// Source : https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/
// Author : Hao Chen
// Date   : 2021-03-28

/***************************************************************************************************** 
 *
 * 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: prem = [0,1] initially.
 * After the 1^st operation, prem = [0,1]
 * So it takes only 1 operation.
 * 
 * Example 2:
 * 
 * Input: n = 4
 * Output: 2
 * Explanation: prem = [0,1,2,3] initially.
 * After the 1^st operation, prem = [0,2,1,3]
 * After the 2^nd operation, prem = [0,1,2,3]
 * So it takes only 2 operations.
 * 
 * Example 3:
 * 
 * Input: n = 6
 * Output: 4
 * 
 * Constraints:
 * 
 * 	2 <= n <= 1000
 * 	n is even.
 ******************************************************************************************************/

class Solution {
private: 
    bool check(vector<int>& a) {
        for(int i=0; i<a.size(); i++) {
            if (a[i] != i) return false;
        }
        return true;
    }
public:
    int reinitializePermutation(int n) {
        vector<int> perm(n);
        vector<int> arr(n);
        for(int i=0; i<n; i++) {
            perm[i] = i;
        }
        int cnt = 0;
        while(1){
            cnt++;
            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];
            }
            if (check(arr)) break;
            perm = arr;
        }
        
        return cnt;
        
    }
};