Formatted question description: https://leetcode.ca/all/1806.html

1806. Minimum Number of Operations to Reinitialize a Permutation

Level

Medium

Description

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: nums = [0,1] initially.

After the 1st operation, nums = [0,1]

So it takes only 1 operation.

Example 2:

Input: n = 4

Output: 2

Explanation: nums = [0,1,2,3] initially.

After the 1st operation, nums = [0,2,1,3]

After the 2nd operation, nums = [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

Simply simulate the process. Create two arrays init of perm, each of which has length n, and initialize init[i] = perm[i] = i for 0 <= i < n. Then update perm each time until perm equals init, and return the number of operations.

class Solution {
    public int reinitializePermutation(int n) {
        int[] init = new int[n];
        int[] perm = new int[n];
        for (int i = 0; i < n; i++) {
            init[i] = i;
            perm[i] = i;
        }
        int ops = 0;
        while (true) {
            ops++;
            update(n, perm);
            if (Arrays.equals(perm, init))
                break;
        }
        return ops;
    }

    public void update(int n, int[] perm) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            if (i % 2 == 0)
                arr[i] = perm[i / 2];
            else
                arr[i] = perm[n / 2 + (i - 1) / 2];
        }
        System.arraycopy(arr, 0, perm, 0, n);
    }
}

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