Medium
You are given an even integer n. You initially have a permutation perm of size n where perm[i] == i (0-indexed).
In one operation, you will create a new array arr, and for each i:
i % 2 == 0, then arr[i] = perm[i / 2].i % 2 == 1, then arr[i] = perm[n / 2 + (i - 1) / 2].You will then assign arr to perm.
Return the minimum non-zero number of operations you need to perform on perm to return the permutation to its initial value.
Example 1:
Input: n = 2
Output: 1
Explanation: perm = [0,1] initially.
After the 1st operation, perm = [0,1]
So it takes only 1 operation.
Example 2:
Input: n = 4
Output: 2
Explanation: perm = [0,1,2,3] initially.
After the 1st operation, perm = [0,2,1,3]
After the 2nd operation, perm = [0,1,2,3]
So it takes only 2 operations.
Example 3:
Input: n = 6
Output: 4
Constraints:
2 <= n <= 1000n is even.public class Solution {
    public int reinitializePermutation(int n) {
        final int factor = n - 1;
        if (factor < 2) {
            return 1;
        }
        int powerOfTwo = 2;
        int ops = 1;
        while (powerOfTwo != 1) {
            powerOfTwo = ((powerOfTwo << 1) % factor);
            ops++;
        }
        return ops;
    }
}