Add solution #2009

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,685 LeetCode solutions in JavaScript
+# 1,686 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1541,6 +1541,7 @@
 2006|[Count Number of Pairs With Absolute Difference K](./solutions/2006-count-number-of-pairs-with-absolute-difference-k.js)|Easy|
 2007|[Find Original Array From Doubled Array](./solutions/2007-find-original-array-from-doubled-array.js)|Medium|
 2008|[Maximum Earnings From Taxi](./solutions/2008-maximum-earnings-from-taxi.js)|Medium|
+2009|[Minimum Number of Operations to Make Array Continuous](./solutions/2009-minimum-number-of-operations-to-make-array-continuous.js)|Hard|
 2011|[Final Value of Variable After Performing Operations](./solutions/2011-final-value-of-variable-after-performing-operations.js)|Easy|
 2016|[Maximum Difference Between Increasing Elements](./solutions/2016-maximum-difference-between-increasing-elements.js)|Easy|
 2017|[Grid Game](./solutions/2017-grid-game.js)|Medium|

solutions/2009-minimum-number-of-operations-to-make-array-continuous.js

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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/
+ * Difficulty: Hard
+ *
+ * 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:
+ * - All elements in nums are unique.
+ * - 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.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minOperations = function(nums) {
+  const n = nums.length;
+  const uniqueNums = [...new Set(nums)].sort((a, b) => a - b);
+  let result = n;
+
+  for (let i = 0, j = 0; i < uniqueNums.length; i++) {
+    while (j < uniqueNums.length && uniqueNums[j] - uniqueNums[i] <= n - 1) {
+      j++;
+    }
+    result = Math.min(result, n - (j - i));
+  }
+
+  return result;
+};