Given an array arr[] of N positive integers. The task is to find the number of pairs whose Bitwise AND value is a power of 2.
Examples:
Input: arr[] = {2, 1, 3, 4}
Output: 2
Explanation:
There are 2 pairs (2, 3) and (1, 3) in this array whose Bitwise AND values are:
1. (2 & 3) = 1 = (20)
2. (1 & 3) = 1 = (20).Input: arr[] = {6, 4, 2, 3}
Output: 4
Explanation:
There are 4 pairs (6, 4), (6, 2), (6, 3), (2, 3) whose Bitwise and is power of 2.
Approach: For each possible pair in the given array, the idea to check whether Bitwise AND of each pairs of elements is perfect power of 2 or not. If “Yes” then count this pair Else check for the next pair.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if x is power of 2 bool check(int x) { // Returns true if x is a power of 2 return x && (!(x & (x - 1))); } // Function to return the // number of valid pairs int count(int arr[], int n) { int cnt = 0; // Iterate for all possible pairs for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // Bitwise and value of // the pair is passed if (check(arr[i] & arr[j])) cnt++; } } // Return the final count return cnt; } // Driver Code int main() { // Given array int arr[] = { 6, 4, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); // Function Call cout << count(arr, n); return 0; } |
Java
// Java program for the above approach class GFG{ // Method to check if x is power of 2 static boolean check(int x) { // First x in the below expression // is for the case when x is 0 return x != 0 && ((x & (x - 1)) == 0); } // Function to return the // number of valid pairs static int count(int arr[], int n) { int cnt = 0; // Iterate for all possible pairs for(int i = 0; i < n - 1; i++) { for(int j = i + 1; j < n; j++) { // Bitwise and value of // the pair is passed if (check(arr[i] & arr[j])) cnt++; } } // Return the final count return cnt; } // Driver Code public static void main(String[] args) { // Given array arr[] int arr[] = new int[]{ 6, 4, 2, 3 }; int n = arr.length; // Function call System.out.print(count(arr, n)); } } // This code is contributed by Pratima Pandey |
Python3
# Python3 program for the above approach # Function to check if x is power of 2 def check(x): # Returns true if x is a power of 2 return x and (not(x & (x - 1))) # Function to return the # number of valid pairs def count(arr, n): cnt = 0 # Iterate for all possible pairs for i in range(n - 1): for j in range(i + 1, n): # Bitwise and value of # the pair is passed if check(arr[i] & arr[j]): cnt = cnt + 1 # Return the final count return cnt # Given array arr = [ 6, 4, 2, 3 ] n = len(arr) # Function Call print(count(arr, n)) # This code is contributed by divyeshrabadiya07 |
C#
// C# program for the above approach using System; class GFG{ // Method to check if x is power of 2 static bool check(int x) { // First x in the below expression // is for the case when x is 0 return x != 0 && ((x & (x - 1)) == 0); } // Function to return the // number of valid pairs static int count(int []arr, int n) { int cnt = 0; // Iterate for all possible pairs for(int i = 0; i < n - 1; i++) { for(int j = i + 1; j < n; j++) { // Bitwise and value of // the pair is passed if (check(arr[i] & arr[j])) cnt++; } } // Return the final count return cnt; } // Driver Code public static void Main() { // Given array arr[] int []arr = new int[]{ 6, 4, 2, 3 }; int n = arr.Length; // Function call Console.Write(count(arr, n)); } } // This code is contributed by Code_Mech |
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
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