Toggle the last m bits


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.


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;



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