feat: update lc problems

Author: yanglbme

solution/0000-0099/0050.Pow(x, n)/README_EN.md

@@ -35,6 +35,7 @@
 <ul>
 	<li><code>-100.0 &lt; x &lt; 100.0</code></li>
 	<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup>-1</code></li>
+	<li><code>n</code> is an integer.</li>
 	<li><code>-10<sup>4</sup> &lt;= x<sup>n</sup> &lt;= 10<sup>4</sup></code></li>
 </ul>
 

solution/0300-0399/0320.Generalized Abbreviation/README_EN.md

@@ -8,6 +8,7 @@
 
 <ul>
 	<li>For example, <code>&quot;abcde&quot;</code> can be abbreviated into:
+
     <ul>
     	<li><code>&quot;a3e&quot;</code> (<code>&quot;bcd&quot;</code> turned into <code>&quot;3&quot;</code>)</li>
     	<li><code>&quot;1bcd1&quot;</code> (<code>&quot;a&quot;</code> and <code>&quot;e&quot;</code> both turned into <code>&quot;1&quot;</code>)</li>
@@ -21,6 +22,7 @@
     	<li><code>&quot;22de&quot;</code> (<code>&quot;ab&quot;</code> turned into <code>&quot;2&quot;</code> and <code>&quot;bc&quot;</code> turned into <code>&quot;2&quot;</code>) is invalid as the substring chosen overlap.</li>
     </ul>
     </li>
+
 </ul>
 
 <p>Given a string <code>word</code>, return <em>a list of all the possible <strong>generalized abbreviations</strong> of</em> <code>word</code>. Return the answer in <strong>any order</strong>.</p>

solution/0300-0399/0363.Max Sum of Rectangle No Larger Than K/README_EN.md

@@ -30,7 +30,7 @@
 <ul>
 	<li><code>m == matrix.length</code></li>
 	<li><code>n == matrix[i].length</code></li>
-	<li><code>1 &lt;= m, n &lt;= 200</code></li>
+	<li><code>1 &lt;= m, n &lt;= 100</code></li>
 	<li><code>-100 &lt;= matrix[i][j] &lt;= 100</code></li>
 	<li><code>-10<sup>5</sup> &lt;= k &lt;= 10<sup>5</sup></code></li>
 </ul>

solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README.md

@@ -1,14 +1,14 @@
-# [462. 最少移动次数使数组元素相等 II](https://leetcode.cn/problems/minimum-moves-to-equal-array-elements-ii)
+# [462. 最小操作次数使数组元素相等 II](https://leetcode.cn/problems/minimum-moves-to-equal-array-elements-ii)
 
 [English Version](/solution/0400-0499/0462.Minimum%20Moves%20to%20Equal%20Array%20Elements%20II/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code> ，返回使所有数组元素相等需要的最少移动数。</p>
+<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code> ，返回使所有数组元素相等需要的最小操作数。</p>
 
-<p>在一步操作中，你可以使数组中的一个元素加 <code>1</code> 或者减 <code>1</code> 。</p>
+<p>在一次操作中，你可以使数组中的一个元素加 <code>1</code> 或者减 <code>1</code> 。</p>
 
 <p>&nbsp;</p>
 
@@ -18,7 +18,7 @@
 <strong>输入：</strong>nums = [1,2,3]
 <strong>输出：</strong>2
 <strong>解释：</strong>
-只需要两步操作（每步操作指南使一个元素加 1 或减 1）：
+只需要两次操作（每次操作指南使一个元素加 1 或减 1）：
 [<strong><em>1</em></strong>,2,3]  =&gt;  [2,2,<strong><em>3</em></strong>]  =&gt;  [2,2,2]
 </pre>
 

solution/0600-0699/0621.Task Scheduler/README.md

@@ -132,7 +132,6 @@ public:
 
 ### **Go**
 
-
 ```go
 func leastInterval(tasks []byte, n int) int {
 	cnt := make([]int, 26)

solution/0600-0699/0698.Partition to K Equal Sum Subsets/README_EN.md

@@ -131,7 +131,7 @@ class Solution {
     private int[] nums;
     private int n;
     private int s;
-    
+
     public boolean canPartitionKSubsets(int[] nums, int k) {
         for (int v : nums) {
             s += v;

solution/1400-1499/1475.Final Prices With a Special Discount in a Shop/README_EN.md

@@ -4,20 +4,22 @@
 
 ## Description
 
-<p>Given the array <code>prices</code> where <code>prices[i]</code> is the price of the <code>ith</code> item in a shop. There is a special discount for items in the shop, if you buy the <code>ith</code> item, then you will receive a discount equivalent to <code>prices[j]</code> where <code>j</code> is the <strong>minimum</strong>&nbsp;index such that <code>j &gt; i</code> and <code>prices[j] &lt;= prices[i]</code>, otherwise, you will not receive any discount at all.</p>
+<p>You are given an integer array <code>prices</code> where <code>prices[i]</code> is the price of the <code>i<sup>th</sup></code> item in a shop.</p>
 
-<p><em>Return an array where the <code>ith</code> element is the final price you will pay for the <code>ith</code> item of the shop considering the special discount.</em></p>
+<p>There is a special discount for items in the shop. If you buy the <code>i<sup>th</sup></code> item, then you will receive a discount equivalent to <code>prices[j]</code> where <code>j</code> is the minimum index such that <code>j &gt; i</code> and <code>prices[j] &lt;= prices[i]</code>. Otherwise, you will not receive any discount at all.</p>
+
+<p>Return an integer array <code>answer</code> where <code>answer[i]</code> is the final price you will pay for the <code>i<sup>th</sup></code> item of the shop, considering the special discount.</p>
 
 <p>&nbsp;</p>
 <p><strong>Example 1:</strong></p>
 
 <pre>
 <strong>Input:</strong> prices = [8,4,6,2,3]
 <strong>Output:</strong> [4,2,4,2,3]
-<strong>Explanation:</strong>&nbsp;
-For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.&nbsp;
-For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.&nbsp;
-For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.&nbsp;
+<strong>Explanation:</strong> 
+For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.
+For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.
+For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.
 For items 3 and 4 you will not receive any discount at all.
 </pre>
 
@@ -41,7 +43,7 @@ For items 3 and 4 you will not receive any discount at all.
 
 <ul>
 	<li><code>1 &lt;= prices.length &lt;= 500</code></li>
-	<li><code>1 &lt;= prices[i] &lt;= 10^3</code></li>
+	<li><code>1 &lt;= prices[i] &lt;= 1000</code></li>
 </ul>
 
 ## Solutions

solution/1500-1599/1518.Water Bottles/README.md

@@ -1,58 +1,50 @@
-# [1518. 换酒问题](https://leetcode.cn/problems/water-bottles)
+# [1518. 换水问题](https://leetcode.cn/problems/water-bottles)
 
 [English Version](/solution/1500-1599/1518.Water%20Bottles/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>小区便利店正在促销，用 <code>numExchange</code> 个空酒瓶可以兑换一瓶新酒。你购入了 <code>numBottles</code> 瓶酒。</p>
+<p>超市正在促销，你可以用 <code>numExchange</code> 个空水瓶从超市兑换一瓶水。最开始，你一共购入了 <code>numBottles</code> 瓶水。</p>
 
-<p>如果喝掉了酒瓶中的酒，那么酒瓶就会变成空的。</p>
+<p>如果喝掉了水瓶中的水，那么水瓶就会变成空的。</p>
 
-<p>请你计算 <strong>最多</strong> 能喝到多少瓶酒。</p>
+<p>给你两个整数 <code>numBottles</code> 和 <code>numExchange</code> ，返回你 <strong>最多</strong> 可以喝到多少瓶水。</p>
 
 <p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
-<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1518.Water%20Bottles/images/sample_1_1875.png" style="height: 240px; width: 480px;"></strong></p>
+<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1518.Water%20Bottles/images/sample_1_1875.png" style="height: 240px; width: 480px;" /></strong></p>
 
-<pre><strong>输入：</strong>numBottles = 9, numExchange = 3
+<pre>
+<strong>输入：</strong>numBottles = 9, numExchange = 3
 <strong>输出：</strong>13
-<strong>解释：</strong>你可以用 <code>3</code> 个空酒瓶兑换 1 瓶酒。
-所以最多能喝到 9 + 3 + 1 = 13 瓶酒。
+<strong>解释：</strong>你可以用 <code>3</code> 个空瓶兑换 1 瓶水。
+所以最多能喝到 9 + 3 + 1 = 13 瓶水。
 </pre>
 
 <p><strong>示例 2：</strong></p>
 
-<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1518.Water%20Bottles/images/sample_2_1875.png" style="height: 240px; width: 790px;"></p>
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1518.Water%20Bottles/images/sample_2_1875.png" style="height: 240px; width: 790px;" /></p>
 
-<pre><strong>输入：</strong>numBottles = 15, numExchange = 4
+<pre>
+<strong>输入：</strong>numBottles = 15, numExchange = 4
 <strong>输出：</strong>19
-<strong>解释：</strong>你可以用 <code>4</code> 个空酒瓶兑换 1 瓶酒。
-所以最多能喝到 15 + 3 + 1 = 19 瓶酒。
+<strong>解释：</strong>你可以用 <code>4</code> 个空瓶兑换 1 瓶水。
+所以最多能喝到 15 + 3 + 1 = 19 瓶水。
 </pre>
 
-<p><strong>示例 3：</strong></p>
-
-<pre><strong>输入：</strong>numBottles = 5, numExchange = 5
-<strong>输出：</strong>6
-</pre>
-
-<p><strong>示例 4：</strong></p>
-
-<pre><strong>输入：</strong>numBottles = 2, numExchange = 3
-<strong>输出：</strong>2
-</pre>
+<p>&nbsp;</p>
 
 <p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code>1 &lt;=&nbsp;numBottles &lt;= 100</code></li>
-	<li><code>2 &lt;=&nbsp;numExchange &lt;= 100</code></li>
+	<li><code>1 &lt;= numBottles &lt;= 100</code></li>
+	<li><code>2 &lt;= numExchange &lt;= 100</code></li>
 </ul>
 
 ## 解法

solution/2400-2499/2412.Minimum Money Required Before Transactions/README.md

@@ -31,7 +31,7 @@
 <b>输出：</b>3
 <strong>解释：</strong>
 - 如果交易执行的顺序是 [[3,0],[0,3]] ，完成所有交易需要的最少钱数是 3 。
-- 如果交易执行的顺序是 [[0,3],[3,0]] ，完成所有交易需要的最少钱数是 1 。
+- 如果交易执行的顺序是 [[0,3],[3,0]] ，完成所有交易需要的最少钱数是 0 。
 所以，刚开始钱数为 3 ，任意顺序下交易都可以全部完成。
 </pre>
 

solution/2400-2499/2414.Length of the Longest Alphabetical Continuous Substring/README.md

@@ -49,6 +49,7 @@
 利用双指针 $i$ 和 $j$，分别指向当前连续子字符串的起始位置和结束位置。遍历字符串 $s$，如果当前字符 $s[j]$ 比 $s[j-1]$ 大，则 $j$ 向右移动一位，否则更新 $i$ 为 $j$，并更新最长连续子字符串的长度。
 
 时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**