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++ 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 OR value is B[j]
if (__builtin_popcount(A[i] | A[j]) == B[j]) {
cnt++;
}
return cnt;
} // Driver code int 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 implementation of the approach 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 OR 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[] = { 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 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# implementation of the 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 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 code public 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 */ |
<script> // JavaScript implementation of the approach // Function to return the count of pairs // which satisfy the given condition function 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 code let 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> |
7
Time Complexity: O(N2)
Auxiliary Space: O(1)