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Turn off the rightmost set bit | Set 2

  • Difficulty Level : Easy
  • Last Updated : 26 Mar, 2021

Given a Number N. The task is to unset the rightmost set bit of N.
Examples: 
 

Input: N = 5 
Output: 4
Explanation:
101(5) -> 100(4)

Input : N = 78
Output : 76
Explanation:
1001110(78) -> 1001100(76)

 

Naive Approach: 
 

  • A simple approach to flip the rightmost set bit is to find the position of the rightmost set bit in the number by a bitwise right shift operation to check for the first occurrence of 1. 
     
  • Then, flip the bit in this position. Flipping can be done by applying XOR of the given number and the number with the bit in this position set.
    Important property to flip bits: 
     
0 ^ 1 = 1
1 ^ 1 = 0

Xor with 1 flips the bit.
  •  
  • Setting only a given bit can be done by taking 2 to the power of position to set a particular bit. 
     

Below is the implementation of the above approach.
 

C++




// C++ program to unset the rightmost
// set bit
#include <bits/stdc++.h>
using namespace std;
 
// Unsets the rightmost set bit 
// of n and returns the result
void FlipBits(int n)
{
 
    for (int bit = 0; bit < 32; bit++)
    {
        // Checking whether bit position is
        // set or not
        if ((n >> bit) & 1)
        {
            // If bit position is found set,
            // we flip this bit by xoring
            // given number and number with
            // bit position set
            n = n ^ (1ll << bit);
            break;
        }
    }
     
    cout<<"The number after unsetting the"
    cout<<" rightmost set bit "<< n; 
 
}
// Driver code
int main()
{
     
    int N = 12;
     
    FlipBits(N);
    return 0;
}

Java




// Java program to unset the rightmost
// set bit
import java.util.*;
 
class GFG{
 
// Unsets the rightmost set bit
// of n and returns the result
static void FlipBits(int n)
{
    for(int bit = 0; bit < 32; bit++)
    {
        
       // Checking whether bit position 
       // is set or not
       if ((n >> bit) % 2 > 0)
       {
            
           // If bit position is found set,
           // we flip this bit by xoring
           // given number and number with
           // bit position set
           n = n ^ (1 << bit);
           break;
       }
    }
    System.out.print("The number after unsetting the");
    System.out.print(" rightmost set bit " + n);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 12;
     
    FlipBits(N);
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 program to unset the rightmost
# set bit
 
# Unsets the rightmost set bit
# of n and returns the result
def FlipBits(n):
     
    for bit in range(32):
         
        # Checking whether bit position is
        # set or not
        if ((n >> bit) & 1):
             
            # If bit position is found set,
            # we flip this bit by xoring
            # given number and number with
            # bit position set
            n = n ^ (1 << bit)
            break
     
    print("The number after unsetting the", end = " ")
    print("rightmost set bit", n)
 
# Driver code
if __name__ == '__main__':
     
    N = 12;
     
    FlipBits(N)
 
# This code is contributed by BhupendraSingh

C#




// C# program to unset the rightmost
// set bit
using System;
 
class GFG{
 
// Unsets the rightmost set bit
// of n and returns the result
static void FlipBits(int n)
{
    for(int bit = 0; bit < 32; bit++)
    {
        
       // Checking whether bit position
       // is set or not
       if ((n >> bit) % 2 > 0)
       {
            
           // If bit position is found set,
           // we flip this bit by xoring
           // given number and number with
           // bit position set
           n = n ^ (1 << bit);
           break;
       }
    }
    Console.Write("The number after unsetting the ");
    Console.Write("rightmost set bit " + n);
}
 
// Driver code
static void Main()
{
    int N = 12;
     
    FlipBits(N);
}
}
 
// This code is contributed by shivanisinghss2110

Javascript




<script>
    // Javascript program to unset the rightmost
    // set bit
     
    // Unsets the rightmost set bit
    // of n and returns the result
    function FlipBits(n)
    {
 
        for (let bit = 0; bit < 32; bit++)
        {
            // Checking whether bit position is
            // set or not
            if (((n >> bit) & 1) > 0)
            {
                // If bit position is found set,
                // we flip this bit by xoring
                // given number and number with
                // bit position set
                n = n ^ (1 << bit);
                break;
            }
        }
 
        document.write("The number after unsetting the");
        document.write(" rightmost set bit " + n);
 
    }
     
    // Driver code
    let N = 12;
      
    FlipBits(N);
 
// This code is contributed by suresh07.
</script>
Output: 



The number after unsetting the rightmost set bit 8

 

Efficient Approach: 
We can flip the LSB in O(1). Though the naive approach also works fine, there also exists a one-liner solution to the problem. We can do the following: 
 

  1. Obtain 2’s complement of the number (simply -n gives 2’s complement) 
     
  2. Perform bitwise AND operation between number and it’s 2’s complement. 
     
  3. Subtract the above result from the given number. 
     
  4. The final result has LSB flipped. 
     

Explaination 
 

             N = 1001110 (78)
2's compliment = 0110010
      N & (-N) = 0000010
N - (N & (-N)) = 1001100 (76)

 

C++




// C++ program to unset the rightmost
// set bit
#include <bits/stdc++.h>
using namespace std;
   
// Unsets the rightmost set bit 
// of n and returns the result 
int FlipBits(unsigned int n) 
    return n -= (n & (-n));
   
// Driver Code 
int main() 
    int N = 12; 
    cout<<"The number after unsetting the"
    cout<<" rightmost set bit: "<<FlipBits(N); 
    return 0; 

Java




// Java program to unset the rightmost set bit
import java.util.*;
class GFG{
     
// Unsets the rightmost set bit
// of n and returns the result
static int FlipBits(int n)
{
    return n -= (n & (-n));
}
     
// Driver Code
public static void main(String[] args)
{
    int N = 12;
     
    System.out.print("The number after unsetting the ");
    System.out.print("rightmost set bit: " + FlipBits(N));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to unset the rightmost set bit
 
# Unsets the rightmost set bit
# of n and returns the result
def FlipBits(n):
     
    n -= (n & (-n));
    return n;
     
# Driver Code
if __name__ == '__main__':
     
    N = 12;
 
    print("The number after unsetting the", end = "");
    print(" rightmost set bit: ", FlipBits(N));
 
# This code is contributed by Rohit_ranjan

C#




// C# program to unset the rightmost set bit
using System;
 
class GFG{
     
// Unsets the rightmost set bit
// of n and returns the result
static int FlipBits(int n)
{
    return n -= (n & (-n));
}
     
// Driver Code
public static void Main(String[] args)
{
    int N = 12;
     
    Console.Write("The number after" +
                  "unsetting the ");
    Console.Write("rightmost set bit: " +
                  FlipBits(N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
    // Javascript program to unset the rightmost set bit
     
    // Unsets the rightmost set bit 
    // of n and returns the result 
    function FlipBits(n) 
    
        return n -= (n & (-n));
    
 
    let N = 12; 
    document.write("The number after unsetting the"); 
    document.write(" rightmost set bit: " + FlipBits(N));
     
</script>
Output: 
The number after unsetting the rightmost set bit: 8

 




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