# 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.

Given a sorted array of single positive integers, 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.

Example

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

Source

## ALGOL 68

Translation of: Lua

<lang algol68># returns the subscripts of a pair of numbers in a that sum to sum, a is assumed to be sorted #

1. 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 # ;
```
1. 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] #</lang>

Output:
```[1, 3]
[0, 6]
[]
[2, 3]
```

## AppleScript

Translation of: JavaScript

Nesting concatMap yields the cartesian product of the list with itself, and functions passed to map have access to the (1-based) 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.

<lang AppleScript>-- summingPairIndices :: Int -> [Int] -> Int on summingPairIndices(n, xs)

```   script
-- productFilter :: [a] -> [b] -> a, b
on productFilter(xs, ys)

script product
on lambda(x, ix)

script filter
on lambda(y, iy)

if iy ≤ ix then
{} -- ignore - mirror-image pairs
else
if n = (x + y) then
-- Solution found, and
-- AppleScript indices are 1-based
Template:Ix - 1, iy - 1
else
{} -- eliminated from map by concat
end if
end if

end lambda
end script

concatMap(filter, ys)
end lambda
end script

concatMap(product, xs)
end productFilter
end script

result's productFilter(xs, xs)
```

end summingPairIndices

-- TEST on run

```   summingPairIndices(21, [0, 2, 11, 19, 90])

--> Template:1, 3

```

end run

-- GENERIC LIBRARY FUNCTIONS

-- concatMap :: (a -> [b]) -> [a] -> [b] on concatMap(f, xs)

```   script append
on lambda(a, b)
a & b
end lambda
end script

foldl(append, {}, map(f, xs))
```

end concatMap

-- 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 lambda(v, item i of xs, i, xs)
end repeat
return v
end tell
```

end foldl

-- 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 lambda(item i of xs, i, xs)
end repeat
return lst
end tell
```

end map

-- 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 lambda : f
end script
end if
```

end mReturn</lang>

Output:

<lang AppleScript>Template:1, 3</lang>

## AWK

<lang AWK>

1. 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("[]")
```

} </lang>

Output:
```[2,4]
[]
```

## C#

<lang csharp>public static int[] TwoSum(int[] numbers, int sum) {

```   var map = new Dictionary<int, int>();
for (var i = 0; i < numbers.Length; i++)
{
var key = sum - numbers[i];
if (map.ContainsKey(key))
return new[] { map[key], i };
}
return Array.Empty<int>();
```

}</lang>

Output:
`[1,3]`

## Dart

<lang> 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");
```

} }

</lang>

## Elixir

<lang 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)</lang>

Output:
```[1, 3]
[]
```

## FreeBASIC

<lang 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</lang>

Output:
```The numbers with indices 1 and 3 sum to 21
```

<lang Haskell> 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] </lang>

Output:
```[1,3]
```

Or, using concatMap, sequence (for a cartesian product) and nubBy, and listing multiple solutions when they exist:

summingPairIndices :: Int -> [Int] -> Int summingPairIndices n xs =

``` (join . (flip elemIndices xs <\$>)) <\$>
nubBy
(\[x, y] zs -> and (flip elem zs <\$> [x, y]))
(concatMap
(\[x, y] ->
[ [x, y] | x /= y && x + y == n ]) \$
sequence [xs, xs])

```

main :: IO () main = print (summingPairIndices 21 [0, 2, 11, 19, 90, 10])</lang>

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.

<lang unicon>#

1. twosum.icn, find two array elements that add up to a given sum
2. Dedicated to the public domain

```   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

1. 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</lang>

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: <lang J> =+/~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</lang>

And, check if any of them are our desired value: <lang J> 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</lang>

Except, we want indices here, so let's toss the structure so we can get those: <lang J> ,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</lang>

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:

<lang J> \$21=+/~0 2 11 19 90 5 5

```  5 5#:I.,21=+/~0 2 11 19 90
```

1 3 3 1</lang>

Or, taking advantage of being able to use verbs to represent combining their results, when we use three of them: <lang J> (\$ #: I.@,)21=+/~0 2 11 19 90 1 3 3 1</lang>

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:

<lang J> (\$ <@#: I.@,)21=+/~0 2 11 19 90 ┌───┬───┐ │1 3│3 1│ └───┴───┘

```  a:,~(\$ <@#: I.@,)21=+/~0 2 11 19 90
```

┌───┬───┬┐ │1 3│3 1││ └───┴───┴┘

```  {.a:,~(\$ <@#: I.@,)21=+/~0 2 11 19 90
```

┌───┐ │1 3│ └───┘

```  ;{.a:,~(\$ <@#: I.@,)21=+/~0 2 11 19 90
```

1 3</lang>

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

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

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

1 3</lang>

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

And, let's finish the job, give this a name, and test it out: <lang J> twosum=: ;@{.@(a:,~ (\$ <@#: I.@,)@([ = +/~@]))

```  21 twosum 0 2 11 19 90
```

1 3</lang>

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
```

)</lang>

And, testing:

<lang J> 21 two_sum 0 2 11 19 90 1 3</lang>

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.
```

)</lang>

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

<lang J>twosum=: ;@{.@(a:,~ (\$ <@#: I.@,)@([ = +/~@]))</lang>

Alternative approach

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

<lang j> 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</lang> The array is symmetrical so we can zero one half to remove duplicate pairs and then retrieve the remaining indicies using sparse array functionality. <lang j>zeroLowerTri=: * [: </~ i.@# getIdx=: 4 \$. \$. twosum_alt=: getIdx@zeroLowerTri@(= +/~)</lang>

Testing ... <lang j> 21 twosum_alt 0 2 11 19 90 1 3</lang>

## Java

Translation of: Lua

<lang java>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;
}
```

}</lang>

`[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.

<lang JavaScript>(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]);
```

})(); </lang>

Output:

<lang JavaScript>1,3</lang>

### ES6

First quickly composing a function from generic primitives like concatMap, cartesianProduct, nubBy.

<lang JavaScript>(() => {

```   'use strict';
```
```   let summingPairIndices = (n, xs) => nubBy(
([x, y], [x1, y1]) => x === x1 && y === y1,
concatMap(
([x, y]) =>
x === y ? [] : (x + y === n ? x, y : []),
cartesianProduct(xs, xs)
).map(xs => xs.sort())
).map(([x, y]) => [xs.indexOf(y), xs.indexOf(x)]);
```

```   // GENERIC LIBRARY FUNCTIONS
```
```   // concatMap :: (a -> [b]) -> [a] -> [b]
let concatMap = (f, xs) => [].concat.apply([], xs.map(f)),
```
```       // cartesianProduct :: [a] -> [b] -> [(a, b)]
cartesianProduct = (xs, ys) =>
concatMap(x => concatMap(y => x, y, ys), xs),
```
```       // nubBy :: (a -> a -> Bool) -> [a] -> [a]
nubBy = (p, xs) => {
let x = xs.length ? xs[0] : undefined;
```
```           return x !== undefined ? [x].concat(
nubBy(p, xs.slice(1).filter(y => !p(x, y)))
) : [];
};
```

```   return summingPairIndices(21, [0, 2, 11, 19, 90]);
```

})(); </lang>

Output:

<lang JavaScript>[1,3]</lang>

and then fusing it down to something smaller and a little more efficient, expressed purely in terms of concatMap

<lang JavaScript>(() => {

```   'use strict';
```
```   // concatMap :: (a -> [b]) -> [a] -> [b]
let concatMap = (f, xs) => [].concat.apply([], xs.map(f));
```
```   // summingPairIndices :: -> Int -> [Int] -> [(Int, Int)]
let summingPairIndices = (n, xs) =>
```
```           // Javascript map functions have access to the array index
// in their second argument.
concatMap((x, ix) => concatMap((y, iy) =>
iy <= ix ? [] : (//  Ignoring mirror-image and duplicate tuples by
//  skipping the 'lower triangle' (y <= x)
//  of the cartesian product grid
x + y === n ? [
[ix, iy]
] : []
), xs), xs);
```
```   return summingPairIndices(21, [0, 2, 11, 19, 90]);
```

})(); </lang>

Output:

<lang JavaScript>1,3</lang>

## Kotlin

<lang scala>// 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")
}
```

}</lang>

Output:
```The numbers with indices 1 and 3 sum to 21
```

## Lua

Lua uses one-based indexing. <lang lua>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), ","))</lang>

Output:
`2,4`

## ooRexx

<lang oorexx>a=.array~of(0, 2, 11, 19, 90) x=21 do i=1 To a~items

``` If a[i]>x Then Leave
Do j=i+1 To a~items
s=a[i]
s+=a[j]
Select
When s=x Then Leave i
When s>x Then Leave j
Otherwise Nop
End
End
End
```

If s=x Then Do

``` i-=1            /* array a's index starts with 1, so adjust */
j-=1
Say '['i','j']'
End
```

Else

``` Say '[] - no items found'</lang>
```
Output:
`[1,3]`

## 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 ). <lang pascal>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.</lang>

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

<lang perl> sub two_sum{

``` my @arr = @{shift @_};
my \$num = shift;
my \$i = 0;
my \$j = \$#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;
}
}
```

}

my @numbers = (0, 2, 11, 19, 90); my (\$n1, \$n2) = two_sum(\@numbers, 21); print "\$n1 \$n2\n"; (\$n1, \$n2) = two_sum(\@numbers, 25); print "\$n1 \$n2\n";

</lang>

## Perl 6

Translation of: zkl

<lang perl6>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;</lang>

Output:
```(1 3)
Nil```

## Phix

<lang 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)</lang>

Translation of: Perl 6

<lang Phix>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</lang>

Output:

Phix uses 1-based indexes

```{2,4}
```

## Python

Translation of: Perl 6

<lang python> 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))

</lang>

## Racket

<lang 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)))</lang>
```

## REXX

### version 1

<lang rexx>list=0 2 11 19 90 Do i=0 By 1 Until list=

``` Parse Var list a.i list
End
```

n=i-1 x=21 do i=1 To n

``` If a.i>x Then Leave
Do j=i+1 To n
s=a.i
s=s+a.j
Select
When s=x Then Leave i
When s>x Then Leave j
Otherwise Nop
End
End
End
```

If s=x Then

``` Say '['i','j']'
```

Else

``` Say '[] - no items found'</lang>
```
Output:
`[1,3]`

### version 2

This REXX version uses optimization in using integers that take advantage that the numbers are an ordered list of positive integers.

All solutions are listed (if any).

This REXX solution makes allowances for the inclusion of zero, even though it is
specifically excluded by the task's description, however zero is included in the

If the list isn't an ordered list of positive integers, then the   while   clause (below) should be removed,   and then this REXX version will work with any list   (including negative numbers and also with any real number. <lang rexx>/*REXX pgm find 2 integers in a ordered list of positive integers that sum to a number. */ list=0 2 11 19 90 ; say 'order list: ' list " (using a zero index.)" sum=21  ; say 'target sum: ' sum

```             do #=0  for words(list)  while word(list, #+1)<=sum  /*only use possibles.*/
@.#=word(list, #+1)
end  /*#*/
```

say /* [↓] look for sum of 2 int = the sum*/

```             do   a=0   for #                   /*scan up to the last possible integer.*/
do b=a+1  to #-1;  if @.a+@.b==sum  then say 'a solution:  ['a","  b']'
end   /*y*/                      /* [↑]  2 elements sum to target?  Tell*/
end     /*x*/                      /*stick a fork in it,  we're all done. */</lang>
```

output

```order list:  0 2 11 19 90           (using a zero index.)
target sum:  21

a solution:  [1, 3]
```

## Ruby

<lang 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)</lang>

Output:
```[1, 3]
[]
```

When the size of the Array is bigger, the following is more suitable. <lang ruby>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</lang>

## Sidef

Translation of: Perl 6

<lang ruby>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)</lang>

Output:
```[1, 3]
[]
```

## VBA

<lang vb>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</lang>

Output:
```(1, 3)
(0, 6)
()
(2, 3)

```

## zkl

The sorted O(n) no external storage solution: <lang zkl>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;
}
```

}</lang> <lang zkl>twoSum2(21,T(0,2,11,19,90)).println(); twoSum2(25,T(0,2,11,19,90)).println();</lang>

Output:
```L(1,3)
False
```

The unsorted O(n!) all solutions solution: <lang zkl>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)) })
```

}</lang> <lang zkl>twoSum2(21,T(0,2,11,19,90,21)).println(); twoSum2(25,T(0,2,11,19,90,21)).println();</lang>

Output:
```L(L(0,5),L(1,3))
L()
```