Elimination Game

There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.

Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.

We keep repeating the steps again, alternating left to right and right to left, until a single number remains.

Find the last number that remains starting with a list of length n.

Example:

Input:
n = 9,
1 2 3 4 5 6 7 8 9
2 4 6 8
2 6
6

Output:
6

This is the second time I work on the problem, and finally I find a solution that is easy to understand. We know that every time we eliminate half of the numbers, the difference between two consecutive numbers doubles. Second, the head of the numbers will only change when we eliminate numbers from left or from right and we have odd number of numbers to eliminate. Then we can track the head, update it based on the above rule until we have only one number left.

For example,

1 2 3 4 5 6 7 8 9 10 => first round, head to 2, step to 2
2 4 6 8 10 => head to 4, step to 4
4 8 => return 4

public int lastRemaining(int n) {
        int head = 1;
        int step = 1;
        boolean left = true;
        while (n > 1) {
            if (left || n % 2 == 1) {
                head += step;
            }
            step *= 2;
            n /= 2;
            left = !left;
        }
        return head;
    }