feat: solve NO.2009

BaffinLee · BaffinLee · commit d0d25217ac13 · 2023-10-10T23:38:24.000+08:00
diff --git a/2001-2100/2009. Minimum Number of Operations to Make Array Continuous.md b/2001-2100/2009. Minimum Number of Operations to Make Array Continuous.md
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+# 2009. Minimum Number of Operations to Make Array Continuous
+
+- Difficulty: Hard.
+- Related Topics: Array, Binary Search.
+- Similar Questions: Longest Repeating Character Replacement, Continuous Subarray Sum, Moving Stones Until Consecutive II, Minimum One Bit Operations to Make Integers Zero, Minimum Adjacent Swaps for K Consecutive Ones.
+
+## Problem
+
+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****.
+
+ 
+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 <= 105`
+	
+- `1 <= nums[i] <= 109`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minOperations = function(nums) {
+    nums.sort((a, b) => a - b);
+
+    var uniqueNums = [];
+    for (var i = 0; i < nums.length; i++) {
+        if (nums[i] !== nums[i - 1]) uniqueNums.push(nums[i]);
+    }
+
+    var left = 0;
+    var maxUniqueNumsInRange = 0;
+    for (var right = 0; right < uniqueNums.length; right++) {
+        while (uniqueNums[right] - uniqueNums[left] >= nums.length) left++;
+        maxUniqueNumsInRange = Math.max(maxUniqueNumsInRange, right - left + 1);
+    }
+
+    return nums.length - maxUniqueNumsInRange;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * log(n)).
+* Space complexity : O(n).
