Merge branch 'dev-3' of https://github.com/CodeHarborHub/codeharborhub into dev-3

Author: ajay-dhangar

dsa-solutions/lc-solutions/2000-2099/2009-minimum-number-of-operations-to-make-array-continuous.md

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+---
+id: minimum-number-of-operations-to-make-array-continuous
+title: Minimum Number of Operations to Make Array Continuous
+sidebar_label: 2009 Minimum Number of Operations to Make Array Continuous
+tags:
+  - Array
+  - Sorting
+  - Sliding Window
+  - LeetCode
+description: "This is a solution to the Minimum Number of Operations to Make Array Continuous problem on LeetCode."
+sidebar_position: 2009
+---
+
+## Problem Description
+
+You are given an integer array `nums`. In one operation, you can replace any element in `nums` with any integer.
+
+`nums` is considered continuous if both of the following conditions are fulfilled:
+
+1. All elements in `nums` are unique.
+2. The difference between the maximum element and the minimum element in `nums` equals `nums.length - 1`.
+
+For example, `nums = [4, 2, 5, 3]` is continuous, but `nums = [1, 2, 3, 5, 6]` is not continuous.
+
+Return the minimum number of operations to make `nums` continuous.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: nums = [4,2,5,3]
+Output: 0
+Explanation: nums is already continuous.
+```
+
+**Example 2:**
+
+```
+Input: nums = [1,2,3,5,6]
+Output: 1
+Explanation: One possible solution is to change the last element to 4.
+The resulting array is [1,2,3,5,4], which is continuous.
+```
+
+**Example 3:**
+
+```
+Input: nums = [1,10,100,1000]
+Output: 3
+Explanation: One possible solution is to:
+
+Change the second element to 2.
+Change the third element to 3.
+Change the fourth element to 4.
+The resulting array is [1,2,3,4], which is continuous.
+```
+
+
+### Constraints
+
+- `1 <= nums.length <= 10^5`
+- `1 <= nums[i] <= 10^9`
+
+### Approach
+
+To solve this problem:
+1. Sort the array `nums` to facilitate finding the longest continuous subsequence.
+2. Use a sliding window to find the longest subsequence that satisfies the conditions of being continuous.
+3. Calculate the minimum number of operations as the difference between the length of `nums` and the length of the longest continuous subsequence found.
+
+#### C++ Solution
+
+```cpp
+class Solution {
+public:
+    int minOperations(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        nums.erase(unique(nums.begin(), nums.end()), nums.end());
+        
+        int n = nums.size();
+        int ans = n;
+        for (int i = 0, j = 0; i < n; ++i) {
+            while (j < n && nums[j] <= nums[i] + n - 1) {
+                ++j;
+            }
+            ans = min(ans, n - (j - i));
+        }
+        
+        return ans;
+    }
+};
+```