1806

1806. Minimum Number of Operations to Reinitialize a Permutation (Medium)

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.

Related Topics:
Array, Greedy

Solution 1.

// OJ: https://leetcode.com/contest/weekly-contest-234/problems/minimum-number-of-operations-to-reinitialize-a-permutation/
// Author: github.com/lzl124631x
// 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;
    }
};