feat: add solution 2009. Minimum Number of Operations to Make Array Continuous

Author: tzuyi0817

solutions/2009. Minimum Number of Operations to Make Array Continuous/README.md

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+# [2009. Minimum Number of Operations to Make Array Continuous](https://leetcode.com/problems/minimum-number-of-operations-to-make-array-continuous)
+
+## Description
+
+<div class="elfjS" data-track-load="description_content"><p>You are given an integer array <code>nums</code>. In one operation, you can replace <strong>any</strong> element in <code>nums</code> with <strong>any</strong> integer.</p>
+
+<p><code>nums</code> is considered <strong>continuous</strong> if both of the following conditions are fulfilled:</p>
+
+<ul>
+	<li>All elements in <code>nums</code> are <strong>unique</strong>.</li>
+	<li>The difference between the <strong>maximum</strong> element and the <strong>minimum</strong> element in <code>nums</code> equals <code>nums.length - 1</code>.</li>
+</ul>
+
+<p>For example, <code>nums = [4, 2, 5, 3]</code> is <strong>continuous</strong>, but <code>nums = [1, 2, 3, 5, 6]</code> is <strong>not continuous</strong>.</p>
+
+<p>Return <em>the <strong>minimum</strong> number of operations to make </em><code>nums</code><em> </em><strong><em>continuous</em></strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre><strong>Input:</strong> nums = [4,2,5,3]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong>&nbsp;nums is already continuous.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre><strong>Input:</strong> nums = [1,2,3,5,6]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong>&nbsp;One possible solution is to change the last element to 4.
+The resulting array is [1,2,3,5,4], which is continuous.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre><strong>Input:</strong> nums = [1,10,100,1000]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong>&nbsp;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.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+## Solutions
+
+**Solution: `Sliding Window`**
+
+- Time complexity: <em>O(nlogn)</em>
+- Space complexity: <em>O(n)</em>
+
+<p>&nbsp;</p>
+
+### **JavaScript**
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const minOperations = function (nums) {
+  const n = nums.length;
+  const uniqueNums = [...new Set(nums)];
+  let left = 0;
+
+  uniqueNums.sort((a, b) => a - b);
+
+  for (let index = 0; index < uniqueNums.length; index++) {
+    if (uniqueNums[index] - uniqueNums[left] > n - 1) {
+      left += 1;
+    }
+  }
+
+  return n - (uniqueNums.length - left);
+};
+```

solutions/2009. Minimum Number of Operations to Make Array Continuous/solution.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+const minOperations = function (nums) {
+  const n = nums.length;
+  const uniqueNums = [...new Set(nums)];
+  let left = 0;
+
+  uniqueNums.sort((a, b) => a - b);
+
+  for (let index = 0; index < uniqueNums.length; index++) {
+    if (uniqueNums[index] - uniqueNums[left] > n - 1) {
+      left += 1;
+    }
+  }
+
+  return n - (uniqueNums.length - left);
+};