// 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& a) { for(int i=0; i perm(n); vector arr(n); for(int i=0; i