feat: add weekly contest 427 and biweekly contest 145

Author: yanglbme

solution/0300-0399/0355.Design Twitter/README.md

@@ -61,6 +61,7 @@ twitter.getNewsFeed(1);  // 用户 1 获取推文应当返回一个列表，其
 	<li><code>0 &lt;= tweetId &lt;= 10<sup>4</sup></code></li>
 	<li>所有推特的 ID 都互不相同</li>
 	<li><code>postTweet</code>、<code>getNewsFeed</code>、<code>follow</code> 和 <code>unfollow</code> 方法最多调用 <code>3 * 10<sup>4</sup></code> 次</li>
+	<li>用户不能关注自己</li>
 </ul>
 
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solution/0300-0399/0355.Design Twitter/README_EN.md

@@ -60,6 +60,7 @@ twitter.getNewsFeed(1);  // User 1's news feed should return a list with 1 t
 	<li><code>0 &lt;= tweetId &lt;= 10<sup>4</sup></code></li>
 	<li>All the tweets have <strong>unique</strong> IDs.</li>
 	<li>At most <code>3 * 10<sup>4</sup></code> calls will be made to <code>postTweet</code>, <code>getNewsFeed</code>, <code>follow</code>, and <code>unfollow</code>.</li>
+	<li>A user cannot follow himself.</li>
 </ul>
 
 <!-- description:end -->

solution/0700-0799/0782.Transform to Chessboard/README.md

@@ -93,7 +93,7 @@ tags:
 
 若 $n$ 为奇数，那么最终的合法棋盘只有一种可能。如果第一行中 $0$ 的数目大于 $1$，那么最终一盘的第一行只能是“01010...”，否则就是“10101...”。同样算出次数作为答案。
 
-时间复杂度 $O(n^2)$。
+时间复杂度 $O(n^2)$，其中 $n$ 是棋盘的大小。空间复杂度 $O(1)$。
 
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solution/0700-0799/0782.Transform to Chessboard/README_EN.md

@@ -68,7 +68,24 @@ The second move swaps the second and third row.
 
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-### Solution 1
+### Solution 1: Pattern Observation + State Compression
+
+In a valid chessboard, there are exactly two types of "rows".
+
+For example, if one row on the chessboard is "01010011", then any other row can only be "01010011" or "10101100". Columns also satisfy this property.
+
+Additionally, each row and each column has half $0$s and half $1$s. Suppose the chessboard is $n \times n$:
+
+-   If $n = 2 \times k$, then each row and each column has $k$ $1$s and $k$ $0$s.
+-   If $n = 2 \times k + 1$, then each row has $k$ $1$s and $k + 1$ $0$s, or $k + 1$ $1$s and $k$ $0$s.
+
+Based on the above conclusions, we can determine whether a chessboard is valid. If valid, we can calculate the minimum number of moves required.
+
+If $n$ is even, there are two possible valid chessboards, where the first row is either "010101..." or "101010...". We calculate the minimum number of swaps required for these two possibilities and take the smaller value as the answer.
+
+If $n$ is odd, there is only one possible valid chessboard. If the number of $0$s in the first row is greater than the number of $1$s, then the first row of the final chessboard must be "01010..."; otherwise, it must be "10101...". We calculate the number of swaps required and use it as the answer.
+
+The time complexity is $O(n^2)$, where $n$ is the size of the chessboard. The space complexity is $O(1)$.
 
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solution/0800-0899/0809.Expressive Words/README.md

@@ -31,7 +31,7 @@ tags:
 <p><strong>示例：</strong></p>
 
 <pre>
-<strong>输入：</strong>
+<strong>输入：</strong> 
 s = "heeellooo"
 words = ["hello", "hi", "helo"]
 <strong>输出：</strong>1

solution/0800-0899/0809.Expressive Words/README_EN.md

@@ -41,7 +41,7 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;heeellooo&quot;, words = [&quot;hello&quot;, &quot;hi&quot;, &quot;helo&quot;]
 <strong>Output:</strong> 1
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 We can extend &quot;e&quot; and &quot;o&quot; in the word &quot;hello&quot; to get &quot;heeellooo&quot;.
 We can&#39;t extend &quot;helo&quot; to get &quot;heeellooo&quot; because the group &quot;ll&quot; is not size 3 or more.
 </pre>

solution/0800-0899/0810.Chalkboard XOR Game/README.md

@@ -35,7 +35,7 @@ tags:
 <pre>
 <strong>输入:</strong> nums = [1,1,2]
 <strong>输出:</strong> false
-<strong>解释:</strong>
+<strong>解释:</strong> 
 Alice 有两个选择: 擦掉数字 1 或 2。
 如果擦掉 1, 数组变成 [1, 2]。剩余数字按位异或得到 1 XOR 2 = 3。那么 Bob 可以擦掉任意数字，因为 Alice 会成为擦掉最后一个数字的人，她总是会输。
 如果 Alice 擦掉 2，那么数组变成[1, 1]。剩余数字按位异或得到 1 XOR 1 = 0。Alice 仍然会输掉游戏。

solution/0800-0899/0810.Chalkboard XOR Game/README_EN.md

@@ -34,9 +34,9 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [1,1,2]
 <strong>Output:</strong> false
-<strong>Explanation:</strong>
-Alice has two choices: erase 1 or erase 2.
-If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose.
+<strong>Explanation:</strong> 
+Alice has two choices: erase 1 or erase 2. 
+If she erases 1, the nums array becomes [1, 2]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 2 = 3. Now Bob can remove any element he wants, because Alice will be the one to erase the last element and she will lose. 
 If Alice erases 2 first, now nums become [1, 1]. The bitwise XOR of all the elements of the chalkboard is 1 XOR 1 = 0. Alice will lose.
 </pre>
 

solution/0900-0999/0999.Available Captures for Rook/README.md

@@ -24,7 +24,7 @@ tags:
 
 <p>注意：车不能穿过其它棋子，比如象和卒。这意味着如果有其它棋子挡住了路径，车就不能够吃掉棋子。</p>
 
-<p>返回白车将能 <strong>吃掉</strong> 的 <strong>卒的数量</strong>。</p>
+<p>返回白车 <strong>攻击</strong>&nbsp;范围内 <strong>兵的数量</strong>。</p>
 
 <p>&nbsp;</p>
 

solution/1000-1099/1004.Max Consecutive Ones III/README.md

@@ -21,7 +21,7 @@ tags:
 
 <!-- description:start -->
 
-<p>给定一个二进制数组&nbsp;<code>nums</code>&nbsp;和一个整数 <code>k</code>，如果可以翻转最多 <code>k</code> 个 <code>0</code> ，则返回 <em>数组中连续 <code>1</code> 的最大个数</em> 。</p>
+<p>给定一个二进制数组&nbsp;<code>nums</code>&nbsp;和一个整数 <code>k</code>，假设最多可以翻转 <code>k</code> 个 <code>0</code> ，则返回执行操作后 <em>数组中连续 <code>1</code> 的最大个数</em> 。</p>
 
 <p>&nbsp;</p>