New Problem Solution -"Minimum Number of Operations to Reinitialize a Permutation"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | 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|

algorithms/cpp/minimumNumberOfOperationsToReinitializeAPermutation/MinimumNumberOfOperationsToReinitializeAPermutation.cpp

@@ -0,0 +1,77 @@
+// 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;
+        
+    }
+};