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[] = {5, 3, 2, 4, 6, 1}, B[] = {2, 2, 1, 4, 2, 3}
Output: 7
All possible pairs are (5, 5), (3, 3), (2, 2),
(2, 6), (4, 6), (6, 6) and (6, 1).
Input: A[] = {4, 3, 5, 6, 7}, B[] = {1, 3, 2, 4, 5}
Output: 4
Approach: Iterate through all the possible pairs (i, j) and check the count of set bits in their OR 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 OR value is B[j] if (__builtin_popcount(A[i] | A[j]) == B[j]) { cnt++; } return cnt;}// Driver codeint main(){ int A[] = { 5, 3, 2, 4, 6, 1 }; int B[] = { 2, 2, 1, 4, 2, 3 }; int size = sizeof(A) / sizeof(A[0]); cout << solve(A, B, size); return 0;} |
Java
// Java implementation of the approachclass GFG {// Function to return the count of pairs// which satisfy the given conditionstatic 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 OR value is B[j] if (Integer.bitCount(A[i] | A[j]) == B[j]) { cnt++; } return cnt;}// Driver codepublic static void main(String args[]){ int A[] = { 5, 3, 2, 4, 6, 1 }; int B[] = { 2, 2, 1, 4, 2, 3 }; int size = A.length; System.out.println(solve(A, B, size));}}// This code is contributed by 29AjayKumar |
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 OR value is B[j] if (bin(A[i] | A[j]).count('1') == B[j]) : cnt += 1; return cnt # Driver code if __name__ == "__main__" : A = [ 5, 3, 2, 4, 6, 1 ]; B = [ 2, 2, 1, 4, 2, 3 ]; size = len(A); print(solve(A, B, size)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System;class GFG {// Function to return the count of pairs// which satisfy the given conditionstatic 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 OR value is B[j] if (bitCount(A[i] | A[j]) == B[j]) { cnt++; } return cnt;}static int bitCount(long x){ // To store the count // of set bits int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits;}// Driver codepublic static void Main(String []args){ int []A = { 5, 3, 2, 4, 6, 1 }; int []B = { 2, 2, 1, 4, 2, 3 }; int size = A.Length; Console.WriteLine(solve(A, B, size));}}/* This code is contributed by PrinciRaj1992 */ |
Javascript
<script>// JavaScript implementation of the approach// Function to return the count of pairs// which satisfy the given conditionfunction solve(A,B,n){ let cnt = 0; for (let i = 0; i < n; i++) for (let j = i; j < n; j++) // Check if the count of set bits // in the OR value is B[j] if (bitCount(A[i] | A[j]) == B[j]) { cnt++; } return cnt;}function bitCount(x){ // To store the count // of set bits let setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits;}// Driver codelet A=[5, 3, 2, 4, 6, 1 ];let B=[2, 2, 1, 4, 2, 3 ];let size = A.length;document.write(solve(A, B, size));// This code is contributed by rag2127</script> |
Output:
7
Time Complexity: O(N2)
Auxiliary Space: O(1)
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