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

# 1806. Minimum Number of Operations to Reinitialize a Permutation

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);
}
}

• // OJ: https://leetcode.com/contest/weekly-contest-234/problems/minimum-number-of-operations-to-reinitialize-a-permutation/
// Time: O(logN)
// Space: O(1)
class Solution {
public:
int reinitializePermutation(int n) {
int i = 1, cnt = 0;
while (true) {
++cnt;
if (i < n / 2) {
i *= 2;
} else  {
i = n - i - 1;
i *= 2;
i = n - i - 1;
}
if (i == 1) break;
}
return cnt;
}
};

• # 1806. Minimum Number of Operations to Reinitialize a Permutation
# https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/

class Solution:
def reinitializePermutation(self, n: int) -> int:
arr = [i for i in range(n)]
desired = [i for i in range(n)]

count = 0

while count == 0 or arr != desired:
tmp = [0] * n

for i in range(n):
if i % 2 == 0:
tmp[i] = arr[i // 2]
else:
tmp[i] = arr[n // 2 + (i-1) // 2]

arr = tmp
count += 1

return count