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New Problem Solution -"Minimum Number of Operations to Reinitialize a…
… Permutation"
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...rationsToReinitializeAPermutation/MinimumNumberOfOperationsToReinitializeAPermutation.cpp
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// Source : https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/ | ||
// Author : Hao Chen | ||
// Date : 2021-03-28 | ||
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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: 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. | ||
******************************************************************************************************/ | ||
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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; | ||
} | ||
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return cnt; | ||
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} | ||
}; |