Find how many pair elements sums up to N

Question

^{Based on a recent question in StackOverflow}

Write a program/function that takes in one ARRAY/list and returns the number of pair-values which sums up to a given TOTAL.

Explaining

TOTAL=4, ARRAY=[1,3,7,1,-3] => RESULT=2

The result was 2 because 1 + 3 and 7 + -3 are valid pairs which sums up to TOTAL=4.

• When one pair is formed, you should not count any of its values again
• Also, you are not allowed to add any value that is not present in the ARRAY.

This is code-golf, shortest in bytes wins

All values will be between your language's int min-max.

Another cases

9, [1,2,3,5,4,6,7,-10] => 3          // [2,7]  [3,6]  [5,4]
0, [] => 0                           //
0, [1,-2,0,-1,2] => 2                // [1,-1]  [-2,2]
5, [5,5,5,0,-1,1,-5] => 1            // [5,0]
3, [7,-7,8,3,9,1,-1,5] => 0          //
2, [1,1] => 1                        // [1,1]
2, [1,1,1,1,1] => 2                  // [1,1]  [1,1]
4, [1,3,3,3] => 1                    // [1,3]

Answer 1

Pyth, 9 bytes

/sM{.cSE2

Test suite

Takes the number and array newline separated.

Explanation:

/sM{.cSE2
/sM{.cSE2Q    Implicit variable introduction.
              Implicit: Q = eval(input())
       E      Take the array as input.
      S       Sort the input.
    .c  2     Form all ordered pairs.
   {          Deduplicate.
 sM           Map each to its sum.
/        Q    Count the number of occurrences of Q in the resulting list.

Answer 2

R, 48 bytes

function(x,y)sum(colSums(combn(unique(x),2))==y)

Answer 3

Pyth - 13 bytes

l{SMfqQsT.PE2

Test Suite.

Answer 4

Julia, 48 bytes

x->n->count(p->sum(p)==n,combinations(∪(x),2))

This is an anonymous function that accepts an array and an integer and returns an integer. To call it, assign it to a variable and call like f(x)(n).

We get all combinations of size 2 from the unique values in the input array and for each pair test whether the sum is equal to the input integer. We count all such cases and return the result.

Try it online! (includes all test cases)

Answer 5

Matlab(38)

 @(a,b)nnz(sum(nchoosek([b,.5],2)')==a)

• the 0.5 part is a dodge for second test case

---

cropped by 2 bytes following Alex' comment.

Answer 6

JavaScript (ES6), 112 67 59 52 bytes

(a,b)=>{c=0;for(i of a)a.map(j=>c+=i+j==b);return c}

Because of amazing for-of loops of ES6, I was able to shave off 42 bytes from my previous solution!

Update : Thanks to @user6188402 for helping me shave off 3 bytes!

Update : Thanks to @Bálint for helping me shave off 15 bytes!

Try it Online!

Note : This fails for repeating numbers.