Given a non-negative number n. The problem is to set the rightmost unset bit in the binary representation of n
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 : 2 Output : 3
One method is discussed in this article
This post discusses another method.
Let the input number be n. n+1 would have all the bits flipped after the rightmost unset bit (including the unset bit). So, doing n|(n+1) would give us the required result.
The number after setting the rightmost unset bit 7
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- Set the rightmost unset bit
- Get the position of rightmost unset bit
- Set the rightmost off bit
- Turn off the rightmost set bit
- Position of rightmost set bit
- Turn off the rightmost set bit | Set 2
- Position of rightmost different bit
- Number formed by the rightmost set bit in N
- Unset the last m bits
- Set the Left most unset bit
- Position of rightmost bit with first carry in sum of two binary
- Position of rightmost common bit in two numbers
- Unset bits in the given range
- Check whether the bit at given position is set or unset
- Count unset bits in a range
- Check whether all the bits are unset in the given range or not
- Count unset bits of a number
- Check whether all the bits are unset in the given range
- Absolute difference between set and unset bit count in N
- Number formed by flipping all bits to the left of rightmost set bit
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