Count pairs in array such that one element is power of another
Given an array arr[], the task is to count the pairs in the array such that one element is the power of another in each pair.
Examples:
Input: arr[] = {16, 2, 3, 9}
Output: 2
The 2 pairs are (16, 2) and (3, 9)
Input: arr[] = {2, 3, 5, 7}
Output: 0
Approach:
- After taking the array as input, first we need to find out all the possible pairs in that array.
- So, find out the pairs from the array
- Then for each pair, check whether an element is a power of another. If it is, then increase the required count by one.
- When all the pairs have been checked, return or print the count of such pair.
Below is the implementation of the above approach:
C++
// C++ program to count pairs in array // such that one element is power of another #include <bits/stdc++.h> using namespace std; // Function to check if given number number y // is power of x bool isPower(int x, int y) { // log function to calculate value int res1 = log(y) / log(x); double res2 = log(y) / log(x); // compare to the result1 // or result2 both are equal return (res1 == res2); } // Function to find pairs from array int countPower(int arr[], int n) { int res = 0; // Iterate through all pairs for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) // Increment count if one is // the power of other if (isPower(arr[i], arr[j]) || isPower(arr[j], arr[i])) res++; return res; } // Driver code int main() { int a[] = { 16, 2, 3, 9 }; int n = sizeof(a) / sizeof(a[0]); cout << countPower(a, n); return 0; } |
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Java
// Java program to count pairs in array // such that one element is power of another class GFG { // Function to check if given number number y // is power of x static boolean isPower(int x, int y) { // log function to calculate value int res1 = (int)(Math.log(y) / Math.log(x)); double res2 = Math.log(y) / Math.log(x); // compare to the result1 // or result2 both are equal return (res1 == res2); } // Function to find pairs from array static int countPower(int arr[], int n) { int res = 0; // Iterate through all pairs for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) // Increment count if one is // the power of other if (isPower(arr[i], arr[j]) || isPower(arr[j], arr[i])) res++; return res; } // Driver code public static void main (String[] args) { int a[] = { 16, 2, 3, 9 }; int n =a.length; System.out.println(countPower(a, n)); } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 program to count pairs in array # such that one element is power of another from math import log # Function to check if given number number y # is power of x def isPower(x, y) : # log function to calculate value res1 = log(y) // log(x); res2 = log(y) / log(x); # compare to the result1 # or result2 both are equal return (res1 == res2); # Function to find pairs from array def countPower( arr, n) : res = 0; # Iterate through all pairs for i in range(n) : for j in range(i + 1, n) : # Increment count if one is # the power of other if isPower(arr[i], arr[j]) or isPower(arr[j], arr[i]) : res += 1; return res; # Driver code if __name__ == "__main__" : a = [ 16, 2, 3, 9 ]; n = len(a); print(countPower(a, n)); # This code is contributed by AnkitRai01 |
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C#
// C# program to count pairs in array // such that one element is power of another using System; public class GFG { // Function to check if given number number y // is power of x static bool isPower(int x, int y) { // log function to calculate value int res1 = (int)(Math.Log(y) / Math.Log(x)); double res2 = Math.Log(y) / Math.Log(x); // compare to the result1 // or result2 both are equal return (res1 == res2); } // Function to find pairs from array static int countPower(int []arr, int n) { int res = 0; // Iterate through all pairs for (int i = 0; i < n; i++) for (int j = i + 1; j < n; j++) // Increment count if one is // the power of other if (isPower(arr[i], arr[j]) || isPower(arr[j], arr[i])) res++; return res; } // Driver code public static void Main () { int []a = { 16, 2, 3, 9 }; int n =a.Length; Console.WriteLine(countPower(a, n)); } } // This code is contributed by AnkitRai01 |
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Output:
2
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