Merge branch 'main' of https://github.com/JuneLin2001/problem-solving

Author: JuneLin2001

docs/LeetCode/3010. Divide an Array Into Subarrays With Minimum Cost I/README.mdx

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+---
+title: 3010. Divide an Array Into Subarrays With Minimum Cost I
+description: You are given an array of integers nums of length n. The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3. You need to divide nums into 3 disjoint contiguous subarrays. Return the minimum possible sum of the cost of these subarrays.
+keywords:
+  [
+    LeetCode,
+    3010. Divide an Array Into Subarrays With Minimum Cost I,
+    Easy,
+    Array,
+    Sorting,
+    Enumeration,
+  ]
+tags: [LeetCode, Easy, Array, Sorting, Enumeration]
+---
+
+# [{frontMatter.title}](https://leetcode.com/problems/divide-an-array-into-subarrays-with-minimum-cost-i/)
+
+import Tabs from "@theme/Tabs";
+import TabItem from "@theme/TabItem";
+import CodeBlock from "@theme/CodeBlock";
+import DifficultyBadge from "@site/src/components/Badges/DifficultyBadge";
+import Solution from "!!raw-loader!./solution.js";
+import Description from "./_Description.md";
+import Examples from "./_Examples.md";
+
+<DifficultyBadge difficulty="Easy" />
+
+<Tabs>
+  <TabItem value="description" label="題目描述" default>
+    <details open>
+      <summary>Description</summary>
+        <Description />
+        <Examples />
+    </details> 
+  </TabItem>
+
+  <TabItem value="solution" label="解答">
+   ## Solution
+    <CodeBlock language="js">{Solution}</CodeBlock>
+  </TabItem>
+</Tabs>
+
+## 解題思路
+
+因為 nums 的第一個項目一定會包含在最終答案之中，所以第一步就是先加上 `nums[0]`  
+接著移除 `nums[0]` 後，再對剩下來的進行排序  
+排序後把最小和第二小再加進 `result` 就是答案了！
+
+## 心得
+
+好像有點鑽漏洞的解法

docs/LeetCode/3010. Divide an Array Into Subarrays With Minimum Cost I/_Description.md

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+You are given an array of integers `nums` of length `n`.
+
+The cost of an array is the value of its first element. For example, the cost of `[1,2,3]` is `1` while the cost of `[3,4,1]` is `3`.
+
+You need to divide `nums` into `3` disjoint contiguous subarrays.
+
+Return the minimum possible sum of the cost of these subarrays.

docs/LeetCode/3010. Divide an Array Into Subarrays With Minimum Cost I/_Examples.md

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+#### Example 1:
+
+> **Input:** [1,2,3,12]  
+> **Output:** 6  
+> **Explanation:**  
+> The best possible way to form 3 subarrays is: [1], [2], and [3,12] at a total cost of 1 + 2 + 3 = 6.  
+> The other possible ways to form 3 subarrays are:
+>
+> - [1], [2,3], and [12] at a total cost of 1 + 2 + 12 = 15.
+> - [1,2], [3], and [12] at a total cost of 1 + 3 + 12 = 16.
+
+#### Example 2:
+
+> **Input:** [5,4,3]  
+> **Output:** 12  
+> **Explanation:** The best possible way to form 3 subarrays is: [5], [4],  
+> and [3] at a total cost of 5 + 4 + 3 = 12.  
+> It can be shown that 12 is the minimum cost achievable.
+
+> #### Example 3:
+
+> **Input:** [10,3,1,1]  
+> **Output:** 12  
+> **Explanation:** The best possible way to form 3 subarrays is: [10,3],  
+> [1], and [1] at a total cost of 10 + 1 + 1 = 12.  
+> It can be shown that 12 is the minimum cost achievable.
+
+## Constraints:
+
+- `1 <= dimensions.length <= 100`
+- `dimensions[i].length == 2`
+- `1 <= dimensions[i][0], dimensions[i][1] <= 100`
+
+Input: nums =
+Output: 12
+Explanation:
+Example 3:
+
+Input: nums = [10,3,1,1]
+Output: 12
+Explanation: The best possible way to form 3 subarrays is: [10,3], [1], and [1] at a total cost of 10 + 1 + 1 = 12.
+It can be shown that 12 is the minimum cost achievable.

docs/LeetCode/3010. Divide an Array Into Subarrays With Minimum Cost I/solution.js

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+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumCost = function (nums) {
+  let result = 0;
+  result += nums[0];
+  nums.splice(0, 1);
+
+  nums.sort((a, b) => a - b);
+
+  result += nums[0] + nums[1];
+
+  return result;
+};