Count pairs of elements such that number of set bits in their AND is B[i]
Last Updated :
25 Nov, 2021
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++
#include <bits/stdc++.h>
using namespace std;
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++)
if (__builtin_popcount(A[i] & A[j]) == B[j]) {
cnt++;
}
return cnt;
}
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
public class GFG
{
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++)
{
if (Integer.bitCount(A[i] & A[j]) == B[j])
{
cnt++;
}
}
}
return cnt;
}
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));
}
}
|
Python3
def solve(A, B, n) :
cnt = 0 ;
for i in range (n) :
for j in range (i, n) :
c = A[i] & A[j]
if ( bin (c).count( '1' ) = = B[j]) :
cnt + = 1 ;
return cnt;
if __name__ = = "__main__" :
A = [ 2 , 3 , 1 , 4 , 5 ];
B = [ 2 , 2 , 1 , 4 , 2 ];
size = len (A);
print (solve(A, B, size));
|
C#
using System;
class GFG
{
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++)
{
if (countSetBits(A[i] & A[j]) == B[j])
{
cnt++;
}
}
}
return cnt;
}
static int countSetBits( int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
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));
}
}
|
Javascript
<script>
function solve(A, B, n)
{
var cnt = 0;
for ( var i = 0; i < n; i++)
{
for ( var j = i; j < n; j++)
{
if (countSetBits(A[i] & A[j]) == B[j])
{
cnt++;
}
}
}
return cnt;
}
function countSetBits(n)
{
var count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
var A = [2, 3, 1, 4, 5];
var B = [2, 2, 1, 4, 2];
var size = A.length;
document.write(solve(A, B, size));
</script>
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Time Complexity: O(n2)
Auxiliary Space: O(1)
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