Divisible Sum Pairs

You are given an array of n integers, ar = ar[0], ar[1], …, ar[n - 1], and a positive integer, k. Find and print the number of (i, j) pairs where i < j and ar[i] + ar[j] is divisible by k.

For example, ar = [1,2,3,4,5,6] and k = 5. Our three pairs meeting the criteria are [1,4],[2,3] and [4,6].

Function Description

Complete the divisibleSumPairs function in the editor below. It should return the integer count of pairs meeting the criteria.

divisibleSumPairs has the following parameter(s):

• n: the integer length of array ar
• ar: an array of integers
• k: the integer to divide the pair sum by

Input Format

The first line contains 2 space-separated integers, n and k. The second line contains n space-separated integers describing the values of ar[ar[0], ar[1],…,ar[n-1]].

Constraints

• 2 <= n <= 100
• 1 <= k <= 100
• 1 <= ar[i] <= 100

Output Format

Print the number of (i, j) pairs where i < j and a[i] + a[j] is evenly divisible by k.

Sample Input

1
2

6 3
1 3 2 6 1 2

Sample Output

1

5

Explanation

Here are the 5 valid pairs when k = 3:

• (0,2) -> ar[0] + ar[2] = 1 + 2 = 3
• (0,5) -> ar[0] + ar[5] = 1 + 2 = 3
• (1,3) -> ar[1] + ar[3] = 3 + 6 = 9
• (2,4) -> ar[2] + ar[4] = 2 + 1 = 3
• (4,5) -> ar[4] + ar[5] = 1 + 2 = 3

Solution

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9

// Complete the divisibleSumPairs function below.
function divisibleSumPairs(n, k, ar) {
    let count = 0;
    let focus = ar.shift();

    (ar || []).reduce((target, item) => !((focus + item) % k) && count++, 0);

    return !(ar || []).length ? count : count + divisibleSumPairs(n, k, ar);
}