Toggle the last m bits

2

Given a non-negative number n. The problem is to toggle the last m bits in the binary representation of n. A toggle operation flips a bit 0 to 1 and a bit 1 to 0.

Constraint: 1 <= m <= n.

Examples:

Input : n = 21, m = 2
Output : 22
(21)10 = (10101)2
(22)10 = (10110)2
The last two bits in the binary
representation of 21 are toggled.

Input : n = 107, m = 4
Output : 100

Approach: Following are the steps:

  1. Calculate num = (1 << m) – 1. This will produce a number num having m number of bits and all will be set.
  2. Now, perform n = n ^ num. This will toggle the last m bits in n.
// C++ implementation to toggle the last m bits
#include <bits/stdc++.h>

using namespace std;

// function to toggle the last m bits
unsigned int toggleLastMBits(unsigned int n,
                             unsigned int m)
{
    // calculating a number 'num' having 'm' bits
    // and all are set
    unsigned int num = (1 << m) - 1;

    // toggle the last m bits and return the number
    return (n ^ num);
}

// Driver program to test above
int main()
{
    unsigned int n = 107;
    unsigned int m = 4;
    cout << toggleLastMBits(n, m);
    return 0;
}

Output:

100

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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