feat: add biweekly contest 124 (doocs#2348)

Author: yanglbme

.github/workflows/deploy.yml

@@ -54,18 +54,18 @@ jobs:
           github_token: ${{ secrets.GITHUB_TOKEN }}
           publish_dir: ./site
 
-      - name: Sync to gitee.com
-        uses: wearerequired/git-mirror-action@master
-        env:
-            SSH_PRIVATE_KEY: ${{ secrets.RSA_PRIVATE_KEY }}
-        with:
-            source-repo: git@github.com:doocs/leetcode.git
-            destination-repo: git@gitee.com:Doocs/leetcode.git
+      # - name: Sync to gitee.com
+      #   uses: wearerequired/git-mirror-action@master
+      #   env:
+      #       SSH_PRIVATE_KEY: ${{ secrets.RSA_PRIVATE_KEY }}
+      #   with:
+      #       source-repo: git@github.com:doocs/leetcode.git
+      #       destination-repo: git@gitee.com:Doocs/leetcode.git
 
-      - name: Build Gitee Pages
-        uses: yanglbme/gitee-pages-action@main
-        with:
-            gitee-username: yanglbme
-            gitee-password: ${{ secrets.GITEE_PASSWORD }}
-            gitee-repo: doocs/leetcode
-            branch: gh-pages
+      # - name: Build Gitee Pages
+      #   uses: yanglbme/gitee-pages-action@main
+      #   with:
+      #       gitee-username: yanglbme
+      #       gitee-password: ${{ secrets.GITEE_PASSWORD }}
+      #       gitee-repo: doocs/leetcode
+      #       branch: gh-pages

solution/0300-0399/0333.Largest BST Subtree/README.md

@@ -1,4 +1,4 @@
-# [333. 最大 BST 子树](https://leetcode.cn/problems/largest-bst-subtree)
+# [333. 最大二叉搜索子树](https://leetcode.cn/problems/largest-bst-subtree)
 
 [English Version](/solution/0300-0399/0333.Largest%20BST%20Subtree/README_EN.md)
 

solution/0300-0399/0373.Find K Pairs with Smallest Sums/README_EN.md

@@ -27,14 +27,6 @@
 <strong>Explanation:</strong> The first 2 pairs are returned from the sequence: [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]
 </pre>
 
-<p><strong class="example">Example 3:</strong></p>
-
-<pre>
-<strong>Input:</strong> nums1 = [1,2], nums2 = [3], k = 3
-<strong>Output:</strong> [[1,3],[2,3]]
-<strong>Explanation:</strong> All possible pairs are returned from the sequence: [1,3],[2,3]
-</pre>
-
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

solution/0600-0699/0629.K Inverse Pairs Array/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>For an integer array <code>nums</code>, an <strong>inverse pair</strong> is a pair of integers <code>[i, j]</code> where <code>0 &lt;= i &lt; j &lt; nums.length</code> and <code>nums[i] &gt; nums[j]</code>.</p>
 
-<p>Given two integers n and k, return the number of different arrays consist of numbers from <code>1</code> to <code>n</code> such that there are exactly <code>k</code> <strong>inverse pairs</strong>. Since the answer can be huge, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+<p>Given two integers n and k, return the number of different arrays consisting of numbers from <code>1</code> to <code>n</code> such that there are exactly <code>k</code> <strong>inverse pairs</strong>. Since the answer can be huge, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/0900-0999/0935.Knight Dialer/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>The chess knight has a <strong>unique movement</strong>,&nbsp;it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an <strong>L</strong>). The possible movements of chess knight are shown in this diagaram:</p>
+<p>The chess knight has a <strong>unique movement</strong>,&nbsp;it may move two squares vertically and one square horizontally, or two squares horizontally and one square vertically (with both forming the shape of an <strong>L</strong>). The possible movements of chess knight are shown in this diagram:</p>
 
 <p>A chess knight can move as indicated in the chess diagram below:</p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0935.Knight%20Dialer/images/chess.jpg" style="width: 402px; height: 402px;" />

solution/1000-1099/1071.Greatest Common Divisor of Strings/README_EN.md

@@ -4,7 +4,7 @@
 
 ## Description
 
-<p>For two strings <code>s</code> and <code>t</code>, we say &quot;<code>t</code> divides <code>s</code>&quot; if and only if <code>s = t + ... + t</code> (i.e., <code>t</code> is concatenated with itself one or more times).</p>
+<p>For two strings <code>s</code> and <code>t</code>, we say &quot;<code>t</code> divides <code>s</code>&quot; if and only if <code>s = t + t + t + ... + t + t</code> (i.e., <code>t</code> is concatenated with itself one or more times).</p>
 
 <p>Given two strings <code>str1</code> and <code>str2</code>, return <em>the largest string </em><code>x</code><em> such that </em><code>x</code><em> divides both </em><code>str1</code><em> and </em><code>str2</code>.</p>
 

solution/1000-1099/1074.Number of Submatrices That Sum to Target/README_EN.md

@@ -40,7 +40,7 @@
 <ul>
 	<li><code>1 &lt;= matrix.length &lt;= 100</code></li>
 	<li><code>1 &lt;= matrix[0].length &lt;= 100</code></li>
-	<li><code>-1000 &lt;= matrix[i] &lt;= 1000</code></li>
+	<li><code>-1000 &lt;= matrix[i][j] &lt;= 1000</code></li>
 	<li><code>-10^8 &lt;= target &lt;= 10^8</code></li>
 </ul>
 

solution/1100-1199/1159.Market Analysis II/README_EN.md

@@ -53,11 +53,11 @@ item_id is the primary key (column with unique values) of this table.
 
 <p>&nbsp;</p>
 
-<p>Write a solution&nbsp;to find for each user, the join date and the number of orders they made as a buyer in <code>2019</code>.</p>
+<p>Write a solution to find for each user whether the brand of the second item (by date) they sold is their favorite brand. If a user sold less than two items, report the answer for that user as no. It is guaranteed that no seller sells more than one item in a day.</p>
 
 <p>Return the result table in <strong>any order</strong>.</p>
 
-<p>The&nbsp;result format is in the following example.</p>
+<p>The result format is in the following example.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>

solution/1400-1499/1403.Minimum Subsequence in Non-Increasing Order/README_EN.md

@@ -24,7 +24,7 @@
 <pre>
 <strong>Input:</strong> nums = [4,4,7,6,7]
 <strong>Output:</strong> [7,7,6] 
-<strong>Explanation:</strong> The subsequence [7,7] has the sum of its elements equal to 14 which is not strictly greater than the sum of elements not included (14 = 4 + 4 + 6). Therefore, the subsequence [7,6,7] is the minimal satisfying the conditions. Note the subsequence has to be returned in non-decreasing order.  
+<strong>Explanation:</strong> The subsequence [7,7] has the sum of its elements equal to 14 which is not strictly greater than the sum of elements not included (14 = 4 + 4 + 6). Therefore, the subsequence [7,6,7] is the minimal satisfying the conditions. Note the subsequence has to be returned in non-increasing order.  
 </pre>
 
 <p>&nbsp;</p>

solution/2100-2199/2149.Rearrange Array Elements by Sign/README_EN.md

@@ -48,6 +48,9 @@ So nums is rearranged to [1,-1].
 	<li><code>nums</code> consists of <strong>equal</strong> number of positive and negative integers.</li>
 </ul>
 
+<p>&nbsp;</p>
+It is not required to do the modifications in-place.
+
 ## Solutions
 
 ### Solution 1

solution/2700-2799/2792.Count Nodes That Are Great Enough/README.md

@@ -1,4 +1,4 @@
-# [2792. 计算满足条件的节点数](https://leetcode.cn/problems/count-nodes-that-are-great-enough)
+# [2792. 计算足够大的节点数](https://leetcode.cn/problems/count-nodes-that-are-great-enough)
 
 [English Version](/solution/2700-2799/2792.Count%20Nodes%20That%20Are%20Great%20Enough/README_EN.md)
 
@@ -13,7 +13,7 @@
 	<li>它的值 <strong>大于</strong> 其子树中 <strong>至少</strong> <code>k</code> 个节点的值。</li>
 </ul>
 
-<p>返回满足条件的节点数。</p>
+<p>返回足够大的节点数。</p>
 
 <p>如果 <code>u == v</code> 或者 <code>v</code> 是 <code>u</code> 的祖先，则节点 <code>u</code> 在节点 <code>v</code> 的 <strong>子树</strong> 中。</p>
 

solution/2800-2899/2899.Last Visited Integers/README_EN.md

@@ -4,46 +4,89 @@
 
 ## Description
 
-<p>Given a <strong>0-indexed</strong> array of strings <code>words</code> where <code>words[i]</code> is either a positive integer represented as a string or the string <code>&quot;prev&quot;</code>.</p>
+<p>Given an integer array <code>nums</code> where <code>nums[i]</code> is either a positive integer or <code>-1</code>.</p>
 
-<p>Start iterating from the beginning of the array; for every <code>&quot;prev&quot;</code> string seen in <code>words</code>, find the <strong>last visited integer</strong> in <code>words</code> which is defined as follows:</p>
+<p>We need to find for each <code>-1</code> the respective positive integer, which we call the last visited integer.</p>
+
+<p>To achieve this goal, let&#39;s define two empty arrays: <code>seen</code> and <code>ans</code>.</p>
+
+<p>Start iterating from the beginning of the array <code>nums</code>.</p>
 
 <ul>
-	<li>Let <code>k</code> be the number of consecutive <code>&quot;prev&quot;</code> strings seen so far (containing the current string). Let <code>nums</code> be the <strong>0-indexed </strong>array of <strong>integers</strong> seen so far and <code>nums_reverse</code> be the reverse of <code>nums</code>, then the integer at <code>(k - 1)<sup>th</sup></code> index of <code>nums_reverse</code> will be the <strong>last visited integer</strong> for this <code>&quot;prev&quot;</code>.</li>
-	<li>If <code>k</code> is <strong>greater</strong> than the total visited integers, then the last visited integer will be <code>-1</code>.</li>
+	<li>If a positive integer is encountered, prepend it to the <strong>front</strong> of <code>seen</code>.</li>
+	<li>If <code>-1</code>&nbsp;is encountered, let <code>k</code> be the number of <strong>consecutive</strong> <code>-1</code>s seen so far (including the current <code>-1</code>),
+	<ul>
+		<li>If <code>k</code> is less than or equal to the length of <code>seen</code>, append the <code>k</code>-th element of <code>seen</code> to <code>ans</code>.</li>
+		<li>If <code>k</code> is strictly greater than the length of <code>seen</code>, append <code>-1</code> to <code>ans</code>.</li>
+	</ul>
+	</li>
 </ul>
 
-<p>Return <em>an integer array containing the last visited integers.</em></p>
+<p>Return <em>the array </em><code>ans</code>.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
 
-<pre>
-<strong>Input:</strong> words = [&quot;1&quot;,&quot;2&quot;,&quot;prev&quot;,&quot;prev&quot;,&quot;prev&quot;]
-<strong>Output:</strong> [2,1,-1]
-<strong>Explanation:</strong> 
-For &quot;prev&quot; at index = 2, last visited integer will be 2 as here the number of consecutive &quot;prev&quot; strings is 1, and in the array reverse_nums, 2 will be the first element.
-For &quot;prev&quot; at index = 3, last visited integer will be 1 as there are a total of two consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, and 1 is the second last visited integer.
-For &quot;prev&quot; at index = 4, last visited integer will be -1 as there are a total of three consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, but the total number of integers visited is two.
-</pre>
+<div class="example-block" style="
+    border-color: var(--border-tertiary);
+    border-left-width: 2px;
+    color: var(--text-secondary);
+    font-size: .875rem;
+    line-height: 1.25rem;
+    margin-bottom: 1rem;
+    margin-top: 1rem;
+    overflow: visible;
+    padding-left: 1rem;
+">
+<p><strong>Input:</strong> <code>nums = [1,2,-1,-1,-1]</code></p>
+
+<p><strong>Output:</strong> <code>[2,1,-1]</code></p>
+
+<p><strong>Explanation:</strong> Start with <code>seen = []</code> and <code>ans = []</code>.</p>
+
+<ol>
+	<li>Process <code>nums[0]</code>: The first element in nums is <code>1</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [1]</code>.</li>
+	<li>Process <code>nums[1]</code>: The next element is <code>2</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [2, 1]</code>.</li>
+	<li>Process <code>nums[2]</code>: The next element is <code>-1</code>. This is the first occurrence of <code>-1</code>, so <code>k == 1</code>. We look for the first element in seen. We append <code>2</code> to <code>ans</code>. Now, <code>ans == [2]</code>.</li>
+	<li>Process <code>nums[3]</code>: Another <code>-1</code>. This is the second consecutive <code>-1</code>, so <code>k == 2</code>. The second element in <code>seen</code> is <code>1</code>, so we append <code>1</code> to <code>ans</code>. Now, <code>ans == [2, 1]</code>.</li>
+	<li>Process <code>nums[4]</code>: Another <code>-1</code>, the third in a row, making <code>k = 3</code>. However, <code>seen</code> only has two elements (<code>[2, 1]</code>). Since <code>k</code> is greater than the number of elements in <code>seen</code>, we append <code>-1</code> to <code>ans</code>. Finally, <code>ans == [2, 1, -1]</code>.</li>
+</ol>
+</div>
 
 <p><strong class="example">Example 2:</strong></p>
 
-<pre>
-<strong>Input:</strong> words = [&quot;1&quot;,&quot;prev&quot;,&quot;2&quot;,&quot;prev&quot;,&quot;prev&quot;]
-<strong>Output:</strong> [1,2,1]
-<strong>Explanation:</strong>
-For &quot;prev&quot; at index = 1, last visited integer will be 1.
-For &quot;prev&quot; at index = 3, last visited integer will be 2.
-For &quot;prev&quot; at index = 4, last visited integer will be 1 as there are a total of two consecutive &quot;prev&quot; strings including this &quot;prev&quot; which are visited, and 1 is the second last visited integer.
-</pre>
+<div class="example-block" style="
+    border-color: var(--border-tertiary);
+    border-left-width: 2px;
+    color: var(--text-secondary);
+    font-size: .875rem;
+    line-height: 1.25rem;
+    margin-bottom: 1rem;
+    margin-top: 1rem;
+    overflow: visible;
+    padding-left: 1rem;
+">
+<p><strong>Input:</strong> <code>nums = [1,-1,2,-1,-1]</code></p>
+
+<p><strong>Output:</strong> <code>[1,2,1]</code></p>
+
+<p><strong>Explanation:</strong> Start with <code>seen = []</code> and <code>ans = []</code>.</p>
+
+<ol>
+	<li>Process <code>nums[0]</code>: The first element in nums is <code>1</code>. We prepend it to the front of <code>seen</code>. Now, <code>seen == [1]</code>.</li>
+	<li>Process <code>nums[1]</code>: The next element is <code>-1</code>. This is the first occurrence of <code>-1</code>, so <code>k == 1</code>. We look for the first element in <code>seen</code>, which is <code>1</code>. Append <code>1</code> to <code>ans</code>. Now, <code>ans == [1]</code>.</li>
+	<li>Process <code>nums[2]</code>: The next element is <code>2</code>. Prepend this to the front of <code>seen</code>. Now, <code>seen == [2, 1]</code>.</li>
+	<li>Process <code>nums[3]</code>: The next element is <code>-1</code>. This <code>-1</code> is not consecutive to the first <code>-1</code> since <code>2</code> was in between. Thus, <code>k</code> resets to <code>1</code>. The first element in <code>seen</code> is <code>2</code>, so append <code>2</code> to <code>ans</code>. Now, <code>ans == [1, 2]</code>.</li>
+	<li>Process <code>nums[4]</code>: Another <code>-1</code>. This is consecutive to the previous <code>-1</code>, so <code>k == 2</code>. The second element in <code>seen</code> is <code>1</code>, append <code>1</code> to <code>ans</code>. Finally, <code>ans == [1, 2, 1]</code>.</li>
+</ol>
+</div>
 
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>1 &lt;= words.length &lt;= 100</code></li>
-	<li><code>words[i] == &quot;prev&quot;</code> or <code>1 &lt;= int(words[i]) &lt;= 100</code></li>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>nums[i] == -1</code> or <code>1 &lt;= nums[i]&nbsp;&lt;= 100</code></li>
 </ul>
 
 ## Solutions