Javascript Program for Pairs such that one is a power multiple of other

You are given an array A[] of n-elements and a positive integer k (k > 1). Now you have find the number of pairs Ai, Aj such that Ai = Aj*(k^x) where x is an integer. 
Note: (Ai, Aj) and (Aj, Ai) must be count once.
Examples :

Input : A[] = {3, 6, 4, 2},  k = 2
Output : 2
Explanation : We have only two pairs 
(4, 2) and (3, 6)

Input : A[] = {2, 2, 2},   k = 2
Output : 3
Explanation : (2, 2), (2, 2), (2, 2) 
that are (A1, A2), (A2, A3) and (A1, A3) are 
total three pairs where Ai = Aj * (k^0) 

To solve this problem, we first sort the given array and then for each element Ai, we find number of elements equal to value Ai * k^x for different value of x till Ai * k^x is less than or equal to largest of Ai. 
Algorithm: 

    // sort the given array
    sort(A, A+n);

    // for each A[i] traverse rest array
    for (int i=0; i ≤ n-1; i++)
    {
        for (int j=i+1; j ≤ n-1; j++)
        {
            // count Aj such that Ai*k^x = Aj
            int x = 0;

            // increase x till Ai * k^x ≤ 
            // largest element
            while ((A[i]*pow(k, x)) ≤ A[j])
            {
                if ((A[i]*pow(k, x)) == A[j])
                {              
                     ans++;
                     break;
                }
                x++;
            }        
        }   
    }
    // return answer
    return ans;

Javascript

<script>

// Javascript Program to find pairs count

// function to count the required pairs
function countPairs(A, n, k) {
  var ans = 0;

  // sort the given array
  A.sort((a,b)=>a-b)

  // for each A[i] traverse rest array
  for (var i = 0; i < n; i++) {
    for (var j = i + 1; j < n; j++) {

      // count Aj such that Ai*k^x = Aj
      var x = 0;

      // increase x till Ai * k^x <= largest element
      while ((A[i] * Math.pow(k, x)) <= A[j]) {
        if ((A[i] * Math.pow(k, x)) == A[j]) {
          ans++;
          break;
        }
        x++;
      }
    }
  }
  return ans;
}

// driver program
var A = [3, 8, 9, 12, 18, 4, 24, 2, 6];
var n = A.length;
var k = 3;
document.write( countPairs(A, n, k));

// This code is contributed by rutvik_56.
</script>

Output :

6

Time Complexity: O(n*n), as nested loops are used
Auxiliary Space: O(1), as no extra space is used

Please refer complete article on Pairs such that one is a power multiple of other for more details!