**Objective: **Given a number, write and algorithm to find the right most unset bit or zero bit in it (In binary representation).

This problem is similar to: Find the right most set bit of a number

**Example:**

Number : 11
Binary representation: 1 0 1 1
Position of right most unset bit: 2
Number : 6
Binary representation: 1 1 0
Position of right most unset bit: 0
Number : 13
Binary representation: 1 1 0 1
Position of right most unset bit: 1

**Approach:**

If N is a number then the expression below will give the right most unset bit.

~N & (N +1)

- Let’s dig little deep to see how this expression will work.
- We know that N & ~N = 0
- If we add 1 from the number, it will make most unset bit to 1, if there are any set bits in the right side of unset bit, those bit will become zero.
- (example : 1 0 1 1 + 0 0 0 1 = 1 1 0 0)
- So if we negate the original number it will make the right most unset bit to 1.
- Now ~N & (N+1) will make all the bits 0 but the right most unset bit of a number.

**Example**:

Say N =11, so N = 1 0 1 1
N + 1 = 1 1 0 0
~N = 0 1 0 1
~N & (N+1) = 0 1 0 0 => 2^{nd} bit (assuming right most index is 0)

**Code**:

**Output**:

Right most Unset bit :2.0

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.

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