1806. Minimum Number of Operations to Reinitialize a Permutation

Problem

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:

• If i % 2 == 0, then arr[i] = perm[i / 2].

• If 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 <= 1000

• n​​​​​​ is even.

Solution (Java)

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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).