1806. Minimum Number of Operations to Reinitialize a Permutation: AC

src/solution/mod.rs

@@ -1361,3 +1361,4 @@ mod s1801_number_of_orders_in_the_backlog;
 mod s1802_maximum_value_at_a_given_index_in_a_bounded_array;
 mod s1803_count_pairs_with_xor_in_a_range;
 mod s1805_number_of_different_integers_in_a_string;
+mod s1806_minimum_number_of_operations_to_reinitialize_a_permutation;

src/solution/s1806_minimum_number_of_operations_to_reinitialize_a_permutation.rs

@@ -0,0 +1,90 @@
+/**
+ * [1806] Minimum Number of Operations to Reinitialize a Permutation
+ *
+ * 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: perm = [0,1] initially.
+ * After the 1^st operation, perm = [0,1]
+ * So it takes only 1 operation.
+ *
+ * Example 2:
+ *
+ * Input: n = 4
+ * Output: 2
+ * Explanation: perm = [0,1,2,3] initially.
+ * After the 1^st operation, perm = [0,2,1,3]
+ * After the 2^nd operation, perm = [0,1,2,3]
+ * So it takes only 2 operations.
+ *
+ * Example 3:
+ *
+ * Input: n = 6
+ * Output: 4
+ *
+ *  
+ * Constraints:
+ *
+ * 	2 <= n <= 1000
+ * 	n​​​​​​ is even.
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/
+// discuss: https://leetcode.com/problems/minimum-number-of-operations-to-reinitialize-a-permutation/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn reinitialize_permutation(n: i32) -> i32 {
+        let mut count = 1;
+        let mut i = n / 2;
+
+        while i != 1 {
+            if i % 2 == 0 {
+                i /= 2;
+            } else {
+                i = n / 2 + (i - 1) / 2;
+            }
+            count += 1;
+        }
+
+        count
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1806_example_1() {
+        let n = 2;
+
+        let result = 1;
+
+        assert_eq!(Solution::reinitialize_permutation(n), result);
+    }
+
+    #[test]
+    fn test_1806_example_2() {
+        let n = 4;
+
+        let result = 2;
+
+        assert_eq!(Solution::reinitialize_permutation(n), result);
+    }
+}