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.

1806. Minimum Number of Operations to Reinitialize a Permutation
``````struct Solution;

impl Solution {
fn reinitialize_permutation(n: i32) -> i32 {
if n == 2 {
return 1;
}
let mut res = 0;
let mut i = 1;
while res == 0 || i != 1 {
if i % 2 == 0 {
i /= 2;
} else {
i = (n - 1 + i) / 2;
}
res += 1;
}
res
}
}

#[test]
fn test() {
let n = 2;
let res = 1;
assert_eq!(Solution::reinitialize_permutation(n), res);
let n = 4;
let res = 2;
assert_eq!(Solution::reinitialize_permutation(n), res);
let n = 6;
let res = 4;
assert_eq!(Solution::reinitialize_permutation(n), res);
}
``````