Given two arrays A[] and B[] of N elements each. The task is to find the number of index pairs (i, j) such that i ? j and F(A[i] & A[j]) = B[j] where F(X) is the count of set bits in the binary representation of X.
Examples:
Input: A[] = {2, 3, 1, 4, 5}, B[] = {2, 2, 1, 4, 2}
Output: 4
All possible pairs are (3, 3), (3, 1), (1, 1) and (5, 5)
Input: A[] = {1, 2, 3, 4, 5}, B[] = {2, 2, 2, 2, 2}
Output: 2
Approach: Iterate through all the possible pairs (i, j) and check the count of set bits in their AND value. If the count is equal to B[j] then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the count of pairs// which satisfy the given conditionint solve(int A[], int B[], int n){ int cnt = 0; for (int i = 0; i < n; i++) for (int j = i; j < n; j++) // Check if the count of set bits // in the AND value is B[j] if (__builtin_popcount(A[i] & A[j]) == B[j]) { cnt++; } return cnt;}// Driver codeint main(){ int A[] = { 2, 3, 1, 4, 5 }; int B[] = { 2, 2, 1, 4, 2 }; int size = sizeof(A) / sizeof(A[0]); cout << solve(A, B, size); return 0;} |
Java
// Java implementation of the approachpublic class GFG { // Function to return the count of pairs // which satisfy the given condition static int solve(int A[], int B[], int n) { int cnt = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) // Check if the count of set bits // in the AND value is B[j] { if (Integer.bitCount(A[i] & A[j]) == B[j]) { cnt++; } } } return cnt; } // Driver code public static void main(String[] args) { int A[] = {2, 3, 1, 4, 5}; int B[] = {2, 2, 1, 4, 2}; int size = A.length; System.out.println(solve(A, B, size)); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approach # Function to return the count of pairs # which satisfy the given condition def solve(A, B, n) : cnt = 0; for i in range(n) : for j in range(i, n) : # Check if the count of set bits # in the AND value is B[j] c = A[i] & A[j] if (bin(c).count('1') == B[j]) : cnt += 1; return cnt; # Driver code if __name__ == "__main__" : A = [ 2, 3, 1, 4, 5 ]; B = [ 2, 2, 1, 4, 2 ]; size = len(A); print(solve(A, B, size)); # This code is contributed # by AnkitRai01 |
C#
// C# Implementation of the above approachusing System;class GFG { // Function to return the count of pairs // which satisfy the given condition static int solve(int []A, int []B, int n) { int cnt = 0; for (int i = 0; i < n; i++) { for (int j = i; j < n; j++) // Check if the count of set bits // in the AND value is B[j] { if (countSetBits(A[i] & A[j]) == B[j]) { cnt++; } } } return cnt; } // Function to get no of set // bits in binary representation // of positive integer n static int countSetBits(int n) { int count = 0; while (n > 0) { count += n & 1; n >>= 1; } return count; } // Driver code public static void Main(String[] args) { int []A = {2, 3, 1, 4, 5}; int []B = {2, 2, 1, 4, 2}; int size = A.Length; Console.WriteLine(solve(A, B, size)); } } // This code is contributed by Princi Singh |
Javascript
<script>// Javascript Implementation of the above approach // Function to return the count of pairs // which satisfy the given condition function solve(A, B, n) { var cnt = 0; for (var i = 0; i < n; i++) { for (var j = i; j < n; j++) // Check if the count of set bits // in the AND value is B[j] { if (countSetBits(A[i] & A[j]) == B[j]) { cnt++; } } } return cnt; } // Function to get no of set // bits in binary representation // of positive integer n function countSetBits(n) { var count = 0; while (n > 0) { count += n & 1; n >>= 1; } return count; } // Driver code var A = [2, 3, 1, 4, 5]; var B = [2, 2, 1, 4, 2];var size = A.length; document.write(solve(A, B, size)); // This code is contributed by rutvik_56.</script> |
Output:
4
Time Complexity: O(n2)
Auxiliary Space: O(1)
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