refactor: separate mdx files

Author: JuneLin2001

docs/LeetCode/11. Container With Most Water/README.mdx

@@ -2,93 +2,76 @@
 title: 11. Container With Most Water
 description: You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).Find two lines that together with the x-axis form a container, such that the container contains the most water.Return the maximum amount of water a container can store.Notice that you may not slant the container.
 keywords:
-  [
-    LeetCode,
-    11. Container With Most Water,
-    Medium,
-    Array,
-    Two Pointers,
-    Greedy,
-  ]
+  [LeetCode, 11. Container With Most Water, Medium, Array, Two Pointers, Greedy]
+tags: [LeetCode, Medium, Array, Two Pointers, Greedy]
 ---
 
 import Tabs from "@theme/Tabs";
 import TabItem from "@theme/TabItem";
-import CodeBlock from '@theme/CodeBlock';
+import CodeBlock from "@theme/CodeBlock";
 import DifficultyBadge from "@site/src/components/Badges/DifficultyBadge";
-import Solution from '!!raw-loader!./solution.js';
+import Solution from "!!raw-loader!./solution.js";
+import Description from "./_Description.md";
+import Examples from "./_Examples.md";
 
 # [{frontMatter.title}](https://leetcode.com/problems/container-with-most-water)
 
 <DifficultyBadge difficulty="Medium" />
 
 <Tabs>
   <TabItem value="description" label="題目描述" default>
-  ## Description
-    You are given an integer array height of length `n`. There are n vertical lines drawn such that the two endpoints of the `ith` line are `1(i, 0)` and `(i, height[i])`.  
-
-    Find two lines that together with the x-axis form a container, such that the container contains the most water.  
-
-    Return the maximum amount of water a container can store.  
-
-    **Notice** that you may not slant the container.  
-
-
-  #### Example 1:
-    ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
-  > **Input:** height = [1,8,6,2,5,4,8,3,7]  
-  **Output:** 49    
-  **Explanation:** The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
-
-  #### Example 2:
-  > **Input:** height = [1,1]  
-  **Output:** 1  
-
+    <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
-    
+
     1. 先定義好 `left pointer` 和 `right pointer` 分別指向 height 的頭與尾，以及 mostWater 紀錄目前的最大水量值。
     ```js
       let left = 0;
       let right = height.length - 1;
       let mostWater = 0;
     ```
     2. 重複執行程式直到 right 和 left 重合。
-   ```js
-    while (right > left) {
 
-    }
-   ```
-    3. 使用 waterCurrently 算出目前水量，因為最大水兩高度是由兩邊中較短的決定的（假設一邊是 8 一邊是 5，水最高只能到 5，再多就會從水槽流出，而 right - left 是寬度，`長 * 寬`就是目前水量了。  
+```js
+while (right > left) {}
+```
+
+    3. 使用 waterCurrently 算出目前水量，因為最大水兩高度是由兩邊中較短的決定的（假設一邊是 8 一邊是 5，水最高只能到 5，再多就會從水槽流出，而 right - left 是寬度，`長 * 寬`就是目前水量了。
         若 waterCurrently > mostWater 則更新。
 
-   ```js
-    const waterCurrently = Math.min(height[left], height[right]) * (right - left); 
+```js
+const waterCurrently = Math.min(height[left], height[right]) * (right - left);
 
-    mostWater = Math.max(mostWater, waterCurrently);
-   ```
+mostWater = Math.max(mostWater, waterCurrently);
+```
 
     4. 如果左邊較短的話就把 `left pointer` 向右移動，反之則把 `right point` 向左移。
-   ```js
-    if (height[left] < height[right]) {
-      left++;
-    } else {
-      right--;
-    }
-   ```
-    5. 最後回傳 mostWater 就是答案了。
-   ```js
-     return mostWater;
-   ```
 
+```js
+if (height[left] < height[right]) {
+  left++;
+} else {
+  right--;
+}
+```
 
+    5. 最後回傳 mostWater 就是答案了。
+
+```js
+return mostWater;
+```
 
 ## 心得
+
     練習 Two Pointers，題目光看描述覺得很抽象，但有圖片輔助說明有好懂了不少！

docs/LeetCode/11. Container With Most Water/_Description.md

@@ -0,0 +1 @@
+Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

docs/LeetCode/11. Container With Most Water/_Examples.md

@@ -0,0 +1,12 @@
+#### Example 1:
+
+    ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
+
+> **Input:** height = [1,8,6,2,5,4,8,3,7]  
+>  **Output:** 49  
+>  **Explanation:** The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
+
+#### Example 2:
+
+> **Input:** height = [1,1]  
+>  **Output:** 1

docs/LeetCode/11. Container With Most Water/solution.js

@@ -22,5 +22,3 @@ var maxArea = function (height) {
 
   return mostWater;
 };
-
-console.log(maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7]));

docs/LeetCode/1346. Check If N and Its Double Exist/README.mdx

@@ -1,75 +1,58 @@
 ---
 title: 1346. Check If N and Its Double Exist
 description: Given an array arr of integers, check if there exist two indices i and j such that
-keywords: [LeetCode, 1346. Check If N and Its Double Exist, Easy, Array ,Hash Table, Two Pointers, Binary Search, Sorting]
+keywords:
+  [
+    LeetCode,
+    1346. Check If N and Its Double Exist,
+    Easy,
+    Array,
+    Hash Table,
+    Two Pointers,
+    Binary Search,
+    Sorting,
+  ]
+tags: [LeetCode, Easy, Array, Hash Table, Two Pointers, Binary Search, Sorting]
 ---
 
 # [{frontMatter.title}](https://leetcode.com/problems/check-if-n-and-its-double-exist)
 
 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>
-    ## Description
-
-Given an array `arr` of integers, check if there exist two indices `i` and `j` such that :  
-
- - `i != j`
- - `0 <= i, j < arr.length`
- - `arr[i] == 2 * arr[j]`
-
-#### Example 1:
-> **Input:**: arr = [10,2,5,3]  
-**Output:**: true  
-**Explanation:**: For i = 0 and j = 2, arr[i] == 10 == 2 * 5 == 2 * arr[j]
-
-#### Example 2:
->**Input:** arr = [3,1,7,11]  
-**Output:** false  
-**Explanation:** There is no i and j that satisfy the conditions.
-
-#### Constraints:
- - `1 <= numBottles <= 100`
- - `2 <= numExchange <= 100`
-
+    <details open>
+      <summary>Description</summary>
+        <Description />
+        <Examples />
+    </details> 
   </TabItem>
+
   <TabItem value="solution" label="解答">
    ## Solution
-```js
-/**
- * @param {number[]} arr
- * @return {boolean}
- */
-var checkIfExist = function (arr) {
-  const existSet = new Set();
-
-  for (let i = 0; i < arr.length; i++) {
-    if (existSet.has(arr[i] * 2) || existSet.has(arr[i] / 2)) {
-      return true;
-    }
-
-    existSet.add(arr[i]);
-  }
-
-  return false;
-};
-```
+    <CodeBlock language="js">{Solution}</CodeBlock>
   </TabItem>
 </Tabs>
 
-
 ## 解題思路
+
 使用一個 Set 來不重複的記錄下已經過的值，接著遍歷一遍 `arr`，共有兩種狀況會判斷為 true
- 1. 若當下的數字 `arr[i]` 的兩倍存在於 `Set` 之中的話就代表當下的數字符合 `j` 且 Set 中抓到的那個數字符合 `i`
- 2. 若當下的數字 `arr[i]` 的一半存在於 `Set` 之中的話就代表當下的數字符合 `i` 且 Set 中抓到的那個數字符合 `j`  
 
-檢查完兩者後若都不符合就把當前的 `arr[i]` 加進 `Set` 中  
+1.  若當下的數字 `arr[i]` 的兩倍存在於 `Set` 之中的話就代表當下的數字符合 `j` 且 Set 中抓到的那個數字符合 `i`
+2.  若當下的數字 `arr[i]` 的一半存在於 `Set` 之中的話就代表當下的數字符合 `i` 且 Set 中抓到的那個數字符合 `j`
+
+檢查完兩者後若都不符合就把當前的 `arr[i]` 加進 `Set` 中
 
 遍歷完都不符合的話就是 false 了
 
 ## 心得
-直覺想到用 Set，結果成功了好耶！ 
+
+直覺想到用 Set，結果成功了好耶！

docs/LeetCode/1346. Check If N and Its Double Exist/_Description.md

@@ -0,0 +1,5 @@
+Given an array `arr` of integers, check if there exist two indices `i` and `j` such that :
+
+- `i != j`
+- `0 <= i, j < arr.length`
+- `arr[i] == 2 * arr[j]`

docs/LeetCode/1346. Check If N and Its Double Exist/_Examples.md

@@ -0,0 +1,16 @@
+#### Example 1:
+
+> **Input:**: arr = [10,2,5,3]  
+> **Output:**: true  
+> **Explanation:**: For i = 0 and j = 2, arr[i] == 10 == 2 _ 5 == 2 _ arr[j]
+
+#### Example 2:
+
+> **Input:** arr = [3,1,7,11]  
+> **Output:** false  
+> **Explanation:** There is no i and j that satisfy the conditions.
+
+#### Constraints:
+
+- `1 <= numBottles <= 100`
+- `2 <= numExchange <= 100`