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. 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
 
Recommended: Please try your approach on {IDE} first, before moving on to the solution.
 

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

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