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: 4

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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