Merge pull request #3611 from Ishitamukherjee2004/new-branch

Solution of 2717 (LC No.) is added

Author: ajay-dhangar

dsa-solutions/lc-solutions/2700-2799/2717-Semi_ordered-Permutation.md

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+---
+id: semi-ordered-permutation
+title: Semi-Ordered Permutation
+sidebar_label: 2717-Semi_ordered-Permutation
+tags:
+  - Array
+  - Simulation
+
+description: "This is a solution to the  2717."
+---
+
+## Problem Description
+You are given a 0-indexed permutation of `n` integers `nums`.
+
+A permutation is called semi-ordered if the first number equals `1` and the last number equals `n`. You can perform the below operation as many times as you want until you make `nums` a semi-ordered permutation:
+
+Pick two adjacent elements in `nums`, then swap them.
+Return the minimum number of operations to make `nums` a semi-ordered permutation.
+
+A permutation is a sequence of integers from `1` to `n` of length `n` containing each number exactly once.
+
+
+### Example
+
+**Example 1:**
+
+
+```
+Input: nums = [2,1,4,3]
+Output: 2
+Explanation: We can make the permutation semi-ordered using these sequence of operations: 
+1 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3].
+2 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4].
+It can be proved that there is no sequence of less than two operations that make nums a semi-ordered permutation.
+```
+**Example 2:**
+```
+Input: nums = [2,4,1,3]
+Output: 3
+Explanation: We can make the permutation semi-ordered using these sequence of operations:
+1 - swap i = 1 and j = 2. The permutation becomes [2,1,4,3].
+2 - swap i = 0 and j = 1. The permutation becomes [1,2,4,3].
+3 - swap i = 2 and j = 3. The permutation becomes [1,2,3,4].
+It can be proved that there is no sequence of less than three operations that make nums a semi-ordered permutation.
+```
+### Constraints
+
+- `2 <= nums.length == n <= 50`
+
+## Solution Approach
+
+### Intuition:
+
+To efficiently determine the Semi-Ordered Permutation
+## Solution Implementation
+
+### Code In Different Languages:
+
+<Tabs>
+  <TabItem value="JavaScript" label="JavaScript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```javascript
+class Solution {
+  semiOrderedPermutation(nums) {
+    let n = nums.length;
+    let left = 0, right = n - 1;
+    let operations = 0;
+    while (left < right) {
+      if (nums[left] === left + 1) {
+        left++;
+      } else if (nums[right] === n - right) {
+        right--;
+      } else {
+        [nums[left], nums[right]] = [nums[right], nums[left]];
+        operations++;
+      }
+    }
+    return operations;
+  }
+}
+
+    ```
+
+  </TabItem>
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@Ishitamukherjee2004"/> 
+   ```typescript
+  class Solution {
+  semiOrderedPermutation(nums: number[]): number {
+    let n = nums.length;
+    let left = 0, right = n - 1;
+    let operations = 0;
+    while (left < right) {
+      if (nums[left] === left + 1) {
+        left++;
+      } else if (nums[right] === n - right) {
+        right--;
+      } else {
+        [nums[left], nums[right]] = [nums[right], nums[left]];
+        operations++;
+      }
+    }
+    return operations;
+  }
+}
+
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python"> 
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```python
+class Solution:
+    def semiOrderedPermutation(self, nums: List[int]) -> int:
+        n = len(nums)
+        left, right = 0, n - 1
+        operations = 0
+        while left < right:
+            if nums[left] == left + 1:
+                left += 1
+            elif nums[right] == n - right:
+                right -= 1
+            else:
+                nums[left], nums[right] = nums[right], nums[left]
+                operations += 1
+        return operations
+
+    ```
+
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```java
+   
+public class Solution {
+    public int semiOrderedPermutation(int[] nums) {
+        int n = nums.length;
+        int left = 0, right = n - 1;
+        int operations = 0;
+        while (left < right) {
+            if (nums[left] == left + 1) {
+                left++;
+            } else if (nums[right] == n - right) {
+                right--;
+            } else {
+                int temp = nums[left];
+                nums[left] = nums[right];
+                nums[right] = temp;
+                operations++;
+            }
+        }
+        return operations;
+    }
+}
+
+
+    ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+   ```cpp
+  class Solution {
+public:
+    int semiOrderedPermutation(vector<int>& nums) {
+    int n = nums.size();
+    int left = 0, right = n - 1;
+    int operations = 0;
+    while (left < right) {
+        if (nums[left] == left + 1) {
+            left++;
+        } else if (nums[right] == n - right) {
+            right--;
+        } else {
+            swap(nums[left], nums[right]);
+            operations++;
+        }
+    }
+    
+    return operations;
+}
+
+};
+   ```
+</TabItem> 
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $$O(n)$$
+- Space Complexity: $$O(1)$$
+- The time complexity is $$O(n)$$ where n is the length of the input array nums. This is because the method iterates through the array once, performing a constant amount of work for each element.
+- The space complexity is $$O(1)$$ because we are not using any extra space.