Lucky alive person in a circle | Set - 2

Given that N person (numbered 1 to N) standing to form a circle. They all have the gun in their hand which is pointed to their leftmost Partner. 

Everyone shoots such that 1 shoots 2, 3 shoots 4, 5 shoots 6 .... (N-1)the shoot N (if N is even otherwise N shoots 1). 
Again on the second iteration, they shoot the rest of the remains as above mentioned logic (now for n as even, 1 will shoot to 3, 5 will shoot to 7, and so on). 

The task is to find which person is the luckiest(didn't die).

Examples: 

Input: N = 3 
Output: 3 
As N = 3 then 1 will shoot 2, 3 will shoot 1 hence 3 is the luckiest person.

Input: N = 8 
Output: 1 
Here as N = 8, 1 will shoot 2, 3 will shoot 4, 5 will shoot 6, 7 will shoot 8, Again 1 will shoot 3, 5 will shoot 7, Again 1 will shoot 5 and hence 1 is the luckiest person.

This problem has already been discussed in Lucky alive person in a circle | Code Solution to sword puzzle. In this post, a different approach is discussed.

Approach:

1. Take the Binary Equivalent of N.
2. Find its 1's compliment and convert its equal decimal number N`.
3. find |N - N`|.

Below is the implementation of the above approach:  

3

Alternate Shorter Implementation : 
The approach used here is same.

3

Alternate Implementation in O(1) : The approach used here is same, but the operations used are of constant time.

1

Another approach in O(1) : On the basis of the pattern that forms in given question, which is displayed in following table.

3