Program to Clear the Rightmost Set Bit of a Number in C++ and Python

CPP Programming, Python, Python / By Vikram Chiluka

In the previous article, we have discussed about C++ Program to Check if it is Sparse Matrix or Not. Let us learn Program to Clear the Rightmost Set Bit of a Number in C++ Program and Python.

Binary Representation of a Number:

Binary is a base-2 number system in which a number is represented by two states: 0 and 1. We can also refer to it as a true and false state. A binary number is constructed in the same way that a decimal number is constructed.

Examples:

Examples1:

Input:

given number=19

Output:

The given number before removing right most set bit : 
19
The given number after removing right most set bit : 
18

Examples2:

Input:

given number =18

Output:

The given number before removing right most set bit : 
18
The given number after removing right most set bit : 
16

Examples3:

Input:

given number=512

Output:

The given number before removing right most set bit : 
512
The given number after removing right most set bit : 
0

Program to Clear the Rightmost Set Bit of a Number in C++ and Python

There are several ways to clear the rightmost set Bit of a Number in C++ and Python some of them are:

Using Bitwise Operators in C++
Using Bitwise Operators in Python

Drive into Python Programming Examples and explore more instances related to python concepts so that you can become proficient in generating programs in Python Programming Language.

Method #1: Using Bitwise Operators in C++

Approach:

There is a function called clear rightmost set bit that is defined.
It accepts n as an argument and returns n with its rightmost set bit cleared.
This is accomplished by computing n & (n – 1) and returning the result.
(n – 1) equals n when all the rightmost successive 0s are flipped and the initial rightmost 1 is flipped.
As a result, n & (n – 1) equals n with the rightmost 1 cleared.

Below is the implementation of above approach:

#include <bits/stdc++.h>
using namespace std;
// function which removes the right most set bit in the
// given number
int clearRight(int numb)
{
    // clearing the right most set bit from
    // the given number and store it in the result
    int reslt = (numb) & (numb - 1);
    // returing the calculated result
    return reslt;
}

// main function
int main()
{
    // given number
    int numb = 19;

    cout << "The given number before removing right most "
            "set bit : "
         << numb << endl;
    // passing the given number to clearRight function
    // to remove the clear the rightmost setbit
    cout << "The given number after removing right most "
            "set bit : "
         << clearRight(numb) << endl;
    return 0;
}

Output:

The given number before removing right most set bit : 19
The given number after removing right most set bit : 18

Method #2: Using Bitwise Operators in Python

Approach:

There is a function called clear rightmost set bit that is defined.
It accepts n as an argument and returns n with its rightmost set bit cleared.
This is accomplished by computing n & (n – 1) and returning the result.
(n – 1) equals n when all the rightmost successive 0s are flipped and the initial rightmost 1 is flipped.
As a result, n & (n – 1) equals n with the rightmost 1 cleared.
We will implement the same function in python

Below is the implementation:

# function which removes the right most set bit in the
# given number


def clearRight(numb):
    # clearing the right most set bit from
    # the given number and store it in the result
    reslt = (numb) & (numb - 1)
    # returing the calculated result
    return reslt
# Driver Code


# given number
numb = 19

print("The given number before removing right most "
      "set bit : ")
print(numb)
# passing the given number to clearRight function
# to remove the clear the rightmost setbit
print("The given number after removing right most set bit : ")
print(clearRight(numb))

Output:

The given number before removing right most set bit : 
19
The given number after removing right most set bit : 
18

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