README.md

1806. Minimum Number of Operations to Reinitialize a Permutation

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.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn reinitialize_permutation(n: i32) -> i32 {
        let n = n as usize;
        let mut perm = (0..n).collect::<Vec<_>>();

        for ret in 1..=n as i32 {
            let mut arr = vec![0; n];

            for i in 0..n {
                if i % 2 == 0 {
                    arr[i] = perm[i / 2];
                } else {
                    arr[i] = perm[n / 2 + (i - 1) / 2];
                }
            }

            perm = arr;

            if perm.iter().enumerate().all(|(i, &x)| i == x) {
                return ret;
            }
        }

        unreachable!()
    }
}