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Two sum

Two sum is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
This task has been clarified. Its programming examples are in need of review to ensure that they still fit the requirements of the task.

Given a sorted array of integers (with possibly duplicates), is it possible to find a pair of integers from that array that sum up to a given sum? If so, return indices of the two integers or an empty array if not. The solution is not necessarily unique.

Example

Given numbers = [0, 2, 11, 19, 90], sum = 21,
Because numbers + numbers = 2 + 19 = 21,
return [1, 3].

Source

11l

Translation of: Python
F two_sum(arr, num)
V i = 0
V j = arr.len - 1
L i < j
I arr[i] + arr[j] == num
R [i, j]
I arr[i] + arr[j] < num
i++
E
j--
R [Int]()

V numbers = [0, 2, 11, 19, 90]
print(two_sum(numbers, 21))
print(two_sum(numbers, 25))
Output:
[1, 3]
[]

Action!

PROC PrintArray(INT ARRAY a INT len)
INT i

Put('[)
FOR i=0 TO len-1
DO
PrintI(a(i))
IF i<len-1 THEN
Put(' )
FI
OD
Put(']) PutE()
RETURN

PROC PrintPairs(INT ARRAY a INT len,sum)
INT i,j,p1,p2,s,count

count=0
FOR i=0 TO len-2
DO
p1=a(i)
FOR j=i+1 TO len-1
DO
p2=a(j)
s=p1+p2
IF s=sum THEN
PrintF("(%I,%I) ",i,j)
count==+1
ELSEIF s>sum THEN
EXIT
FI
OD
OD
IF count=0 THEN
Print("none")
FI
PutE()
RETURN

PROC Test(INT ARRAY a INT len,sum)
Print("Array: ") PrintArray(a,len)
Print("Sum: ") PrintIE(sum)
Print("Pairs: ") PrintPairs(a,len,sum)
PutE()
RETURN

PROC Main()
INT ARRAY a=[0 2 11 19 90]
INT ARRAY b=[0 2 3 3 4 11 17 17 18 19 90]

Test(a,5,21)
Test(a,5,22)
Test(b,11,21)
RETURN
Output:
Array: [0 2 11 19 90]
Sum:   21
Pairs: (1,3)

Array: [0 2 11 19 90]
Sum:   22
Pairs: none

Array: [0 2 3 3 4 11 17 17 18 19 90]
Sum:   21
Pairs: (1,9) (2,8) (3,8) (4,6) (4,7)

Aime

integer i, u, v;
index x;
list l;

l_bill(l, 0, 0, 2, 11, 19, 90);

for (i, u in l) {
x[u] = i;
if (i_jack(v, x, 21 - u)) {
o_(v, " ", i, "\n");
break;
}
}
Output:
1 3

ALGOL 68

Translation of: Lua
# returns the subscripts of a pair of numbers in a that sum to sum, a is assumed to be sorted #
# if there isn't a pair of numbers that summs to sum, an empty array is returned #
PRIO TWOSUM = 9;
OP TWOSUM = ( []INT a, INT sum )[]INT:
BEGIN
BOOL found := FALSE;
INT i := LWB a;
INT j := UPB a;
INT s = a[ i ] + a[ j ];
IF s = sum THEN
found := TRUE
ELIF s < sum THEN
i +:= 1
ELSE
j -:= 1
FI
OD;
IF found THEN ( i, j ) ELSE () FI
END # TWOSUM # ;

# test the TWOSUM operator #
PROC print twosum = ( []INT a, INT sum )VOID:
BEGIN
[]INT pair = a[ AT 0 ] TWOSUM sum;
IF LWB pair > UPB pair THEN
# no pair with the required sum #
print( ( "[]", newline ) )
ELSE
# have a pair #
print( ( "[", whole( pair[ LWB pair ], 0 ), ", ", whole( pair[ UPB pair ], 0 ), "]", newline ) )
FI
END # print twosum # ;
print twosum( ( 0, 2, 11, 19, 90 ), 21 ); # should be [1, 3] #
print twosum( ( -8, -2, 0, 1, 5, 8, 11 ), 3 ); # should be [0, 6] (or [1, 4]) #
print twosum( ( -3, -2, 0, 1, 5, 8, 11 ), 17 ); # should be [] #
print twosum( ( -8, -2, -1, 1, 5, 9, 11 ), 0 ) # should be [2, 3] #
Output:
[1, 3]
[0, 6]
[]
[2, 3]

APL

Works with Dyalog APL.

∘.+⍨ ⍺ makes a table that is the outer sum of the left argument (the numbers).

We want to remove the diagonal, to avoid edge cases. We can achieve this by setting all these numbers to an arbitrary decimal value, since two integers can't sum to a decimal.

≢⍺ is the length of the numbers. ⍳≢⍺ creates an array from 0 to the length of the numbers. ∘.=⍳≢⍺ returns an identity matrix of size ≢⍺ (using the outer product with the equality function). ⍸ returns the indices of these numbers. [email protected] sets that list to 0.1.

Then, we just need to find where the right argument (the target) is equal to the matrix, get the indices, and return the first one (⊃).

⎕io ← 0 ⍝ sets index origin to 0
ts ← {⊃⍸ ⍵= [email protected](⍸∘.=⍨⍳≢⍺) ∘.+⍨ ⍺}
⎕ ← 0 2 11 19 90 ts 21 ⍝ should be 1 3

Output:
1 3

AppleScript

Functional

Translation of: JavaScript

Nesting concatMap or (>>=) (flip concatMap) yields the cartesian product of the list with itself. Skipping products where the y index is lower than the x index (see the use of 'drop' below) ignores the 'lower triangle' of the cartesian grid, excluding mirror-image and duplicate number pairs.

Note that this draft assumes that the task and target output are specified in terms of the prevailing convention of zero-based indices.

AppleScript, unusually, happens to make internal use of one-based indices, rigid adherence to which would, of course, in this case, simply produce the wrong result :-)

-------------------------- TWO SUM -------------------------

-- twoSum :: Int -> [Int] -> [(Int, Int)]
on twoSum(n, xs)
set ixs to zip(enumFromTo(0, |length|(xs) - 1), xs)

script ijIndices
on |λ|(ix)
set {i, x} to ix

script jIndices
on |λ|(jy)
set {j, y} to jy

if (x + y) = n then
{{i, j}}
else
{}
end if
end |λ|
end script

|>>=|(drop(i + 1, ixs), jIndices)
end |λ|
end script

|>>=|(ixs, ijIndices)
end twoSum

---------------------------- TEST --------------------------
on run
twoSum(21, [0, 2, 11, 19, 90])

--> {{1, 3}} Single solution.
end run

--------------------- GENERIC FUNCTIONS --------------------

-- (>>=) :: Monad m => m a -> (a -> m b) -> m b
on |>>=|(xs, f)
concat(map(f, xs))
end |>>=|

-- concat :: [[a]] -> [a] | [String] -> String
on concat(xs)
script append
on |λ|(a, b)
a & b
end |λ|
end script

if length of xs > 0 and class of (item 1 of xs) is string then
set empty to ""
else
set empty to {}
end if
foldl(append, empty, xs)
end concat

-- drop :: Int -> a -> a
on drop(n, a)
if n < length of a then
if class of a is text then
text (n + 1) thru -1 of a
else
items (n + 1) thru -1 of a
end if
else
{}
end if
end drop

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo

-- 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

-- length :: [a] -> Int
on |length|(xs)
length of xs
end |length|

-- 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

-- 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
Output:
{{1, 3}}

Idiomatic

Like the "Functional" script above, this returns multiple solutions when they're found. However it assumes a sorted list, as per the clarified task description, which allows some optimisation of the search. Also, the indices returned are 1-based, which is the AppleScript norm.

on twoSum(givenNumbers, givenSum)
script o
property lst : givenNumbers
property output : {}
end script

set listLength to (count o's lst)
repeat with i from 1 to (listLength - 1)
set n1 to item i of o's lst
repeat with j from (i + 1) to listLength
set thisSum to n1 + (item j of o's lst)
if (thisSum = givenSum) then
set end of o's output to {i, j}
else if (thisSum > givenSum) then
exit repeat
end if
end repeat
end repeat

return o's output
end twoSum

-- Test code:
twoSum({0, 2, 11, 19, 90}, 21) -- Task-specified list.
twoSum({0, 3, 11, 19, 90}, 21) -- No matches.
twoSum({-44, 0, 0, 2, 10, 11, 19, 21, 21, 21, 65, 90}, 21) -- Multiple solutions.
Output:
{{2, 4}}
{}
{{1, 11}, {2, 8}, {2, 9}, {2, 10}, {3, 8}, {3, 9}, {3, 10}, {4, 7}, {5, 6}}

AutoHotkey

TwoSum(a, target){
i := 1, j := a.MaxIndex()
while(i < j){
if (a[i] + a[j] = target)
return i ", " j
else if (a[i] + a[j] < target)
i++
else if (a[i] + a[j] > target)
j--
}
}
Examples:
MsgBox % TwoSum([0, 2, 11, 19, 90], 21) ; returns 2, 4 (first index is 1 not 0)
Outputs:
2,4

AWK

# syntax: GAWK -f TWO_SUM.AWK
BEGIN {
numbers = "0,2,11,19,90"
print(two_sum(numbers,21))
print(two_sum(numbers,25))
exit(0)
}
function two_sum(numbers,sum, arr,i,j,s) {
i = 1
j = split(numbers,arr,",")
while (i < j) {
s = arr[i] + arr[j]
if (s == sum) {
return(sprintf("[%d,%d]",i,j))
}
else if (s < sum) {
i++
}
else {
j--
}
}
return("[]")
}

Output:
[2,4]
[]

Befunge

>000pv
>&:0\`#v_00g:1+00p6p
v >\$&50p110p020p
v>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>+50g-!#v_48*10g8p10g1+:00g1-`v >
v >[email protected] v_v
"""""""""""""""""""""""""""""""""""""""""""""""v ">"p4\"v"p02:+1p4\">":g02\$< :
> 20g8p20g1+:00g1-`#v_0 ^1
""""""""""""""""""""""""""""""""""""""""""""""" " 0
>^ l p
i "
a ^
F "
" \
>:#,[email protected] 8
p
> ^

There are a couple of caveats due to limitations of the language. The target cannot be above 127, there can be no more than 47 elements in the list and the list must be delimited by a negative number before the target value as follows:

0 2 11 19 90 -1 21
Output:
1 3

C

#include<stdio.h>

int main()
{
int arr = {0, 2, 11, 19, 90},sum = 21,i,j,check = 0;

for(i=0;i<4;i++){
for(j=i+1;j<5;j++){
if(arr[i]+arr[j]==sum){
printf("[%d,%d]",i,j);
check = 1;
break;
}
}
}

if(check==0)
printf("[]");

return 0;
}

Output :

[1,3]

C#

using System;
using System.Collections.Generic;

public class Program
{
public static void Main(string[] args)
{
int[] arr = { 0, 2, 11, 19, 90 };
const int sum = 21;

var ts = TwoSum(arr, sum);
Console.WriteLine(ts != null ? \$"{ts}, {ts}" : "no result");

}

public static int[] TwoSum(int[] numbers, int sum)
{
var map = new Dictionary<int, int>();
for (int i = 0; i < numbers.Length; i++)
{
// see if the complement is stored
var key = sum - numbers[i];
if (map.ContainsKey(key))
{
return new[] { map[key], i };
}
}
return null;
}
}

Output:
1, 3

C++

Translation of: C#
#include <iostream>
#include <map>
#include <tuple>
#include <vector>

using namespace std;

pair<int, int> twoSum(vector<int> numbers, int sum) {
auto m = map<int, int>();
for (size_t i = 0; i < numbers.size(); ++i) {
// see if the complement is stored
auto key = sum - numbers[i];

if (m.find(key) != m.end()) {
return make_pair(m[key], i);
}
m[numbers[i]] = i;
}

return make_pair(-1, -1);
}

int main() {
auto numbers = vector<int>{ 0, 2, 11, 19, 90 };
const int sum = 21;

auto ts = twoSum(numbers, sum);
if (ts.first != -1) {
cout << "{" << ts.first << ", " << ts.second << "}" << endl;
} else {
cout << "no result" << endl;
}

return 0;
}
Output:
{1,3}

D

import std.stdio;

void main() {
const arr = [0, 2, 11, 19, 90];
const sum = 21;

writeln(arr.twoSum(21));
}

/**
* Searches arr for two indexes whose value adds to sum, and returns those indexes.
* Returns an empty array if no such indexes exist.
* The values of arr are assumed to be sorted.
*/

int[] twoSum(const int[] arr, const int sum) in {
import std.algorithm.sorting : isSorted;
assert(arr.isSorted);
} out(result) {
assert(result.length == 0 || arr[result] + arr[result] == sum);
} body {
int i=0;
int j=arr.length-1;

while (i <= j) {
auto temp = arr[i] + arr[j];
if (temp == sum) {
return [i, j];
}

if (temp < sum) {
i++;
} else {
j--;
}
}

return [];
}
Output:
[1, 3]

Dart

main() {
var a = [1,2,3,4,5];
var s=25,c=0;
var z=(a.length*(a.length-1))/2;
for (var x = 0; x < a.length; x++) {
print(a[x]);
}
for (var x = 0; x < a.length; x++) {
for(var y=x+1;y< a.length; y++)
{
if(a[x]+a[y]==s)
{
print([a[x],a[y]]);
break;
}
else
{
c++;
}
}
}
if(c==z)
{
print("such pair doesn't exist");
}
}

Delphi

Translation of: Python

program Two_Sum;

{\$APPTYPE CONSOLE}

uses
System.SysUtils,
System.Generics.Collections;

function TwoSum(arr: TArray<Integer>; num: Integer; var i, j: integer): boolean;
begin
TArray.Sort<Integer>(arr);
i := 0;
j := Length(arr) - 1;
while i < j do
begin
if arr[i] + arr[j] = num then
exit(True);

if arr[i] + arr[j] < num then
inc(i)
else
Dec(j);
end;
Result := false;
end;

var
i, j: Integer;

begin
if TwoSum([0, 2, 11, 19, 90], 21, i, j) then
Writeln('(', i, ',', j, ')');

if TwoSum([0, 2, 11, 19, 90], 25, i, j) then
Writeln('(', i, ',', j, ')');
end.
Output:
(1,3)

Elixir

defmodule RC do
def two_sum(numbers, sum) do
Enum.with_index(numbers) |>
Enum.reduce_while([], fn {x,i},acc ->
y = sum - x
case Enum.find_index(numbers, &(&1 == y)) do
nil -> {:cont, acc}
j -> {:halt, [i,j]}
end
end)
end
end

numbers = [0, 2, 11, 19, 90]
IO.inspect RC.two_sum(numbers, 21)
IO.inspect RC.two_sum(numbers, 25)
Output:
[1, 3]
[]

F#

// Two Sum : Nigel Galloway December 5th., 2017
let fN n i =
let rec fN n e =
match n with
|n::g when n < i -> match List.mapi(fun g i-> (n,i,g)) g |> List.tryFind(fun (n,g,l)->(n+g)=i) with
|Some (n,g,l) -> [e;e+l+1]
|_ -> fN g (e+1)
|_ -> []
fN n 0
printfn "%A" (fN [0; 2; 11; 19; 90] 21)

Output:
[1; 3]

Factor

USING: combinators fry kernel locals math prettyprint sequences ;
IN: rosetta-code.two-sum

:: two-sum ( seq target -- index-pair )
0 seq length 1 - :> ( x! y! ) [
x y [ seq nth ] [email protected] + :> sum {
{ [ sum target = x y = or ] [ f ] }
{ [ sum target > ] [ y 1 - y! t ] }
[ x 1 + x! t ]
} cond
] loop
x y = { } { x y } ? ;

{ 21 55 11 } [ '[ { 0 2 11 19 90 } _ two-sum . ] call ] each
Output:
{ 1 3 }
{ }
{ 0 2 }

Forth

Works with: Gforth version 0.7.3
CREATE A CELL ALLOT
: A[] ( n -- A[n]) CELLS A @ + @ ;
:NONAME 1- ;
:NONAME R> DROP R> DROP TRUE ;
:NONAME SWAP 1+ SWAP ;
CREATE VTABLE , , ,
: CMP ( n n' -- -1|0|1) - DUP IF DUP ABS / THEN ;
: (TWOSUM) ( addr n n' -- u1 u2 t | f)
>R SWAP A ! 0 SWAP 1- ( lo hi) ( R: n')
BEGIN OVER OVER < WHILE
OVER A[] OVER A[] + [email protected]
CMP 1+ CELLS VTABLE + @ EXECUTE
REPEAT
DROP DROP R> DROP FALSE ;
: TWOSUM ( addr n n' --) [CHAR] [ EMIT
(TWOSUM) IF SWAP 0 .R [CHAR] , EMIT SPACE 0 .R THEN
[CHAR] ] EMIT ;
CREATE TEST0 0 , 2 , 11 , 19 , 90 , DOES> 5 ;
CREATE TEST1 -8 , -2 , 0 , 1 , 5 , 8 , 11 , DOES> 7 ;
TEST0 21 TWOSUM CR
TEST0 25 TWOSUM CR
TEST1 3 TWOSUM CR
TEST1 8 TWOSUM CR
BYE
Output:
[1, 3]
[]
[0, 6]
[2, 5]

Fortran

program twosum
implicit none

integer, parameter, dimension(5) :: list = (/ 0, 2, 11, 19, 90/)
integer, parameter :: target_val = 21
integer :: nelem
integer :: i, j
logical :: success = .false.

nelem = size(list)
outer:do i = 1,nelem
do j = i+1,nelem
success = list(i) + list(j) == target_val
if (success) exit outer
end do
end do outer

if (success) then
!Just some fancy formatting for nicer output
print('("(",2(i3.1,1X),")",3(A1,i3.1))'), i,j, ":", list(i), "+", list(j), "=", target_val
else
print*, "Failed"
end if

end program twosum

Output:
(  2   4 ):  2+ 19= 21

FreeBASIC

' FB 1.05.0 Win64

' "a" is the array of sorted non-negative integers
' "b" is the array to contain the result and is assumed to be empty initially

Sub twoSum (a() As UInteger, b() As Integer, targetSum As UInteger)
Dim lb As Integer = LBound(a)
Dim ub As Integer = UBound(a)
If ub = -1 Then Return '' empty array
Dim sum As UInteger

For i As Integer = lb To ub - 1
If a(i) <= targetSum Then
For j As Integer = i + 1 To ub
sum = a(i) + a(j)
If sum = targetSum Then
Redim b(0 To 1)
b(0) = i : b(1) = j
Return
ElseIf sum > targetSum Then
Exit For
End If
Next j
Else
Exit For
End If
Next i
End Sub

Dim a(0 To 4) As UInteger = {0, 2, 11, 19, 90}
Dim b() As Integer
Dim targetSum As UInteger = 21
twoSum a(), b(), targetSum
If UBound(b) = -1 Then
Print "No two numbers were found whose sum is "; targetSum
Else
Print "The numbers with indices"; b(LBound(b)); " and"; b(UBound(b)); " sum to "; targetSum
End If
Print
Print "Press any number to quit"
Sleep
Output:
The numbers with indices 1 and 3 sum to 21

Go

Translation of: Kotlin
package main

import "fmt"

func twoSum(a []int, targetSum int) (int, int, bool) {
len := len(a)
if len < 2 {
return 0, 0, false
}
for i := 0; i < len - 1; i++ {
if a[i] <= targetSum {
for j := i + 1; j < len; j++ {
sum := a[i] + a[j]
if sum == targetSum {
return i, j, true
}
if sum > targetSum {
break
}
}
} else {
break
}
}
return 0, 0, false
}

func main() {
a := []int {0, 2, 11, 19, 90}
targetSum := 21
p1, p2, ok := twoSum(a, targetSum)
if (!ok) {
fmt.Println("No two numbers were found whose sum is", targetSum)
} else {
fmt.Println("The numbers with indices", p1, "and", p2, "sum to", targetSum)
}
}
Output:
The numbers with indices 1 and 3 sum to 21

Returning first match

twoSum::(Num a,Ord a) => a -> [a] -> [Int]
twoSum num list = sol ls (reverse ls)
where
ls = zip list [0..]
sol [] _ = []
sol _ [] = []
sol xs@((x,i):us) ys@((y,j):vs) = ans
where
s = x + y
ans | s == num = [i,j]
| j <= i = []
| s < num = sol (dropWhile ((<num).(+y).fst) us) ys
| otherwise = sol xs \$ dropWhile ((num <).(+x).fst) vs

main = print \$ twoSum 21 [0, 2, 11, 19, 90]
Output:
[1,3]

Returning all matches

Listing multiple solutions (as zero-based indices) where they exist:

sumTo :: Int -> [Int] -> [(Int, Int)]
sumTo n ns =
let ixs = zip [0 ..] ns
in ixs
>>= ( \(i, x) ->
drop (succ i) ixs
>>= \(j, y) ->
[ (i, j)
| (x + y) == n
]
)

main :: IO ()
main = mapM_ print \$ sumTo 21 [0, 2, 11, 19, 90, 10]

Or, resugaring a little – pulling more into the scope of the list comprehension:

sumTo :: Int -> [Int] -> [(Int, Int)]
sumTo n ns =
let ixs = zip [0 ..] ns
in [ (i, j)
| (i, x) <- ixs
, (j, y) <- drop (succ i) ixs
, (x + y) == n ]

main :: IO ()
main = mapM_ print \$ sumTo 21 [0, 2, 11, 19, 90, 10]
Output:
(1,3)
(2,5)

Icon and Unicon

Translation of: Lua

Icon and Unicon are ordinal languages, first index is one.

fullimag library used to pretty print lists.

#
# twosum.icn, find two array elements that add up to a given sum
# Dedicated to the public domain
#
procedure main(arglist)
sum := pop(arglist) | 21
L := []
if *arglist > 0 then every put(L, integer(!arglist)) & L := sort(L)
else L := [0, 2, 11, 19, 90]

write(sum)
write(fullimage(L))
write(fullimage(twosum(sum, L)))
end

# assume sorted list, only interested in zero or one solution
procedure twosum(sum, L)
i := 1
j := *L
while i < j do {
try := L[i] + L[j]
if try = sum then return [i,j]
else
if try < sum then
i +:= 1
else
j -:= 1
}
return []
end
Output:
\$ unicon -s twosum.icn -x
21
[0,2,11,19,90]
[2,4]

J

So, first off, our basic approach will be to find the sums:

=+/~0 2 11 19 90
0 2 11 19 90
2 4 13 21 92
11 13 22 30 101
19 21 30 38 109
90 92 101 109 180

And, check if any of them are our desired value:

21=+/~0 2 11 19 90
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0

Except, we want indices here, so let's toss the structure so we can get those:

,21=+/~0 2 11 19 90
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
I.,21=+/~0 2 11 19 90
8 16

Except, we really needed that structure - in this case, since we had a five by five table, we want to interpret this result as a base five pair of numbers:

\$21=+/~0 2 11 19 90
5 5
5 5#:I.,21=+/~0 2 11 19 90
1 3
3 1

Or, taking advantage of being able to use verbs to represent combining their results, when we use three of them:

(\$ #: [email protected],)21=+/~0 2 11 19 90
1 3
3 1

But to be more like the other task implementations here, we don't want all the results, we just want zero or one result. We can't just take the first result, though, because that would fill in a 0 0 result if there were none, and 0 0 could have been a valid result which does not make sense for the failure case. So, instead, let's package things up so we can add an empty to the end and take the first of those:

(\$ <@#: [email protected],)21=+/~0 2 11 19 90
┌───┬───┐
1 33 1
└───┴───┘
a:,~(\$ <@#: [email protected],)21=+/~0 2 11 19 90
┌───┬───┬┐
1 33 1││
└───┴───┴┘
{.a:,~(\$ <@#: [email protected],)21=+/~0 2 11 19 90
┌───┐
1 3
└───┘
;{.a:,~(\$ <@#: [email protected],)21=+/~0 2 11 19 90
1 3

Finally, let's start pulling our arguments out using that three verbs combining form:

;{.a:,~(\$ <@#: [email protected],) 21([ = +/[email protected]])0 2 11 19 90
1 3
;{.a:,~21 (\$ <@#: [email protected],)@([ = +/[email protected]])0 2 11 19 90
1 3

a: is not a verb, but we can use a noun as the left verb of three as an implied constant verb whose result is itself:

;{. 21 (a:,~ (\$ <@#: [email protected],)@([ = +/[email protected]]))0 2 11 19 90
1 3

And, let's finish the job, give this a name, and test it out:

twosum=: ;@{[email protected](a:,~ (\$ <@#: [email protected],)@([ = +/[email protected]]))
21 twosum 0 2 11 19 90
1 3

Except that looks like a bit of a mess. A lot of the reason for this is that ascii is ugly to look at. (Another issue, though, is that a lot of people are not used to architecting control flow as expressions.)

So... let's do this over again, using a more traditional implementation where we name intermediate results. (We're going to stick with our architecture, though, because changing the architecture to the more traditional approach would change the space/time tradeoff to require more time.)

sums=. +/~ y
matches=. x = sums
sum_inds=. I. , matches
pair_inds=. (\$matches) #: sum_inds
; {. a: ,~ <"1 pair_inds
)

And, testing:

21 two_sum 0 2 11 19 90
1 3

Or, we could go slightly more traditional and instead of doing that boxing at the end, use an if/else statement:

sums=. +/~ y
matches=. x = sums
sum_inds=. I. , matches
pair_inds=. (\$matches) #: sum_inds
if. #pair_inds do.
{.pair_inds
else.
i.0
end.
)

Then again, most people don't read J anyways, so maybe just stick with the earlier implementation:

twosum=: ;@{[email protected](a:,~ (\$ <@#: [email protected],)@([ = +/[email protected]]))

Alternative approach

An alternative method for identifying and returning non-duplicate indicies of the pairs follows.

21 (= +/~) 0 2 11 19 90
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0

The array is symmetrical so we can zero one half to remove duplicate pairs and then retrieve the remaining indicies using sparse array functionality.

zeroLowerTri=: * [: </~ [email protected]#
getIdx=: 4 \$. \$.
twosum_alt=: [email protected]@(= +/~)

Testing ...

21 twosum_alt 0 2 11 19 90
1 3

Java

Translation of: Lua
import java.util.Arrays;

public class TwoSum {

public static void main(String[] args) {
long sum = 21;
int[] arr = {0, 2, 11, 19, 90};

System.out.println(Arrays.toString(twoSum(arr, sum)));
}

public static int[] twoSum(int[] a, long target) {
int i = 0, j = a.length - 1;
while (i < j) {
long sum = a[i] + a[j];
if (sum == target)
return new int[]{i, j};
if (sum < target) i++;
else j--;
}
return null;
}
}
[1, 3]

JavaScript

ES5

Nesting concatMap yields the cartesian product of the list with itself, and functions passed to Array.map() have access to the array index in their second argument. Returning [] where the y index is lower than or equal to the x index ignores the 'lower triangle' of the cartesian grid, skipping mirror-image and duplicate number pairs. Returning [] where a sum condition is not met similarly acts as a filter – all of the empty lists in the map result are eliminated by the concat.

(function () {
var concatMap = function (f, xs) {
return [].concat.apply([], xs.map(f))
};

return function (n, xs) {
return concatMap(function (x, ix) {
return concatMap(function (y, iy) {
return iy <= ix ? [] : x + y === n ? [
[ix, iy]
] : []
}, xs)
}, xs)
}(21, [0, 2, 11, 19, 90]);
})();

Output:
[[1,3]]

ES6

Composing a solution from generic functions like zip, bind (>>=, or flip concatMap) etc.

(() => {
'use strict';

// SUMTWO ----------------------------------------------------------------

// sumTwo :: Int -> [Int] -> [(Int, Int)]
function sumTwo(n, xs) {
const ixs = zip(enumFromTo(0, length(xs) - 1), xs);
return bind(ixs,
([i, x]) => bind(drop(i + 1, ixs),
([j, y]) => (x + y === n) ? [
[i, j]
] : []
)
);
};

// GENERIC FUNCTIONS -----------------------------------------------------

// bind (>>=) :: [a] -> (a -> [b]) -> [b]
const bind = (xs, f) => [].concat.apply([], xs.map(f));

// drop :: Int -> [a] -> [a]
const drop = (n, xs) => xs.slice(n);

// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);

// length :: [a] -> Int
const length = xs => xs.length;

// show :: a -> String
const show = (...x) =>
JSON.stringify.apply(
null, x.length > 1 ? [x, null, x] : x
);

// zip :: [a] -> [b] -> [(a,b)]
const zip = (xs, ys) =>
xs.slice(0, Math.min(xs.length, ys.length))
.map((x, i) => [x, ys[i]]);

// TEST ------------------------------------------------------------------
return show(
sumTwo(21, [0, 2, 11, 19, 90, 10])
);
})();
Output:
[[1,3],[2,5]]

Jsish

Based on Javascript entry.

/* Two Sum, in Jsish */
function twoSum(target, list) {
var concatMap = function (f, xs) {
return [].concat.apply([], xs.map(f));
};

return function (n, xs) {
return concatMap(function (x, ix) {
return concatMap(function (y, iy) {
return iy <= ix ? [] : x + y === n ? [
[ix, iy]
] : [];
}, xs);
}, xs);
}(target, list);
}

var list = [0, 2, 11, 19, 90];
;list;
;twoSum(21, list);
;list[twoSum(21, list)];
;list[twoSum(21, list)];
Output:
prompt\$ jsish --U twoSum.jsi
list ==> [ 0, 2, 11, 19, 90 ]
twoSum(21, list) ==> [ [ 1, 3 ] ]
list[twoSum(21, list)] ==> 2
list[twoSum(21, list)] ==> 19

jq

Works with: jq

Works with gojq, the Go implementation of jq.

Translation of: Julia
def twosum(\$s):
. as \$v
| {i: 0, j: (\$v|length - 1) }
| until( .i >= .j or \$v[.i] + \$v[.j] == \$s;
if \$v[.i] + \$v[.j] < \$s then .i += 1
else .j -= 1
end)
| if .i >= .j then [] else [.[]] end ; # as required

[0, 2, 11, 19, 90]
| (twosum(21), twosum(25))

Output:
[1,3]
[]

Julia

Works with: Julia version 0.6
Translation of: Python
function twosum(v::Vector, s)
i = 1
j = length(v)
while i < j
if v[i] + v[j] == s
return [i, j]
elseif v[i] + v[j] < s
i += 1
else
j -= 1
end
end
return similar(v, 0)
end

@show twosum([0, 2, 11, 19, 90], 21)
Output:
twosum([0, 2, 11, 19, 90], 21) = [2, 4]

Kotlin

// version 1.1

fun twoSum(a: IntArray, targetSum: Int): Pair<Int, Int>? {
if (a.size < 2) return null
var sum: Int
for (i in 0..a.size - 2) {
if (a[i] <= targetSum) {
for (j in i + 1..a.size - 1) {
sum = a[i] + a[j]
if (sum == targetSum) return Pair(i, j)
if (sum > targetSum) break
}
} else {
break
}
}
return null
}

fun main(args: Array<String>) {
val a = intArrayOf(0, 2, 11, 19, 90)
val targetSum = 21
val p = twoSum(a, targetSum)
if (p == null) {
println("No two numbers were found whose sum is \$targetSum")
} else {
println("The numbers with indices \${p.first} and \${p.second} sum to \$targetSum")
}
}
Output:
The numbers with indices 1 and 3 sum to 21

Liberty BASIC

myArray(0) = 0
myArray(1) = 2
myArray(2) = 11
myArray(3) = 19
myArray(4) = 90

sum = 21

Print twoToSum\$("myArray", sum, 0, 4)
End

Function twoToSum\$(arrayName\$, targetSum, minElement, maxElement)
i = minElement : j = maxElement
While (i < j)
Select Case
Case (Eval(arrayName\$;"(";i;")") + Eval(arrayName\$;"(";j;")")) < targetSum
i = (i + 1)
Case (Eval(arrayName\$;"(";i;")") + Eval(arrayName\$;"(";j;")")) > targetSum
j = (j - 1)
Case Else
twoToSum\$ = "[";i;",";j;"]"
Exit Function
End Select
Wend
twoToSum\$ = "[]"
End Function
Output:
[1,3]

Lua

Lua uses one-based indexing.

function twoSum (numbers, sum)
local i, j, s = 1, #numbers
while i < j do
s = numbers[i] + numbers[j]
if s == sum then
return {i, j}
elseif s < sum then
i = i + 1
else
j = j - 1
end
end
return {}
end

print(table.concat(twoSum({0,2,11,19,90}, 21), ","))
Output:
2,4

Maple

two_sum := proc(arr, sum)
local i,j,temp:
i,j := 1,numelems(arr):
while (i < j) do
temp := arr[i] + arr[j]:
if temp = sum then
return [i,j]:
elif temp < sum then
i := i + 1:
else
j := j-1:
end if:
end do:
return []:
end proc:
L := Array([0,2,2,11,19,19,90]);
two_sum(L, 21);
Output:

Note that Maple does 1 based indexing.

[2,5]

Mathematica / Wolfram Language

twoSum[data_List, sum_] :=
Block[{indices = Subsets[[email protected]@data, {2}]},
Cases[indices, _?([email protected][[#]] == sum &)]]

twoSum[{0, 2, 11, 19, 90}, 21] // TableForm
Output:

2 4 Note, indexing in Mathematica starts at 1

MiniScript

twoSum = function(numbers, sum)
// Make a map of values to their indices in the numbers array
// as we go, so we will know when we've found a match.
map = {}
for i in numbers.indexes
key = sum - numbers[i]
if map.hasIndex(key) then return [map[key], i]
map[numbers[i]] = i
end for
end function

print twoSum([0, 2, 11, 19, 90], 21)

Output:

[1, 3]

Modula-2

MODULE TwoSum;
FROM FormatString IMPORT FormatString;

TYPE
Pair = RECORD
f,s : INTEGER;
END;

PROCEDURE TwoSum(CONST arr : ARRAY OF INTEGER; CONST sum : INTEGER) : Pair;
VAR i,j,temp : INTEGER;
BEGIN
i := 0;
j := HIGH(arr)-1;

WHILE i<=j DO
temp := arr[i] + arr[j];
IF temp=sum THEN
RETURN Pair{i,j};
END;
IF temp<sum THEN
INC(i);
ELSE
DEC(j);
END;
END;

RETURN Pair{-1,-1};
END TwoSum;

VAR
buf : ARRAY[0..63] OF CHAR;
arr : ARRAY[0..4] OF INTEGER;
res : Pair;
BEGIN
arr:=0;
arr:=2;
arr:=11;
arr:=19;
arr:=90;

res := TwoSum(arr, 21);
FormatString("[%i, %i]\n", buf, res.f, res.s);
WriteString(buf);
END TwoSum.

Nim

proc twoSum(src: openarray[int], target: int): array[2, int] =
if src.len < 2:
return

for base in 0 .. (src.len - 2):
for ext in (base + 1) ..< src.len:
if src[base] + src[ext] == target:
result = base
result = ext

proc main =
var data0 = [0, 2, 11, 19, 90]
var res = twoSum(data0, 21)
assert(res == [1, 3])

var data1 = [0, 2, 11, 19, 90]
res = twoSum(data1, 22)
assert(res == [0, 0])

var data2 = 
res = twoSum(data2, 22)
assert(res == [0, 0])

var data3 = [1, 99]
res = twoSum(data3, 100)
assert(res == [0, 1])

var data4 = [1, 99]
res = twoSum(data4, 101)
assert(res == [0, 0])

main()

Objeck

Translation of: Java
class TwoSum {
function : Main(args : String[]) ~ Nil {
sum := 21;
arr := [0, 2, 11, 19, 90];
Print(TwoSum(arr, sum));
}

function : TwoSum(a : Int[], target : Int) ~ Int[] {
i := 0;
j := a->Size() - 1;

while (i < j) {
sum := a[i] + a[j];
if(sum = target) {
r := Int->New;
r := i;
r := j;
return r;
};

if (sum < target) {
i++;
}
else {
j--;
};
};

return Nil;
}

function : Print(r : Int[]) ~ Nil {
'['->Print();
each(i : r) {
r[i]->Print();
if(i + 1 < r->Size()) {
','->Print();
};
};
']'->PrintLine();
}
}

Output:

[1,3]

OCaml

Translation of: C
let get_sums ~numbers ~sum =
let n = Array.length numbers in
let res = ref [] in
for i = 0 to n - 2 do
for j = i + 1 to n - 1 do
if numbers.(i) + numbers.(j) = sum then
res := (i, j) :: !res
done
done;
!res

let () =
let numbers = [| 0; 2; 11; 19; 90 |]
and sum = 21
in
let res = get_sums ~numbers ~sum in

List.iter (fun (i, j) ->
Printf.printf "# Found: %d %d\n" i j
) res

Will return all possible sums, not just the first one found.

Output:
\$ ocaml two_sum.ml
# Found: 1 3

ooRexx

a=.array~of( -5, 26, 0, 2, 11, 19, 90)
x=21
n=0
do i=1 To a~items
Do j=i+1 To a~items
If a[i]+a[j]=x Then Do
Say '['||i-1||','||j-1||']'
n=n+1
End
End
End
If n=0 Then
Say '[] - no items found'
Output:
[0,1]
[3,5]

Pascal

A little bit lengthy. Implemented an unsorted Version with quadratic runtime too and an extra test case with 83667 elements that needs 83667*86666/2 ~ 3.5 billion checks ( ~1 cpu-cycles/check, only if data in cache ).

program twosum;
{\$IFDEF FPC}{\$MODE DELPHI}{\$ELSE}{\$APPTYPE CONSOLE}{\$ENDIF}
uses
sysutils;
type
tSolRec = record
SolRecI,
SolRecJ : NativeInt;
end;
tMyArray = array of NativeInt;
const
// just a gag using unusual index limits
ConstArray :array[-17..-13] of NativeInt = (0, 2, 11, 19, 90);

function Check2SumUnSorted(const A :tMyArray;
sum:NativeInt;
var Sol:tSolRec):boolean;
//Check every possible sum A[max] + A[max-1..0]
//than A[max-1] + A[max-2..0] etc pp.
//quadratic runtime: maximal (max-1)*max/ 2 checks
//High(A) always checked for dynamic array, even const
//therefore run High(A) to low(A), which is always 0 for dynamic array
label
SolFound;
var
i,j,tmpSum: NativeInt;
Begin
Sol.SolRecI:=0;
Sol.SolRecJ:=0;
i := High(A);
while i > low(A) do
Begin
tmpSum := sum-A[i];
j := i-1;
while j >= low(A) do
begin
if tmpSum = a[j] Then
GOTO SolFound;
dec(j);
end;
dec(i);
end;
result := false;
exit;
SolFound:
Sol.SolRecI:=j;Sol.SolRecJ:=i;
result := true;
end;

function Check2SumSorted(const A :tMyArray;
sum:NativeInt;
var Sol:tSolRec):boolean;
var
i,j,tmpSum: NativeInt;
Begin
Sol.SolRecI:=0;
Sol.SolRecJ:=0;
i := low(A);
j := High(A);
while(i < j) do
Begin
tmpSum := a[i] + a[j];
if tmpSum = sum then
Begin
Sol.SolRecI:=i;Sol.SolRecJ:=j;
result := true;
EXIT;
end;
if tmpSum < sum then
begin
inc(i);
continue;
end;
//if tmpSum > sum then
dec(j);
end;
writeln(i:10,j:10);
result := false;
end;

var
Sol :tSolRec;
CheckArr : tMyArray;
MySum,i : NativeInt;

Begin
randomize;
setlength(CheckArr,High(ConstArray)-Low(ConstArray)+1);
For i := High(CheckArr) downto low(CheckArr) do
CheckArr[i] := ConstArray[i+low(ConstArray)];

MySum := 21;
IF Check2SumSorted(CheckArr,MySum,Sol) then
writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum)
else
writeln('No solution found');

//now test a bigger sorted array..
setlength(CheckArr,83667);
For i := High(CheckArr) downto 0 do
CheckArr[i] := i;
MySum := CheckArr[Low(CheckArr)]+CheckArr[Low(CheckArr)+1];
writeln(#13#10,'Now checking array of ',length(CheckArr),
' elements',#13#10);
IF Check2SumUnSorted(CheckArr,MySum,Sol) then
writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum)
else
writeln('No solution found');
//runtime not measurable
IF Check2SumSorted(CheckArr,MySum,Sol) then
writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum)
else
writeln('No solution found');
end.
Output:
[1,3] sum to 21

Now checking array of 83667 elements

[0,1] sum to 1
[0,1] sum to 1

real    0m1.013s

Perl

Translation of: Python
use strict;
use warnings;
use feature 'say';

sub two_sum{
my(\$sum,@numbers) = @_;
my \$i = 0;
my \$j = \$#numbers - 1;
my @indices;
while (\$i < \$j) {
if (\$numbers[\$i] + \$numbers[\$j] == \$sum) { push @indices, (\$i, \$j); \$i++; }
elsif (\$numbers[\$i] + \$numbers[\$j] < \$sum) { \$i++ }
else { \$j-- }
}
return @indices
}

my @numbers = <0 2 11 19 90>;
my @indices = two_sum(21, @numbers);
say join(', ', @indices) || 'No match';

@indices = two_sum(25, @numbers);
say join(', ', @indices) || 'No match';
Output:
1, 3
No match

Phix

function twosum(sequence s, integer t)
for i=1 to length(s) do
for j=i+1 to length(s) do
if s[i]+s[j]=t then
return {i,j}
end if
end for
end for
return {}
end function
?twosum({0, 2, 11, 19, 90},21)
Translation of: Raku
function twosum(sequence numbers, integer total)
integer i=1, j=length(numbers)
while i<j do
switch compare(numbers[i]+numbers[j],total) do
case -1: i += 1
case  0: return {i, j}
case +1: j -= 1
end switch
end while
return {}
end function
Output:

Phix uses 1-based indexes

{2,4}

Phixmonti

include ..\Utilitys.pmt

def two_sum /# arr num -- n #/
var num
1 var i
len var j
true
while
i get swap j get rot + >ps
tps num == if
ps> drop j get swap i get rot 2 tolist false
else
ps> num < if i 1 + var i else j 1 - var j endif true
endif
i j < and
endwhile
len 2 > if drop ( ) endif
enddef

( 0 2 11 19 90 )
21 two_sum ?
25 two_sum ?
Output:
[2, 19]
[]

=== Press any key to exit ===

PicoLisp

(de twosum (Lst N)
(for ((I . A) Lst A (cdr A))
(T
(for ((J . B) (cdr Lst) B (cdr B))
(T (= N (+ (car A) (car B)))
(cons I (inc J)) ) )
@ ) ) )
(println
(twosum (0 2 11 19 90) 21)
(twosum (-3 -2 0 1 5 8 11) 17)
(twosum (-8 -2 -1 1 5 9 11) 0) )
Output:
(2 . 4) NIL (3 . 4)

PowerShell

Lazy, very lazy.

\$numbers = @(0, 2, 11, 19, 90)
\$sum = 21

\$totals = for (\$i = 0; \$i -lt \$numbers.Count; \$i++)
{
for (\$j = \$numbers.Count-1; \$j -ge 0; \$j--)
{
[PSCustomObject]@{
FirstIndex = \$i
SecondIndex = \$j
TargetSum = \$numbers[\$i] + \$numbers[\$j]
}
}
}

\$totals | Where-Object TargetSum -EQ \$sum |
Select-Object -First 1 `
-Property @{
Name = "Sum"
Expression = { \$_.TargetSum }
},
@{
Name = "Indices"
Expression = { @(\$_.FirstIndex, \$_.SecondIndex) }
}

Output:
Sum Indices
--- -------
21 {1, 3}

Python

Translation of: Raku
def two_sum(arr, num):
i = 0
j = len(arr) - 1
while i < j:
if arr[i] + arr[j] == num:
return (i, j)
if arr[i] + arr[j] < num:
i += 1
else:
j -= 1
return None

numbers = [0, 2, 11, 19, 90]
print(two_sum(numbers, 21))
print(two_sum(numbers, 25))

or, in terms of itertools.product:

Works with: Python version 3.7
'''Finding two integers that sum to a target value.'''

from itertools import (product)

# sumTwo :: [Int] -> Int -> [(Int, Int)]
def sumTwo(xs):
'''All the pairs of integers in xs which
sum to n.
'''

def go(n):
ixs = list(enumerate(xs))
return [
(fst(x), fst(y)) for (x, y) in (
product(ixs, ixs[1:])
) if fst(x) < fst(y) and n == snd(x) + snd(y)
]
return lambda n: go(n)

# TEST ----------------------------------------------------

# main :: IO ()
def main():
'''Tests'''

xs = [0, 2, 11, 19, 90, 10]

print(
fTable(
'The indices of any two integers drawn from ' + repr(xs) +
'\nthat sum to a given value:\n'
)(str)(
lambda x: str(x) + ' = ' + ', '.join(
['(' + str(xs[a]) + ' + ' + str(xs[b]) + ')' for a, b in x]
) if x else '(none)'
)(
sumTwo(xs)
)(enumFromTo(10)(25))
)

# GENERIC -------------------------------------------------

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))

# fst :: (a, b) -> a
def fst(tpl):
'''First member of a pair.'''
return tpl

# snd :: (a, b) -> b
def snd(tpl):
'''Second member of a pair.'''
return tpl

# DISPLAY -------------------------------------------------

# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''

def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)

# MAIN ---
if __name__ == '__main__':
main()
Output:
The indices of any two integers drawn from [0, 2, 11, 19, 90, 10]
that sum to a given value:

10 -> [(0, 5)] = (0 + 10)
11 -> [(0, 2)] = (0 + 11)
12 -> [(1, 5)] = (2 + 10)
13 -> [(1, 2)] = (2 + 11)
14 -> (none)
15 -> (none)
16 -> (none)
17 -> (none)
18 -> (none)
19 -> [(0, 3)] = (0 + 19)
20 -> (none)
21 -> [(1, 3), (2, 5)] = (2 + 19), (11 + 10)
22 -> (none)
23 -> (none)
24 -> (none)
25 -> (none)

or, a little more parsimoniously (not generating the entire cartesian product), in terms of concatMap:

Works with: Python version 3.7
'''Finding two integers that sum to a target value.'''

from itertools import chain

# sumTwo :: Int -> [Int] -> [(Int, Int)]
def sumTwo(n, xs):
'''All the pairs of integers in xs which
sum to n.
'''

def go(vs):
return [vs] if n == sum(vs) else []
ixs = list(enumerate(xs))
return list(
bind(ixs)(
lambda ix: bind(ixs[ix:])(
lambda jy: go(tuple(zip(*(ix, jy))))
)
)
)

# TEST ----------------------------------------------------

# main :: IO ()
def main():
'''Tests'''

for n in [21, 25]:
print(
sumTwo(n, [0, 2, 11, 19, 90, 10])
)

# GENERIC -------------------------------------------------

# bind (>>=) :: [a] -> (a -> [b]) -> [b]
def bind(xs):
Two computations sequentially composed,
with any value produced by the first
passed as an argument to the second.
'''

return lambda f: list(
chain.from_iterable(
map(f, xs)
)
)

if __name__ == '__main__':
main()
Output:
[(1, 3), (2, 5)]
[]

Quackery

So… I initially misread the task as "return the two integers" and then realised it was "…the indices of…", but that's OK — it just meant writing an extra word to find the indices, given the numbers.

The last three lines of task are in case the two integers found by twosome are equal - in which case, as find finds the first instance in the array and the array is sorted, we can safely take the index plus one as the second index.

[ 0 peek ]                    is first  (   [ --> n )

[ -1 peek ] is last ( [ --> n )

[ 1 split nip ] is top ( [ --> [ )

[ -1 split drop ] is tail ( [ --> [ )

[ temp put
[ dup size 2 < iff
[ drop [] ] done
dup first over last +
temp share -
dup 0 = iff
[ drop dup first
swap last join ] done
0 < iff top else tail
again ]
temp release ] is twosum ( [ n --> [ )

[ over temp put
twosum
[] swap
witheach
[ temp share find join ]
temp release
dup [] != if
[ dup unpack = if
[ behead 1+ join ] ] ] is task ( [ n --> [ )

' [ 0 2 11 19 20 ] 21 task echo cr
' [ 0 2 11 19 20 ] 25 task echo cr
' [ 0 2 12 12 20 ] 24 task echo cr
Output:
[ 1 3 ]
[ ]
[ 2 3 ]

Racket

#lang racket/base
(define (two-sum v m)
(let inr ((l 0) (r (sub1 (vector-length v))))
(and
(not (= l r))
(let ((s (+ (vector-ref v l) (vector-ref v r))))
(cond [(= s m) (list l r)] [(> s m) (inr l (sub1 r))] [else (inr (add1 l) r)])))))

(module+ test
(require rackunit)
;; test cases
;; no output indicates returns are as expected
(check-equal? (two-sum #( 0 2 11 19 90) 21) '(1 3))
(check-equal? (two-sum #(-8 -2 0 1 5 8 11) 3) '(0 6))
(check-equal? (two-sum #(-3 -2 0 1 5 8 11) 17) #f)
(check-equal? (two-sum #(-8 -2 -1 1 5 9 11) 0) '(2 3)))

Raku

(formerly Perl 6)

Procedural

Translation of: zkl
sub two_sum ( @numbers, \$sum ) {
die '@numbers is not sorted' unless [<=] @numbers;

my ( \$i, \$j ) = 0, @numbers.end;
while \$i < \$j {
given \$sum <=> @numbers[\$i,\$j].sum {
when Order::More { \$i += 1 }
when Order::Less { \$j -= 1 }
when Order::Same { return \$i, \$j }
}
}
return;
}

say two_sum ( 0, 2, 11, 19, 90 ), 21;
say two_sum ( 0, 2, 11, 19, 90 ), 25;
Output:
(1 3)
Nil

Functional

The two versions differ only in how one 'reads' the notional flow of processing: left-to-right versus right-to-left. Both return all pairs that sum to the target value, not just the first (e.g. for input of 0 2 10 11 19 90 would get indices 1/4 and 2/3).

sub two-sum-lr (@a, \$sum) {
# (((^@a X ^@a) Z=> (@a X+ @a)).grep(\$sum == *.value)>>.keys.map:{ .split(' ').sort.join(' ')}).unique
(
(
(^@a X ^@a) Z=> (@a X+ @a)
).grep(\$sum == *.value)>>.keys
.map:{ .split(' ').sort.join(' ')}
).unique
}

sub two-sum-rl (@a, \$sum) {
# unique map {.split(' ').sort.join(' ')}, keys %(grep {.value == \$sum}, ((^@a X ^@a) Z=> (@a X+ @a)))
unique
map {.split(' ').sort.join(' ')},
keys %(
grep {.value == \$sum}, (
(^@a X ^@a) Z=> (@a X+ @a)
)
)
}

my @a = <0 2 11 19 90>;
for 21, 25 {
say two-sum-rl(@a, \$_);
say two-sum-lr(@a, \$_);
}
Output:
(1 3)
(1 3)
()
()

REXX

version 1

/* REXX */
list='-5 26 0 2 11 19 90'
Do i=0 By 1 Until list=''
Parse Var list a.i list
End
n=i
x=21
z=0
do i=0 To n
Do j=i+1 To n
s=a.i+a.j
If s=x Then Do
z=z+1
Say '['i','j']' a.i a.j s
End
End
End
If z=0 Then
Say '[] - no items found'
Else
Say z 'solutions found'
Output:
[0,1] -5 26 21
[3,5] 2 19 21
2 solutions found

version 2

All solutions are listed (if any),   along with a count of the number of solutions.

Also, it's mentioned that the indices are zero─based,   and formatted solutions are shown.

The list of numbers can be in any format,   not just integers.   Also, they need not be unique.

The list of integers   needn't   be sorted.

A   numeric digits 500   statement was added just in case some humongous numbers were entered.

No verification was performed to ensure that all items in the list were numeric.

A little extra code was added to have the output columns aligned.

/*REXX program finds two numbers in a list of numbers that  sum  to a particular target.*/
numeric digits 500 /*be able to handle some larger numbers*/
parse arg targ list /*obtain optional arguments from the CL*/
if targ='' | targ="," then targ= 21 /*Not specified? Then use the defaults*/
if list='' | list="," then list= 0 2 11 19 90 /* " " " " " " */
say 'the list: ' list /*echo the list to the terminal*/
say 'the target sum: ' targ /* " " target sum " " " */
w= 0; sol= 0 /*width; # of solutions found (so far)*/
do #=0 for words(list); _=word(list, #+1) /*examine the list, construct an array.*/
@.#= _; w= max(w, length(_) ) /*assign a number to an indexed array. */
end /*#*/ /*W: the maximum width of any number. */
L= length(#) /*L: " " " " " index. */
@solution= 'a solution: zero─based indices ' /*a SAY literal for space conservation.*/
say /* [↓] look for sum of 2 numbers=target*/
do a=0 for # /*scan up to the last number in array. */
do b=a+1 to #-1; if @.a + @.b\=targ then iterate /*Sum not correct? Skip.*/
sol= sol + 1 /*bump count of the number of solutions*/
say @solution center( "["right(a, L)',' right(b, L)"]", L+L+5) ,
right(@.a, w*4) " + " right(@.b, w) ' = ' targ
end /*b*/ /*show the 2 indices and the summation.*/
end /*a*/
say
if sol==0 then sol= 'None' /*prettify the number of solutions if 0*/
say 'number of solutions found: ' sol /*stick a fork in it, we're all done. */
output   when using the default inputs:
the list:        0 2 11 19 90
the target sum:  21

a solution:  zero─based indices    [1, 3]         2  +  19  =  21

number of solutions found:  1
output   when using the input of:     21     -78 -5 1 0 -1 -4 11 14 23.5 5 +3 2. 18 -2.50 +2 16 19 018 23 24 25 26 199 2 3 17 +18 19 03 3 .18e2
the list:        -78 -5 1 0 -1 -4 11 14 23.5 5 +3 2. 18 -2.50 +2 16 19 018 23 24 25 26 199 2 3 17 +18 19 03 3 .18e2
the target sum:  21

a solution:  zero─based indices    [ 1, 21]                    -5  +     26  =  21
a solution:  zero─based indices    [ 5, 20]                    -4  +     25  =  21
a solution:  zero─based indices    [ 8, 13]                  23.5  +  -2.50  =  21
a solution:  zero─based indices    [ 9, 15]                     5  +     16  =  21
a solution:  zero─based indices    [10, 12]                    +3  +     18  =  21
a solution:  zero─based indices    [10, 17]                    +3  +    018  =  21
a solution:  zero─based indices    [10, 26]                    +3  +    +18  =  21
a solution:  zero─based indices    [10, 30]                    +3  +  .18e2  =  21
a solution:  zero─based indices    [11, 16]                    2.  +     19  =  21
a solution:  zero─based indices    [11, 27]                    2.  +     19  =  21
a solution:  zero─based indices    [12, 24]                    18  +      3  =  21
a solution:  zero─based indices    [12, 28]                    18  +     03  =  21
a solution:  zero─based indices    [12, 29]                    18  +      3  =  21
a solution:  zero─based indices    [14, 16]                    +2  +     19  =  21
a solution:  zero─based indices    [14, 27]                    +2  +     19  =  21
a solution:  zero─based indices    [16, 23]                    19  +      2  =  21
a solution:  zero─based indices    [17, 24]                   018  +      3  =  21
a solution:  zero─based indices    [17, 28]                   018  +     03  =  21
a solution:  zero─based indices    [17, 29]                   018  +      3  =  21
a solution:  zero─based indices    [23, 27]                     2  +     19  =  21
a solution:  zero─based indices    [24, 26]                     3  +    +18  =  21
a solution:  zero─based indices    [24, 30]                     3  +  .18e2  =  21
a solution:  zero─based indices    [26, 28]                   +18  +     03  =  21
a solution:  zero─based indices    [26, 29]                   +18  +      3  =  21
a solution:  zero─based indices    [28, 30]                    03  +  .18e2  =  21
a solution:  zero─based indices    [29, 30]                     3  +  .18e2  =  21

number of solutions found:  26

Ring

# Project : Two Sum

numbers = [0, 2, 11, 19, 90]
sum = 21

see "order list: "
for n=1 to len(numbers)
see " " + numbers[n]
next
see " (using a zero index.)" + nl
for n=1 to len(numbers)
for m=n to len(numbers)
if numbers[n] + numbers[m] = sum
see "target sum: " + sum + nl
see "a solution: ["
see "" + (n-1) + " " + (m-1) + "]" + nl
ok
next
next

Output:

order list:  0 2 11 19 90 (using a zero index.)
target sum:  21
a solution: [1 3]

Ruby

def two_sum(numbers, sum)
numbers.each_with_index do |x,i|
if j = numbers.index(sum - x) then return [i,j] end
end
[]
end

numbers = [0, 2, 11, 19, 90]
p two_sum(numbers, 21)
p two_sum(numbers, 25)
Output:
[1, 3]
[]

When the size of the Array is bigger, the following is more suitable.

def two_sum(numbers, sum)
numbers.each_with_index do |x,i|
key = sum - x
if j = numbers.bsearch_index{|y| key<=>y}
return [i,j]
end
end
[]
end

Rust

use std::cmp::Ordering;

fn two_sum<T>(arr: &[T], sum: T) -> Option<(usize, usize)>
where
T: Add<Output = T> + Ord + Copy,
{
if arr.len() == 0 {
return None;
}

let mut i = 0;
let mut j = arr.len() - 1;

while i < j {
match (arr[i] + arr[j]).cmp(&sum) {
Ordering::Equal => return Some((i, j)),
Ordering::Less => i += 1,
Ordering::Greater => j -= 1,
}
}

None
}

fn main() {
let arr = [0, 2, 11, 19, 90];
let sum = 21;

println!("{:?}", two_sum(&arr, sum));
}
Output:
Some((1, 3))

Scala

import java.util

object TwoSum extends App {
val (sum, arr)= (21, Array(0, 2, 11, 19, 90))
println(util.Arrays.toString(twoSum(arr, sum)))

private def twoSum(a: Array[Int], target: Long): Array[Int] = {
var (i, j) = (0, a.length - 1)
while (i < j) {
val sum = a(i) + a(j)
if (sum == target) return Array[Int](i, j)
if (sum < target) i += 1 else j -= 1
}
null
}

}
Output:
See it running in your browser by ScalaFiddle (JavaScript, non JVM) or by Scastie (JVM).

Sidef

Translation of: Raku
func two_sum(numbers, sum) {
var (i, j) = (0, numbers.end)
while (i < j) {
given (sum <=> numbers[i]+numbers[j]) {
when (-1) { --j }
when (+1) { ++i }
default { return [i, j] }
}
}
return []
}

say two_sum([0, 2, 11, 19, 90], 21)
say two_sum([0, 2, 11, 19, 90], 25)
Output:
[1, 3]
[]

Stata

Notice that array indexes start at 1 in Stata.

function find(a, x) {
i = 1
j = length(a)
while (i<j) {
s = a[i]+a[j]
if (s<x) i++
else if (s>x) j--
else return((i,j))
}
}

find((0,2,11,19,90),21)
1 2
+---------+
1 | 2 4 |
+---------+

Vala

void main() {
int arr[] = { 0, 2, 11, 19, 90 }, sum = 21, i, j, check = 0;

for (i = 0; i < 4; i++) {
for ( j = i+1; j < 5; j++) {
if (arr[i] + arr[j] == sum) {
print("[%d,%d]",i,j);
check = 1;
break;
}
}
}
if (check == 0)
print("[]");
}
Output:
[1,3]

VBA

Option Explicit
Function two_sum(a As Variant, t As Integer) As Variant
Dim i, j As Integer
i = 0
j = UBound(a)
Do While (i < j)
If (a(i) + a(j) = t) Then
two_sum = Array(i, j)
Exit Function
ElseIf (a(i) + a(j) < t) Then i = i + 1
ElseIf (a(i) + a(j) > t) Then j = j - 1
End If
Loop
two_sum = Array()
End Function
Sub prnt(a As Variant)
If UBound(a) = 1 Then
Selection.TypeText Text:="(" & a(0) & ", " & a(1) & ")" & vbCrLf
Else
Selection.TypeText Text:="()" & vbCrLf
End If
End Sub
Sub main()
Call prnt(two_sum(Array(0, 2, 11, 19, 90), 21))
Call prnt(two_sum(Array(-8, -2, 0, 1, 5, 8, 11), 3))
Call prnt(two_sum(Array(-3, -2, 0, 1, 5, 8, 11), 17))
Call prnt(two_sum(Array(-8, -2, -1, 1, 5, 9, 11), 0))
End Sub
Output:
(1, 3)
(0, 6)
()
(2, 3)

Visual Basic .NET

Translation of: C#
Module Module1

Function TwoSum(numbers As Integer(), sum As Integer) As Integer()
Dim map As New Dictionary(Of Integer, Integer)
For index = 1 To numbers.Length
Dim i = index - 1
' see if the complement is stored
Dim key = sum - numbers(i)
If map.ContainsKey(key) Then
Return {map(key), i}
End If
Next
Return Nothing
End Function

Sub Main()
Dim arr = {0, 2, 1, 19, 90}
Const sum = 21

Dim ts = TwoSum(arr, sum)
Console.WriteLine(If(IsNothing(ts), "no result", \$"{ts(0)}, {ts(1)}"))
End Sub

End Module
Output:
1, 3

X86 Assembly

Translation of: Python
Works with: nasm

section .data
outputArr dd 0,0,0
inputArr dd 5,0,2,11,19,90

section .text
global _main
_main:
mov ebp, esp
mov eax, 21 ;num we search for
push inputArr
call func
ret

func:
mov esi, [ebp - 4];get arr address from stack
add esi, 4 ;esi now points to the first element instead of the length
mov edx, 0 ;i
mov ecx, [esi - 4] ;j
dec ecx ;counting starts from 0
looping:
cmp edx, ecx ;while i < j
jge return
mov ebx, [esi + edx * 4]
add ebx, [esi + ecx * 4] ;inputArr[i] + inputArr[j]
cmp ebx, eax ;inputArr[i] + inputArr[j] (==|<|else) eax
je end ;==
jl i ;<
dec ecx ;else j--
jmp looping
i:
inc edx ;i++
jmp looping
end:
mov eax, 2 ;if we find a combination our array has a length of 2
mov [outputArr], eax ;length is in the first 4 byte cell
mov [outputArr + 4], edx ;i
mov [outputArr + 8], ecx ;j
return:
mov eax, outputArr ;address of outputArr is returned in eax
ret

Wren

var twosum = Fn.new { |a, n|
var c = a.count
if (c < 2) return []
for (i in 0...c-1) {
for (j in i+1...c) {
var s = a[i] + a[j]
if (s == n) return [i, j]
if (s > n) break
}
}
return []
}

var a = [0, 2, 11, 19, 90]
System.print("Numbers: %(a)\n")
for (n in [21, 25, 90]) {
var pair = twosum.call(a, n)
if (pair.count == 2) {
System.print("Indices: %(pair) sum to %(n) (%(a[pair]) + %(a[pair]) = %(n))")
} else {
System.print("No pairs of the above numbers sum to %(n).")
}
System.print()
}
Output:
Numbers: [0, 2, 11, 19, 90]

Indices: [1, 3] sum to 21 (2 + 19 = 21)

No pairs of the above numbers sum to 25.

Indices: [0, 4] sum to 90 (0 + 90 = 90)

zkl

The sorted O(n) no external storage solution:

fcn twoSum(sum,ns){
i,j:=0,ns.len()-1;
while(i<j){
if((s:=ns[i] + ns[j]) == sum) return(i,j);
else if(s<sum) i+=1;
else if(s>sum) j-=1;
}
}
twoSum2(21,T(0,2,11,19,90)).println();
twoSum2(25,T(0,2,11,19,90)).println();
Output:
L(1,3)
False

The unsorted O(n!) all solutions solution:

fcn twoSum2(sum,ns){
Utils.Helpers.combosKW(2,ns).filter('wrap([(a,b)]){ a+b==sum }) // lazy combos
.apply('wrap([(a,b)]){ return(ns.index(a),ns.index(b)) })
}
twoSum2(21,T(0,2,11,19,90,21)).println();
twoSum2(25,T(0,2,11,19,90,21)).println();
Output:
L(L(0,5),L(1,3))
L()