Comprueba si B se puede formar permutando los dígitos binarios de A

Dados dos enteros A y B , la tarea es verificar si la representación binaria de B se puede generar permutando los dígitos binarios de A .
Ejemplos: 
 

Entrada: A = 3, B = 9 
Salida: Sí 
Binario(3) = 0011 y Binario(9) = 1001
Entrada: A = 6, B = 7 
Salida: No 
 

Enfoque: la idea es contar el número de bits establecidos en las representaciones binarias de ambos números, ahora, si son iguales, la respuesta es Sí o, de lo contrario, la respuesta es No.
A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if the
// binary representation of b can be
// generated by permuting the
// binary digits of a
bool isPossible(int a, int b)
{
 
    // Find the count of set bits
    // in both the integers
    int cntA = __builtin_popcount(a);
    int cntB = __builtin_popcount(b);
 
    // If both the integers have
    // equal count of set bits
    if (cntA == cntB)
        return true;
    return false;
}
 
// Driver code
int main()
{
    int a = 3, b = 9;
 
    if (isPossible(a, b))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java implementation of the approach
class GFG
{
     
    // recursive function to count set bits
    public static int countSetBits(int n)
    {
 
        // base case
        if (n == 0)
            return 0;
        else
 
            // if last bit set add 1 else add 0
            return (n & 1) + countSetBits(n >> 1);
    }
     
    // Function that returns true if the
    // binary representation of b can be
    // generated by permuting the
    // binary digits of a
    static boolean isPossible(int a, int b)
    {
     
        // Find the count of set bits
        // in both the integers
        int cntA = countSetBits(a);
        int cntB = countSetBits(b);
     
        // If both the integers have
        // equal count of set bits
        if (cntA == cntB)
            return true;
        return false;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int a = 3, b = 9;
     
        if (isPossible(a, b))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by AnkitRai01

# Python3 implementation of the approach
 
# function to count set bits
def bitsoncount(x):
    return bin(x).count('1')
 
# Function that returns true if the
# binary representation of b can be
# generated by permuting the
# binary digits of a
def isPossible(a, b):
 
    # Find the count of set bits
    # in both the integers
    cntA = bitsoncount(a);
    cntB = bitsoncount(b);
 
    # If both the integers have
    # equal count of set bits
    if (cntA == cntB):
        return True
    return False
 
# Driver code
a = 3
b = 9
 
if (isPossible(a, b)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Sanjit Prasad

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
     
    // recursive function to count set bits
    public static int countSetBits(int n)
    {
 
        // base case
        if (n == 0)
            return 0;
        else
 
            // if last bit set.Add 1 else.Add 0
            return (n & 1) + countSetBits(n >> 1);
    }
     
    // Function that returns true if the
    // binary representation of b can be
    // generated by permuting the
    // binary digits of a
    static bool isPossible(int a, int b)
    {
     
        // Find the count of set bits
        // in both the integers
        int cntA = countSetBits(a);
        int cntB = countSetBits(b);
     
        // If both the integers have
        // equal count of set bits
        if (cntA == cntB)
            return true;
        return false;
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int a = 3, b = 9;
     
        if (isPossible(a, b))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by Rajput-Ji

<script>
    // Javascript implementation of the approach
     
    // recursive function to count set bits 
    function countSetBits(n) 
    { 
   
        // base case 
        if (n == 0) 
            return 0; 
        else
   
            // if last bit set add 1 else add 0 
            return (n & 1) + countSetBits(n >> 1); 
    } 
       
    // Function that returns true if the 
    // binary representation of b can be 
    // generated by permuting the 
    // binary digits of a 
    function isPossible(a, b) 
    { 
       
        // Find the count of set bits 
        // in both the integers 
        let cntA = countSetBits(a); 
        let cntB = countSetBits(b); 
       
        // If both the integers have 
        // equal count of set bits 
        if (cntA == cntB) 
            return true; 
        return false; 
    } 
     
    let a = 3, b = 9; 
       
    if (isPossible(a, b)) 
      document.write("Yes"); 
    else
      document.write("No"); 
 
// This code is contributed by divyesh072019.
</script>

Yes

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)