Dadas dos arrays A[] y B[] de N elementos cada una. La tarea es encontrar el número de pares de índices (i, j) tales que i ≤ j y F(A[i] & A[j]) = B[j] donde F(X) es el conteo de bits establecidos en la representación binaria de X .
Ejemplos:
Entrada: A[] = {2, 3, 1, 4, 5}, B[] = {2, 2, 1, 4, 2}
Salida: 4
Todos los pares posibles son (3, 3), (3, 1 ), (1, 1) y (5, 5)
Entrada: A[] = {1, 2, 3, 4, 5}, B[] = {2, 2, 2, 2, 2}
Salida: 2
Enfoque: iterar a través de todos los pares posibles (i, j) y verificar el recuento de bits establecidos en su valor AND. Si el conteo es igual a B[j] entonces incremente el conteo.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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
# 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 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
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>
Producción:
4
Complejidad temporal: O(n 2 )
Espacio Auxiliar: O(1)