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