Set the rightmost unset bit


Given a non-negative number n. The problem is to set the rightmost unset bit in the binary representation of n. If there are no unset bits, then just leave the number as it is.


Input : 21
Output : 23
(21)10 = (10101)2
Rightmost unset bit is at position 2(from right) as 
highlighted in the binary representation of 21.
(23)10 = (10111)2
The bit at position 2 has been set.

Input : 15
Output : 15

Approach: Following are the steps:

  1. If n = 0, return 1.
  2. If all bits of n are set, return n. Refer this post.
  3. Else perform bitwise not on the given number(operation equivalent to 1’s complement). Let it be num = ~n.
  4. Get the position of rightmost set bit of num. Let the position be pos.
  5. Return (1 << (pos – 1)) | n.
// C++ implementation to set the rightmost unset bit
#include <bits/stdc++.h>
using namespace std;

// function to find the position 
// of rightmost set bit
int getPosOfRightmostSetBit(int n)
    return log2(n&-n)+1;

int setRightmostUnsetBit(int n)
    // if n = 0, return 1
    if (n == 0)
        return 1;
    // if all bits of 'n' are set
    if ((n & (n + 1)) == 0)    
        return n;
    // position of rightmost unset bit in 'n'
    // passing ~n as argument
    int pos = getPosOfRightmostSetBit(~n);    
    // set the bit at position 'pos'
    return ((1 << (pos - 1)) | n);

// Driver program to test above
int main()
    int n = 21;
    cout << setRightmostUnsetBit(n);
    return 0;



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