feat: solve No.910

BaffinLee · BaffinLee · commit 11e0102549fb · 2023-08-22T23:55:39.000+08:00
diff --git a/901-1000/910. Smallest Range II.md b/901-1000/910. Smallest Range II.md
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+# 910. Smallest Range II
+
+- Difficulty: Medium.
+- Related Topics: Array, Math, Greedy, Sorting.
+- Similar Questions: .
+
+## Problem
+
+You are given an integer array `nums` and an integer `k`.
+
+For each index `i` where `0 <= i < nums.length`, change `nums[i]` to be either `nums[i] + k` or `nums[i] - k`.
+
+The **score** of `nums` is the difference between the maximum and minimum elements in `nums`.
+
+Return **the minimum **score** of **`nums`** after changing the values at each index**.
+
+ 
+Example 1:
+
+```
+Input: nums = [1], k = 0
+Output: 0
+Explanation: The score is max(nums) - min(nums) = 1 - 1 = 0.
+```
+
+Example 2:
+
+```
+Input: nums = [0,10], k = 2
+Output: 6
+Explanation: Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.
+```
+
+Example 3:
+
+```
+Input: nums = [1,3,6], k = 3
+Output: 3
+Explanation: Change nums to be [4, 6, 3]. The score is max(nums) - min(nums) = 6 - 3 = 3.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= nums.length <= 104`
+	
+- `0 <= nums[i] <= 104`
+	
+- `0 <= k <= 104`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+var smallestRangeII = function(nums, k) {
+    nums.sort((a, b) => a - b);
+    var n = nums.length;
+    var res = nums[n - 1] - nums[0];
+    for (var i = 0; i < n - 1; i++) {
+        var low = Math.min(nums[0] + k, nums[i + 1] - k);
+        var high = Math.max(nums[n - 1] - k, nums[i] + k);
+        res = Math.min(res, high - low);
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).
