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*(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 ? 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;
PHP
<?php// PHP Program to find pairs count// function to count// the required pairsfunction countPairs($A, $n, $k) {$ans = 0;// sort the given arraysort($A);// for each A[i] // traverse rest arrayfor ($i = 0; $i < $n; $i++) { for ($j = $i + 1; $j < $n; $j++) { // count Aj such that Ai*k^x = Aj $x = 0; // increase x till Ai * // k^x <= largest element while (($A[$i] * pow($k, $x)) <= $A[$j]) { if (($A[$i] * pow($k, $x)) == $A[$j]) { $ans++; break; } $x++; } }}return $ans;}// Driver Code$A = array(3, 8, 9, 12, 18, 4, 24, 2, 6);$n = count($A);$k = 3;echo countPairs($A, $n, $k);// This code is contributed by anuj_67.?> |
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!
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