Does the sum of 2 numbers in the list equal the desired sum?

Question

The Task

This is my first challenge so apologies if it is very simple! In this challenge, your task is to write a program or function which takes in a list of integers and a desired sum and outputs a truthy or falsey value based on whether the list contains two numbers that can be summed together to achieve the desired sum.

Input

The input must be acquired in a standard way for your language. The list of integers can be any standard collection type for your language but cannot be a hashed set, sorted set, or any highly functional collection type. Don't assume the input is sorted, any valid integer type is acceptable.

The desired sum is also an integer and must be provided to your program as input.

Output

The output may be a function return, console log, writing to a file, etc... Any valid Truthy or Falsey value is acceptable.

Additional Rules

• The input list may or may not be sorted.
• Duplicate values may occur in the list
• The same item in the list may not be summed to itself to achieve the desired sum (Eg. [3,4,5] with desired sum 6 will expect False because 3 cannot be used twice)
• Standard loopholes are not allowed.

Test Cases

INPUT                 OUTPUT
[3,4,5], 6            Falsey
[1,2,3,4,5,6],11      Truthy
[4,23,8174,42,-1],41  Truthy
[1,1,1,1,1,3],3       Falsey
[2,1,1,1,53,2],4      Truthy
[2,2,2,2],4           Truthy
[3],3                 Falsey
[],-43                Falsey

This is code-golf, so the shortest code in bytes wins!

Answer 1

Brachylog, 3 bytes

⊇Ċ+

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⊇Ċ+   input as STDIN, output as ARG1
      (prove that:)
⊇     the input is a superset of
 Ċ       a list of two elements
  +         which sums up to the output

Answer 2

Jelly, 7 6 5 bytes

ŒcS€ċ

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-1 byte by using the Unordered Pairs atom

ŒcS€ċ  Main Link; left argument is the array, right argument is the required sum
  S€   Take the sum of each...
Œc     ...unordered pair...
    ċ  Check if the sum array contains the right argument

-1 byte thanks to Erik!

Answer 3

05AB1E, 6 bytes

æ2ùOIå

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Explanation

æ        # get powerset of first input
 2ù      # keep only those of size 2
   O     # sum each
    Iå   # check if 2nd input is in the list of sums

Answer 4

Python 2, 43 42 bytes

^{-1 byte thanks to mbomb007}

lambda a,b:any(b-a.pop()in a for x in a*1)

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Answer 5

Mathematica, 29 bytes

!FreeQ[Tr/@#~Subsets~{2},#2]&

Answer 6

MATL, 6 bytes

2XN!sm

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Explanation

2XN   % Implicitly input an array. Combinations of elements taken 2 at a time
      % The result is a matrix where each row is a combination
!s    % Sum of each row
m     % Implicitly input a number. Ismember function. Implicitly display

Answer 7

JavaScript (ES6), 38 43 bytes

a=>n=>[...a].some(_=>a.includes(n-a.pop()))

---

Try It

f=
a=>n=>[...a].some(_=>a.includes(n-a.pop()))
o.innerText=`           f([3,4,5])(6) = ${f([3,4,5])(6)}
    f([1,2,3,4,5,6])(11) = ${f([1,2,3,4,5,6])(11)}
f([4,23,8174,42,-1])(41) = ${f([4,23,8174,42,-1])(41)}
     f([1,1,1,1,1,3])(3) = ${f([1,1,1,1,1,3])(3)}
    f([2,1,1,1,53,2])(4) = ${f([2,1,1,1,53,2])(4)}
         f([2,2,2,2])(4) = ${f([2,2,2,2])(4)}
               f([3])(3) = ${f([3])(3)}
              f([])(-43) = ${f([])(-43)}
     f([1,2,3,4,5,6])(3) = ${f([1,2,3,4,5,6])(3)}`

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Answer 8

Haskell, 39 37 36 bytes

(a:t)#n=n`elem`map(a+)t||t#n
_#n=1<0

Try it online! Example usage: [1,3,2,5] # 3. Returns True or False.

---

Alternative (also 36 bytes)

f(a:t)=map(a+)t++f t
f e=e
(.f).elem

Answer 9

C, 114 113 bytes

i,j,k="F";main(c,v)char**v;{for(--c;++i<c;)for(j=0;++j<c;)k=j^i&atoi(v[i])+atoi(v[j])==atoi(v[c])?"T":k;puts(k);}

Here's the output

C:\eng\golf>a.exe 3 4 5 6
F
C:\eng\golf>a.exe 1 2 3 4 5 6 11
T
C:\eng\golf>a.exe 4 23 8174 42 -1 41
T
C:\eng\golf>a.exe 1 1 1 1 1 3
F
C:\eng\golf>a.exe 2 1 1 1 53 2 4
T
C:\eng\golf>a.exe 2 2 2 2 4
T

Answer 10

Ruby, 59 53 37 bytes

inspiration comes from this answer

->a,b{a.map{|e|a.include?(b-e)}.any?}

will now look where can I chop a few bytes -))

HISTORY

#53 bytes
->a,b{a.combination(2).map{|e|e.reduce(&:+)==b}.any?}

#59 bytes 
->a,b{a.combination(2).map{|e|e.reduce(&:+)}.any?{|x|x==b}}

Answer 11

PHP (>= 7.1), 53 51 50 46 Bytes

<?foreach($_GET[0]as$c)+$$c||${$_GET[1]-$c}=a;

sadly it's quite big but at least better than 2 loops. +a fails with a warning if the sum can be made => non empty output => truthy. Outputs nothing if the sum can not be made => falsey.

Thanks @user63956 !

Answer 12

TAESGL, 8 bytes

Ş⇒AĨB-AĖ

Interpreter

Explanation

Ş⇒        Array.some with anonymous function on implicit input
  AĨ      first input includes
    B-    second input minus
      AĖ  first input popped

Answer 13

Clojure, 53 bytes

#(loop[[v & r]%](if v(or((set r)(- %2 v))(recur r))))

Uses destructuring to take the first and rest of the input list (and subsequent recur of rest), and returns the number "target - value1" if found in the set of remaining numbers (a "truthy"), nil otherwise.

Answer 14

Java 7, 138 bytes

boolean c(int[]a,int s){String r="";for(int l=a.length,i=0,j;i<l;i++)for(j=-1;++j<l;)if(i!=j)r+=(a[i]+a[j])+",";return r.contains(s+",");}

Explanation:

boolean c(int[]a,int s){             // Method with integer-array and integer parameters and boolean return-type
  String r="";                       // Temp String
  for(int l=a.length,i=0,j;i<l;i++)  //  Loop (1) over the input array
    for(j=-1;++j<l;)                 //   Inner loop (2) over the input array again
      if(i!=j)                       //    If the index of both loops aren't equal (so not the same item in the list)
        r+=(a[i]+a[j])+",";          //     Append the sum with a comma to the temp-String
                                     //   End of loop (2) (implicit / single-line body)
                                     //  End of loop (1) (implicit / single-line body)
  return r.contains(s+",");          //  Return whether the temp-String contains the given sum (+ comma)
}                                    // End of method

Test code:

Try it here.

class M{
  static boolean c(int[]a,int s){String r="";for(int l=a.length,i=0,j;i<l;i++)for(j=-1;++j<l;)if(i!=j)r+=(a[i]+a[j])+",";return r.contains(s+",");}

  public static void main(String[] a){
    System.out.println(c(new int[]{1,2,3,4,5,6},11));
    System.out.println(c(new int[]{4,23,8174,42,-1},41));
    System.out.println(c(new int[]{2,1,1,1,53,2},4));
    System.out.println(c(new int[]{2,2,2,2},4));
    System.out.println();
    System.out.println(c(new int[]{1,1,1,1,1,3},3));
    System.out.println(c(new int[]{3,4,5},6));
  }
}

Output:

true
true
true
true

false
false

Answer 15

Mathematica, 38 bytes

MemberQ[Total/@#~Permutations~{2},#2]&

input style [{1,2,3,4,5,6},11]

Answer 16

R, 32 bytes

function(x,l)x%in%combn(l,2,sum)

Returns an anonymous function. combn returns all combinations of l taken m at a time, in this case 2, and then optionally applies a function, in this case sum. Returns an empty logical vector logical(0) for an empty first input, which I define for this problem as falsey, and TRUE or FALSE corresponding to the other elements otherwise.

Try it online!

Answer 17

Python 3, 41 bytes

f=lambda a,b,*c:a-b in c or c and f(a,*c)

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Recursive aproach with flexible input and output. Returns an empty set as falsy.

Answer 18

MATLAB, 28 bytes

@(a,b)any(pdist(a,@plus)==b)

This defines an anonymous function with inputs a: column vector (possibly empty); b: number.

The code also works in Octave except if the first input is empty, which causes an error. Try it online!

Explanation

pdist(a,@fun) applies function fun (in our case, addition) to each combination of two rows of the input matrix a (in our case, a column vector). The output is a column vector. any(...==b) gives true if any of the results equals the input number b.

Answer 19

Pyth - 9 bytes

/msd.cE2E

Try it

/msd.cE2E
    .cE2     # All subsets of length 2
 msd         # The sum of each subset
/       E    # Number of occurrences of the input number

Answer 20

Axiom, 90 92 bytes

f(a,k)==(for i in 1..#a-1 repeat for j in i+1..#a repeat(q:=a.i+a.j;q=k=>return true);false)

test

(42) -> f([3,4,5],6)
   (42)  false
                                                            Type: Boolean
(43) -> f([1,2,3,4,5,6],11)
   (43)  true
                                                            Type: Boolean
(44) -> f([4,23,8174,42,-1],11)
   (44)  false
                                                            Type: Boolean
(45) -> f([4,23,8174,42,-1],41)
   (45)  true
                                                            Type: Boolean
(46) -> f([2,23,8174,42,2],2)
   (46)  false

Answer 21

Python 2, 71 bytes

from itertools import*
lambda L,n:n in map(sum,list(combinations(L,2)))

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Yes, I know there is already a better solution posted.

Answer 22

C#, 96 90 bytes

using System.Linq;a=>n=>Enumerable.Range(0,a.Count).Any(i=>a.Skip(i+1).Any(j=>a[i]+j==n));

Compiles to a Func<List<int>, Func<int, bool>>. Formatted version:

Func<List<int>, Func<int, bool>> f = a => n =>
    Enumerable.Range(0, a.Count).Any(i => a.Skip(i + 1).Any(j => a[i] + j == n));

Old for loop version for 96 bytes:

a=>n=>{for(int i=0,j,l=a.Count;i<l;++i)for(j=i+1;j<l;)if(a[i]+a[j++]==n)return 1>0;return 1<0;};

Answer 23

C#, 103 102 bytes

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Data

• Input Int32[] l A list of Int32 to sum
• Input Int32 r The result to check against
• Output Boolean A result indicating if the sum can be performed

---

Golfed

(int[] l,int r)=>{for(int x=l.Length,y;--x>=0;)for(y=0;y<x;)if(l[x]+l[y++]==r)return 1>0;return 0>1;};

---

Ungolfed

( int[] l, int r ) => {
   for( int x = l.Length, y; --x >= 0; )
      for( y = 0; y < x; )
         if( l[ x ] + l[ y++ ] == r )
            return 1 > 0;

   return 0 > 1;
};

---

Ungolfed readable

Working on it...

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Full code

using System;
using System.Collections.Generic;

namespace Namespace {
   class Program {
      static void Main( String[] args ) {
        Func<Int32[], Int32, Boolean> f = ( int[] l, int r ) => {
            for( int x = l.Length, y; --x >= 0; )
                for( y = 0; y < x; )
                    if( l[ x ] + l[ y++ ] == r )
                        return 1 > 0;

            return 0 > 1;
        };

        List<KeyValuePair<Int32[], Int32>>
            testCases = new List<KeyValuePair<Int32[], Int32>>() {
                new KeyValuePair<Int32[], Int32>( new Int32[] { 3, 4, 5 }, 6 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] { 1, 2, 3, 4, 5, 6 }, 11 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] { 4, 23, 8174, 42, -1 }, 41 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] { 1, 1, 1, 1, 1, 3 }, 3 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] { 2, 1, 1, 1, 53, 2 }, 4 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] { 2, 2, 2, 2 }, 4 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] { 3 }, 3 ),
                new KeyValuePair<Int32[], Int32>( new Int32[] {  }, -43 ),
            };

        foreach( var testCase in testCases ) {
            Console.WriteLine( $" Input: {testCase.Value}; {{ {String.Join(", ", testCase.Key)} }}\nOutput: {f( testCase.Key, testCase.Value )}\n" );
        }

         Console.ReadLine();
      }
   }
}

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Releases

• v1.1 -  -1 byte  - Thanks to TheLethalCoder suggestion.
• v1.0 - 103 bytes - Initial solution.

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Notes

• None

Answer 24

AWK, 77 bytes

{for(s=i=0;++i<NF;)for(j=0;++j<NF;){if(i==j)continue
if($i+$j==$NF)s=1}$0=s}1

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I could save 2 bytes by removing the i= in the for(s=i=0...), but then it would only work for one line of data. Since this isn't the shortest answer, I thought I'd make a general solution :)

Input is whitespace separated numbers with the last value being the desired sum.

Answer 25

Jellyfish, 13 bytes

p
cd!i
i 2
 1

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How it works

print(exists(i,from_base(1,permutations(i,2)))

where i are the inputs.