Número formado por el bit establecido más a la derecha en N

Dado un número entero N , la tarea es encontrar un número entero M formado tomando el bit establecido más a la derecha en N , es decir, el único bit establecido en M será el bit establecido más a la derecha en N y el resto de los bits no estarán establecidos.
Ejemplos: 
 

Entrada: N = 7 
Salida: 1 
7 = 111, el número formado por el último bit establecido es 001, es decir, 1.
Entrada: N = 10 
Salida: 2 
10 = 1010 -> 0010 = 2
Entrada: N = 16 
Salida: 16 
 

Acercarse: 
 

• Almacene x = n & (n – 1) que desactivará el primer bit establecido desde la derecha en n .
• Ahora, actualice n = n ^ x para establecer el bit modificado y desactive todos los demás, que es el número entero requerido.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the integer formed
// by taking the rightmost set bit in n
int firstSetBit(int n)
{
 
    // n & (n - 1) unsets the first set
    // bit from the right in n
    int x = n & (n - 1);
 
    // Take xor with the original number
    // The position of the 'changed bit'
    // will be set and rest will be unset
    return (n ^ x);
}
 
// Driver code
int main()
{
    int n = 12;
 
    cout << firstSetBit(n);
 
    return 0;
}

// Java implementation of the approach
class geeks
{
     
    // Function to return the integer formed
    // by taking the rightmost set bit in n
    public static int firstSetBit(int n)
    {
         
        // n & (n - 1) unsets the first set
        // bit from the right in n
        int x = n & (n-1);
         
        // Take xor with the original number
        // The position of the 'changed bit'
        // will be set and rest will be unset
        return (n ^ x);
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int n = 12;
        System.out.println(firstSetBit(n));
    }
}
 
// This code is contributed by
// sanjeev2552

# Python3 implementation of the approach
 
# Function to return the integer formed
# by taking the rightmost set bit in n
def firstSetBit(n):
 
    # n & (n - 1) unsets the first set
    # bit from the right in n
    x = n & (n - 1)
 
    # Take xor with the original number
    # The position of the 'changed bit'
    # will be set and rest will be unset
    return (n ^ x)
 
# Driver code
n = 12
 
print(firstSetBit(n))
 
# This code is contributed by mohit kumar 29

// C# implementation of the approach
using System;
 
class geeks
{
     
    // Function to return the integer formed
    // by taking the rightmost set bit in n
    public static int firstSetBit(int n)
    {
         
        // n & (n - 1) unsets the first set
        // bit from the right in n
        int x = n & (n-1);
         
        // Take xor with the original number
        // The position of the 'changed bit'
        // will be set and rest will be unset
        return (n ^ x);
    }
     
    // Driver Code
    public static void Main ()
    {
        int n = 12;
        Console.WriteLine(firstSetBit(n));
    }
}
 
// This code is contributed by
// anuj_67..

<script>
// JavaScript implementation of the approach
 
// Function to return the integer formed
// by taking the rightmost set bit in n
function firstSetBit(n)
{
 
    // n & (n - 1) unsets the first set
    // bit from the right in n
    let x = n & (n - 1);
 
    // Take xor with the original number
    // The position of the 'changed bit'
    // will be set and rest will be unset
    return (n ^ x);
}
 
// Driver code
    let n = 12;
    document.write(firstSetBit(n));
 
// This code is contributed by Manoj.
</script>

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