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++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the count of pairs // which satisfy the given condition 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 (__builtin_popcount(A[i] & A[j]) == B[j]) {
cnt++;
}
return cnt;
} // Driver code int 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 implementation of the approach public 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 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# Implementation of the above approach using 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 |
<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> |
4
Time Complexity: O(n2)
Auxiliary Space: O(1)