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

**Example:**

Number : 1
Binary representation: 1
Position of right most set bit: 1
Number : 6
Binary representation: 1 1 0
Position of right most set bit: 2
Number : 11
Binary representation: 1 0 1 1
Position of right most set bit: 1
Number : 24
Binary representation: 1 1 0 0 0
Position of right most set bit: 3

**Approach:**

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

N & ~ (N -1)

- Let’s dig little deep to see how this expression will work.
- We know that N & ~N = 0
- If we subtract 1 from the number, it will be subtracted from the right most set bit and that bit will be become 0.
- So if we negate the remaining number from step above then that bit will become 1.
- Now N & ~(N-1) will make all the bits 0 but the right most set bit of a number.

**Example**:

Say N =10, so N = 1 0 1 0, then ~N = 0 1 1 0 => N & ~ N =0
N – 1 = 1 0 1 0 – 0 0 0 1 = 1 0 0 1
~(N-1) = 0 1 1 0
N & ~(N-1) = 0 0 1 0 => 2^{nd} bit

**Code**:

**Output**:

Right most set bit for 1 is : 1
Position: 1.0

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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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