Two sum
- 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[1] + numbers[3] = 2 + 19 = 21,
return [1, 3].
- Source
Stack Overflow: Find pair of numbers in array that add to given sum
11l
<lang 11l>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))</lang>
- Output:
[1, 3] []
Action!
<lang 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</lang>
- Output:
Screenshot from Atari 8-bit computer
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
<lang 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;
}
}</lang>
- Output:
1 3
ALGOL 68
<lang algol68># 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;
WHILE i < j AND NOT found DO
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] #</lang>
- 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. 0.1@ 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 (⊃). <lang APL> ⎕io ← 0 ⍝ sets index origin to 0 ts ← {⊃⍸ ⍵= 0.1@(⍸∘.=⍨⍳≢⍺) ∘.+⍨ ⍺} ⎕ ← 0 2 11 19 90 ts 21 ⍝ should be 1 3 </lang>
- Output:
1 3
AppleScript
Functional
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 :-)
<lang AppleScript>-------------------------- 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
Template: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]) --> Template: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</lang>
- Output:
<lang AppleScript>Template:1, 3</lang>
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.
<lang applescript>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.</lang>
- Output:
<lang applescript>Template:2, 4 {} {{1, 11}, {2, 8}, {2, 9}, {2, 10}, {3, 8}, {3, 9}, {3, 10}, {4, 7}, {5, 6}}</lang>
AutoHotkey
<lang 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-- } return "not found" }</lang> Examples:<lang AutoHotkey>MsgBox % TwoSum([0, 2, 11, 19, 90], 21) ; returns 2, 4 (first index is 1 not 0)</lang>
Outputs:
2,4
AWK
<lang 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("[]")
} </lang>
- Output:
[2,4] []
Befunge
<lang befunge>>000pv
>&:0\`#v_00g:1+00p6p
v >$&50p110p020p v>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>+50g-!#v_48*10g8p10g1+:00g1-`v > v >10g20g..@ v_v """""""""""""""""""""""""""""""""""""""""""""""v ">"p4\"v"p02:+1p4\">":g02$< :
> 20g8p20g1+:00g1-`#v_0 ^1
""""""""""""""""""""""""""""""""""""""""""""""" " 0 >^ l p
i "
a ^
F "
" \
>:#,_@ 8
p
> ^</lang>
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
<lang C>
- include<stdio.h>
int main() { int arr[5] = {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; } </lang> Output :
[1,3]
C#
<lang csharp>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[0]}, {ts[1]}" : "no result");
Console.ReadLine(); }
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 };
}
map.Add(numbers[i], i);
}
return null;
}
} </lang>
- Output:
1, 3
C++
<lang cpp>#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; }</lang>
- Output:
{1,3}
D
<lang 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[0]] + arr[result[1]] == 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 [];
}</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>
Delphi
<lang Delphi> 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, ')');
Readln;
end.</lang>
- Output:
(1,3)
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] []
F#
<lang fsharp> // 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) </lang>
- Output:
[1; 3]
Factor
<lang 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 ] bi@ + :> 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</lang>
- Output:
{ 1 3 }
{ }
{ 0 2 }
Forth
<lang forth>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[] + R@
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</lang>
- Output:
[1, 3] [] [0, 6] [2, 5]
Fortran
<lang 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 </lang>
- Output:
( 2 4 ): 2+ 19= 21
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
Go
<lang go>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)
}
}</lang>
- Output:
The numbers with indices 1 and 3 sum to 21
Haskell
Returning first match
<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]
Returning all matches
Listing multiple solutions (as zero-based indices) where they exist:
<lang haskell>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]</lang>
Or, resugaring a little – pulling more into the scope of the list comprehension: <lang Haskell>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]</lang>
- Output:
(1,3) (2,5)
Icon and Unicon
Icon and Unicon are ordinal languages, first index is one.
fullimag library used to pretty print lists.
<lang unicon>#
- twosum.icn, find two array elements that add up to a given sum
- Dedicated to the public domain
link fullimag 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</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.)
<lang J>two_sum=:dyad define
sums=. +/~ y
matches=. x = sums
sum_inds=. I. , matches
pair_inds=. ($matches) #: sum_inds
; {. a: ,~ <"1 pair_inds
)</lang>
And, testing:
<lang J> 21 two_sum 0 2 11 19 90 1 3</lang>
Or, we could go slightly more traditional and instead of doing that boxing at the end, use an if/else statement:
<lang J>two_sum=:dyad define
sums=. +/~ y
matches=. x = sums
sum_inds=. I. , matches
pair_inds=. ($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
<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
Composing a solution from generic functions like zip, bind (>>=, or flip concatMap) etc.
<lang JavaScript>(() => {
'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[0], null, x[1]] : 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])
);
})();</lang>
- Output:
[[1,3],[2,5]]
Jsish
Based on Javascript entry. <lang javascript>/* 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)[0][0]];
- list[twoSum(21, list)[0][1]];</lang>
- Output:
prompt$ jsish --U twoSum.jsi list ==> [ 0, 2, 11, 19, 90 ] twoSum(21, list) ==> [ [ 1, 3 ] ] list[twoSum(21, list)[0][0]] ==> 2 list[twoSum(21, list)[0][1]] ==> 19
jq
Works with gojq, the Go implementation of jq.
<lang jq>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)) </lang>
- Output:
[1,3] []
Julia
<lang julia>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)</lang>
- Output:
twosum([0, 2, 11, 19, 90], 21) = [2, 4]
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 $targetSum")
}
}</lang>
- Output:
The numbers with indices 1 and 3 sum to 21
Liberty BASIC
<lang liberty basic>myArray(0) = 0 myArray(1) = 2 myArray(2) = 11 myArray(3) = 19 myArray(4) = 90
sum = 21
Print twoToSum$("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</lang>
- Output:
[1,3]
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
Maple
<lang 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);</lang>
- Output:
Note that Maple does 1 based indexing.
[2,5]
Mathematica / Wolfram Language
<lang Mathematica>ttwoSum[data_List, sum_] :=
Block[{indices = Subsets[Range@Length@data, {2}]},
Cases[indices, _?(Total@data# == sum &)]]
twoSum[{0, 2, 11, 19, 90}, 21] // TableForm</lang>
- Output:
2 4
MiniScript
<lang 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)</lang>
Output:
[1, 3]
Modula-2
<lang modula2>MODULE TwoSum; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,ReadChar;
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]:=0; arr[1]:=2; arr[2]:=11; arr[3]:=19; arr[4]:=90;
res := TwoSum(arr, 21);
FormatString("[%i, %i]\n", buf, res.f, res.s);
WriteString(buf);
ReadChar;
END TwoSum.</lang>
Nim
<lang 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[0] = base
result[1] = 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 = [1] 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()</lang>
Objeck
<lang objeck>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[2];
r[0] := i;
r[1] := 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();
}
} </lang>
Output:
[1,3]
OCaml
<lang ocaml>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</lang>
Will return all possible sums, not just the first one found.
- Output:
$ ocaml two_sum.ml # Found: 1 3
ooRexx
<lang 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' </lang>
- 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 ). <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
//Goto is bad, but fast...
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);
//runtime about 1 second
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
<lang perl>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';</lang>
- 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)
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
<lang 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 ?</lang>
- Output:
[2, 19] [] === Press any key to exit ===
PicoLisp
<lang 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) )</lang>
- Output:
(2 . 4) NIL (3 . 4)
PowerShell
Lazy, very lazy. <lang PowerShell> $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) }
}
</lang>
- Output:
Sum Indices
--- -------
21 {1, 3}
Python
<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>
or, in terms of itertools.product:
<lang python>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[0]
- snd :: (a, b) -> b
def snd(tpl):
Second member of a pair. return tpl[1]
- 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()</lang>
- 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:
<lang python>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[0]] if n == sum(vs[1]) else []
ixs = list(enumerate(xs))
return list(
bind(ixs)(
lambda ix: bind(ixs[ix[0]:])(
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):
List monad injection operator.
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()</lang>
- 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.
<lang Quackery> [ 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</lang>
- Output:
[ 1 3 ] [ ] [ 2 3 ]
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>
Raku
(formerly Perl 6)
Procedural
<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
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).
<lang perl6>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, $_);
}</lang>
- Output:
(1 3) (1 3) () ()
REXX
version 1
<lang rexx>/* 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'</lang>
- 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. <lang rexx>/*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. */</lang>
- 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
<lang 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 </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>
Rust
<lang Rust>use std::cmp::Ordering; use std::ops::Add;
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));
}</lang>
- Output:
Some((1, 3))
Scala
<lang 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
}
}</lang>
- Output:
See it running in your browser by ScalaFiddle (JavaScript, non JVM) or by Scastie (JVM).
Sidef
<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] []
Stata
Notice that array indexes start at 1 in Stata. <lang 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 | +---------+</lang>
Vala
<lang 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("[]");
}</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)
Visual Basic .NET
<lang vbnet>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
map.Add(numbers(i), i)
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</lang>
- Output:
1, 3
X86 Assembly
<lang asm> 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 add esp, 4 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
</lang>
Wren
<lang ecmascript>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[0]]) + %(a[pair[1]]) = %(n))")
} else {
System.print("No pairs of the above numbers sum to %(n).")
}
System.print()
}</lang>
- 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: <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()
- Draft Programming Tasks
- Arithmetic operations
- Clarified and Needing Review
- 11l
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