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New Problem Solution -"Minimum Number of Operations to Reinitialize a…
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… Permutation"
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haoel committed Mar 28, 2021
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| # | Title | Solution | Difficulty |
|---| ----- | -------- | ---------- |
|1806|[Minimum Number of Operations to Reinitialize a Permutation](https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/) | [C++](./algorithms/cpp/minimumNumberOfOperationsToReinitializeAPermutation/MinimumNumberOfOperationsToReinitializeAPermutation.cpp)|Medium|
|1805|[Number of Different Integers in a String](https://leetcode.com/problems/number-of-different-integers-in-a-string/) | [C++](./algorithms/cpp/numberOfDifferentIntegersInAString/NumberOfDifferentIntegersInAString.cpp)|Easy|
|1803|[Count Pairs With XOR in a Range](https://leetcode.com/problems/count-pairs-with-xor-in-a-range/) | [C++](./algorithms/cpp/countPairsWithXorInARange/CountPairsWithXorInARange.cpp)|Hard|
|1802|[Maximum Value at a Given Index in a Bounded Array](https://leetcode.com/problems/maximum-value-at-a-given-index-in-a-bounded-array/) | [C++](./algorithms/cpp/maximumValueAtAGivenIndexInABoundedArray/MaximumValueAtAGivenIndexInABoundedArray.cpp)|Medium|
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// 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;

}
};

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