1806. Minimum Number of Operations to Reinitialize a Permutation

Problem Link

Description

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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

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Python3

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