude functions:
equilibriumIndices :: [Int] -> [Int] equilibriumIndices xs = foldr (\(x, y, i) a -> (if x == y then i : a else a)) [] (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] , [] ])
{{Out}}
[3,6]
[]
[1]
[0,1,2,3,4,5,6]
[0]
[]
=={{header|Icon}} and {{header|Unicon}}==
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
{{out}}
equilibrium indicies of [ -7 1 5 2 -4 3 0 ] = 4 7
J
{{out|Example use}}
```j
equilidx _7 1 5 2 _4 3 0
3 6
Java
{{works with|Java|1.5+}}
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;
}
}
}
{{out}}
3
6
JavaScript
ES5
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)) });
{{Out}}
[[3,6],[],[1],[0,1,2,3,4,5,6],[0],[]]
===ES6 - two pass O(n), return index===
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; }
{{Out}}
3, -1, 1, 0, 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.
# 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);
'''Example 1:'''
[-7, 1, 5, 2, -4, 3, 0] | equilibrium_indices
{{out}} $ jq -M -n -f equilibrium_indices.jq 3 6 '''Example 2:'''
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 )
{{out}} $ jq -M -n -f equilibrium_indices.jq 10000
Julia
{{works with|Julia|0.6}}
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])
{{out}}
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
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
Kotlin
// 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 : ${ei.joinToString(", ")}") } }
{{out}}
The equilibrium indices are : 3, 6
Liberty BASIC
a(0)=-7
a(1)=1
a(2)=5
a(3)=2
a(4)=-4
a(5)=3
a(6)=0
print "EQ Indices are ";EQindex$("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
{{out}}
EQ Indices are 3, 6
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]
=={{header|Mathematica}} / {{header|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.
equilibriumIndex[data_]:=Reap[
Do[If[Total[data[[;; n - 1]]] == Total[data[[n + 1 ;;]]],Sow[n]],
{n, Length[data]}]][[2, 1]]
{{out|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.
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
{{out}}
equilibriumIndex([-7 1 5 2 -4 3 0])
ans =
4 7
NetRexx
{{trans|Java}}
/* 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
{{out}}
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
{{trans|Python}}
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))
Objeck
{{Trans|Java}}
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;
};
}
}
Output:
3
6
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()
Oforth
Oforth collections are 1-based
: 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
] ;
{{out}}
>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.
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))
};
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.
{{out}}
:> ./EquilibriumIndex
List of numbers: -7 1 5 2 -4 3 0
Equilibirum indices: 4 7
alternative
{{works with|Free Pascal}} 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.
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.
{{out}}
List of numbers: -7 1 5 2 -4 3 0
Equilibirum indices: 3 6
List of numbers: 0 0 0 0 0 0 0
Equilibirum indices: 0 1 2 3 4 5 6
Perl
{{trans|Perl 6}}
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
Perl 6
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/);
And here's an FP solution that manages to remain O(n):
sub equilibrium_index(@list) {
my @a = [\+] @list;
my @b = reverse [\+] reverse @list;
^@list Zxx (@a »==« @b);
}
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.
# 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
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:
{
-7 => [ 0, 2 ],
-13 => [ 1 ],
0 => [ 3, 6 ],
-2 => [ 4 ],
-3 => [ 5 ],
}
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.
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
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.
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})
{{out}} (Remember that indices are 1-based in Phix)
{4,7}
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, "; ?>
{{out}}
# results:
3, 6,
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)) ) ) ) )
{{out}}
: (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
{{works with|PowerShell|2}} 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.
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 }
Get-EquilibriumIndex -7, 1, 5, 2, -4, 3, 0
{{out}}
3
6
Prolog
equilibrium_index(List, Index) :-
append(Front, [_|Back], List),
sumlist(Front, Sum),
sumlist(Back, Sum),
length(Front, Len),
Index is Len.
Example:
?- equilibrium_index([-7, 1, 5, 2, -4, 3, 0], Index).
Index = 3 ;
Index = 6 ;
false.
PureBasic
{{trans|Java}}
If OpenConsole()
Define i, c=CountProgramParameters()-1
For i=0 To c
Define j, LSum=0, RSum=0
For j=0 To c
If j<i
LSum+Val(ProgramParameter(j))
ElseIf j>i
RSum+Val(ProgramParameter(j))
EndIf
Next j
If LSum=RSum: PrintN(Str(i)): EndIf
Next i
EndIf
{{out}}
> 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.
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
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:
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
Shorter alternative:
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])
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.
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]
Tests
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))))
{{out|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:
"""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()
{{Out}}
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
(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))
{{out}}
'(3 6)
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.
/*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)*/
'''output''' using the input: -7 1 5 2 -4 3 0
array list: -7 1 5 2 -4 3 0
equilibrium indices: 3 6
'''output''' using the input: 2 9 2
array list: 2 9 2
equilibrium index: 1
'''output''' using the input: 5 4 4 5
array list: 5 4 4 5
equilibrium (none)
'''output''' 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
/* 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 'There''s 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
'''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
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
Output:
equilibrium indices are : 3,6
Ruby
{{works with|Ruby|1.8.7}} ;Functional Style
def eq_indices(list) list.each_index.select do |i| list[0...i].inject(0, :+) == list[i+1..-1].inject(0, :+) end end
;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).
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).inject(:+), 0 result end
;Imperative Style (faster)
def eq_indices(list) left, right = 0, list.inject(0, :+) equilibrium_indices = [] list.each_with_index do |val, i| right -= val equilibrium_indices << i if right == left left += val end equilibrium_indices end
;Test
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
{{out}}
[-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
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) }
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;
{{out}}
Found: 4 7
Sidef
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} \\ []; }
Test:
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}); }
{{out}}
[-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]
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 }
;Example of use
set testData {-7 1 5 2 -4 3 0} puts Equilibria=[join [listEquilibria $testData] ", "]
{{out}}
Equilibria=3, 6
Ursala
#import std
#import int
edex = num@yK33ySPtK33xtS2px; ~&nS+ *~ ==+ ~~r sum:-0
#cast %nL
example = edex <-7,1,5,2,-4,3,0>
{{out}}
<3,6>
VBScript
Solution adopted from http://www.geeksforgeeks.org/equilibrium-index-of-an-array/ .
arr = Array(-7,1,5,2,-4,3,0)
WScript.StdOut.Write equilibrium(arr,UBound(arr))
WScript.StdOut.WriteLine
Function equilibrium(arr,n)
sum = 0
leftsum = 0
'find the sum of the whole array
For i = 0 To UBound(arr)
sum = sum + arr(i)
Next
For i = 0 To UBound(arr)
sum = sum - arr(i)
If leftsum = sum Then
equilibrium = equilibrium & i & ", "
End If
leftsum = leftsum + arr(i)
Next
End Function
{{out}}
Indices: 3, 6,
XPL0
code Ran=1, ChOut=8, IntOut=11;
def Size = 1_000_000;
int I, S, A(Size), Hi(Size), Lo(Size);
[for I:= 0 to Size-1 do A(I):= Ran(100) - 50;
S:= 0;
for I:= 0 to Size-1 do [S:= S+A(I); Lo(I):= S];
S:= 0;
for I:= Size-1 downto 0 do [S:= S+A(I); Hi(I):= S];
for I:= 0 to Size-1 do
if Lo(I) = Hi(I) then [IntOut(0, I); ChOut(0, ^ )];
]
{{out}}
502910 504929 508168
Yorick
Yorick arrays are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays.
func equilibrium_indices(A) {
return where(A(psum) == A(::-1)(psum)(::-1));
}
{{out|Example interactive usage}}
> equilibrium_indices([-7, 1, 5, 2, -4, 3, 0])
[4,7]
zkl
{{trans|Clojure}}
fcn equilibrium(lst){ // two pass
reg acc=List(), left=0,right=lst.sum(0),i=0;
foreach x in (lst){
right-=x;
if(left==right) acc.write(i);
i+=1; left+=x;
}
acc
}
{{trans|D}}
fcn equilibrium(lst){ // lst should immutable, n^2
(0).filter(lst.len(),'wrap(n){ lst[0,n].sum(0) == lst[n+1,*].sum(0) })
}
If the input list is immutable, no new lists are generated (other than accumulating the result).
equilibrium(T(-7, 1, 5, 2, -4, 3, 0)).println();
{{out}}
L(3,6)
ZX Spectrum Basic
{{trans|AWK}}
10 DATA 7,-7,1,5,2,-4,3,0
20 READ n
30 DIM a(n): LET sum=0: LET leftsum=0: LET s$=""
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$