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);
        }
    }
    
  • // 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
            
    
    

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