feat: update lc problems (#3881)

solution/0200-0299/0222.Count Complete Tree Nodes/README.md

@@ -21,9 +21,9 @@ tags:
 
 <p>给你一棵<strong> 完全二叉树</strong> 的根节点 <code>root</code> ，求出该树的节点个数。</p>
 
-<p><a href="https://baike.baidu.com/item/%E5%AE%8C%E5%85%A8%E4%BA%8C%E5%8F%89%E6%A0%91/7773232?fr=aladdin">完全二叉树</a> 的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 <code>h</code> 层，则该层包含 <code>1~ 2<sup>h</sup></code> 个节点。</p>
+<p><a href="https://baike.baidu.com/item/%E5%AE%8C%E5%85%A8%E4%BA%8C%E5%8F%89%E6%A0%91/7773232?fr=aladdin">完全二叉树</a> 的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 <code>h</code> 层（从第 0 层开始），则该层包含 <code>1~&nbsp;2<sup>h</sup></code>&nbsp;个节点。</p>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 <img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0222.Count%20Complete%20Tree%20Nodes/images/complete.jpg" style="width: 372px; height: 302px;" />
@@ -46,17 +46,17 @@ tags:
 <strong>输出：</strong>1
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
 	<li>树中节点的数目范围是<code>[0, 5 * 10<sup>4</sup>]</code></li>
-	<li><code>0 <= Node.val <= 5 * 10<sup>4</sup></code></li>
+	<li><code>0 &lt;= Node.val &lt;= 5 * 10<sup>4</sup></code></li>
 	<li>题目数据保证输入的树是 <strong>完全二叉树</strong></li>
 </ul>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>进阶：</strong>遍历树来统计节点是一种时间复杂度为 <code>O(n)</code> 的简单解决方案。你可以设计一个更快的算法吗？</p>
 

solution/0300-0399/0321.Create Maximum Number/README_EN.md

@@ -57,6 +57,7 @@ tags:
 	<li><code>1 &lt;= m, n &lt;= 500</code></li>
 	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 9</code></li>
 	<li><code>1 &lt;= k &lt;= m + n</code></li>
+	<li><code>nums1</code> and <code>nums2</code> do not have leading zeros.</li>
 </ul>
 
 <!-- description:end -->

solution/0600-0699/0634.Find the Derangement of An Array/README.md

@@ -5,6 +5,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0634.Fi
 tags:
     - 数学
     - 动态规划
+    - 组合数学
 ---
 
 <!-- problem:start -->

solution/0600-0699/0634.Find the Derangement of An Array/README_EN.md

@@ -5,6 +5,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/0600-0699/0634.Fi
 tags:
     - Math
     - Dynamic Programming
+    - Combinatorics
 ---
 
 <!-- problem:start -->

solution/0700-0799/0735.Asteroid Collision/README_EN.md

@@ -18,7 +18,7 @@ tags:
 
 <!-- description:start -->
 
-<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row.</p>
+<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row. The indices of the asteriod in the array represent their relative position in space.</p>
 
 <p>For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.</p>
 

solution/0900-0999/0977.Squares of a Sorted Array/README.md

@@ -23,7 +23,7 @@ tags:
 <ul>
 </ul>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>示例 1：</strong></p>
 
@@ -40,22 +40,22 @@ tags:
 <strong>输出：</strong>[4,9,9,49,121]
 </pre>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>提示：</strong></p>
 
 <ul>
-	<li><code><span>1 <= nums.length <= </span>10<sup>4</sup></code></li>
-	<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
+	<li><code><span>1 &lt;= nums.length &lt;= </span>10<sup>4</sup></code></li>
+	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
 	<li><code>nums</code> 已按 <strong>非递减顺序</strong> 排序</li>
 </ul>
 
-<p> </p>
+<p>&nbsp;</p>
 
 <p><strong>进阶：</strong></p>
 
 <ul>
-	<li>请你<span style="color: rgb(36, 41, 46); font-family: -apple-system, BlinkMacSystemFont, &quot;Segoe UI&quot;, Helvetica, Arial, sans-serif, &quot;Apple Color Emoji&quot;, &quot;Segoe UI Emoji&quot;; font-size: 14px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; background-color: rgb(255, 255, 255); text-decoration-style: initial; text-decoration-color: initial; display: inline !important; float: none;">设计时间复杂度为 <code>O(n)</code> 的算法解决本问题</span></li>
+	<li>请你设计时间复杂度为 <code>O(n)</code> 的算法解决本问题</li>
 </ul>
 
 <!-- description:end -->

solution/1800-1899/1891.Cutting Ribbons/README.md

@@ -29,7 +29,7 @@ tags:
 	<li>切割成四条长度为 <code>1</code>&nbsp;的绳子。</li>
 </ul>
 
-<p>你的任务是找出最大 <code>x</code> 值，要求满足可以裁切出恰好 <code>k</code> 条长度均为 <code>x</code> 的绳子。你可以丢弃裁切后剩余的任意长度的绳子。如果不可能切割出&nbsp;<code>k</code> 条相同长度的绳子，返回 0。</p>
+<p>你的任务是找出最大 <code>x</code> 值，要求满足可以裁切出至少&nbsp;<code>k</code> 条长度均为 <code>x</code> 的绳子。你可以丢弃裁切后剩余的任意长度的绳子。如果不可能切割出&nbsp;<code>k</code> 条相同长度的绳子，返回 0。</p>
 
 <p>&nbsp;</p>
 

solution/2500-2599/2558.Take Gifts From the Richest Pile/README_EN.md

@@ -25,7 +25,7 @@ tags:
 <ul>
 	<li>Choose the pile with the maximum number of gifts.</li>
 	<li>If there is more than one pile with the maximum number of gifts, choose any.</li>
-	<li>Leave behind the floor of the square root of the number of gifts in the pile. Take the rest of the gifts.</li>
+	<li>Reduce the number of gifts in the pile to the floor of the square root of the original number of gifts in the pile.</li>
 </ul>
 
 <p>Return <em>the number of gifts remaining after </em><code>k</code><em> seconds.</em></p>

solution/3300-3399/3391.Design a 3D Binary Matrix with Efficient Layer Tracking/README.md

@@ -6,70 +6,72 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3391.De
 
 <!-- problem:start -->
 
-# [3391. Design a 3D Binary Matrix with Efficient Layer Tracking 🔒](https://leetcode.cn/problems/design-a-3d-binary-matrix-with-efficient-layer-tracking)
+# [3391. 设计一个高效的层跟踪三维二进制矩阵 🔒](https://leetcode.cn/problems/design-a-3d-binary-matrix-with-efficient-layer-tracking)
 
 [English Version](/solution/3300-3399/3391.Design%20a%203D%20Binary%20Matrix%20with%20Efficient%20Layer%20Tracking/README_EN.md)
 
 ## 题目描述
 
 <!-- description:start -->
 
-<p>You are given a <code>n x n x n</code> <strong>binary</strong> 3D array <code>matrix</code>.</p>
+<p>给定一个&nbsp;<code>n x n x n</code>&nbsp;的 <strong>二进制&nbsp;</strong>三维数组&nbsp;<code>matrix</code>。</p>
 
-<p>Implement the <code>matrix3D</code> class:</p>
+<p>实现&nbsp;<code>Matrix3D</code>&nbsp;类：</p>
 
 <ul>
-	<li><code>matrix3D(int n)</code> Initializes the object with the 3D binary array <code>matrix</code>, where <strong>all</strong> elements are initially set to 0.</li>
-	<li><code>void setCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 1.</li>
-	<li><code>void unsetCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 0.</li>
-	<li><code>int largestMatrix()</code> Returns the index <code>x</code> where <code>matrix[x]</code> contains the most number of 1&#39;s. If there are multiple such indices, return the <strong>largest</strong> <code>x</code>.</li>
+	<li><code>Matrix3D(int n)</code>&nbsp;用三维二进制数组&nbsp;<code>matrix</code>&nbsp;初始化对象，其中 <strong>所有</strong>&nbsp;元素都初始化为 0。</li>
+	<li><code>void setCell(int x, int y, int z)</code>&nbsp;将 <code>matrix[x][y][z]</code>&nbsp;的值设为 1。</li>
+	<li><code>void unsetCell(int x, int y, int z)</code>&nbsp;将 <code>matrix[x][y][z]</code>&nbsp;的值设为 0。</li>
+	<li><code>int largestMatrix()</code>&nbsp;返回包含最多 1 的 <code>matrix[x]</code>&nbsp;的下标&nbsp;<code>x</code>。如果这样的对应值有多个，返回&nbsp;<strong>最大的</strong>&nbsp;<code>x</code>。</li>
 </ul>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><strong class="example">示例 1：</strong></p>
 
 <div class="example-block">
-<p><strong>Input:</strong><br />
-<span class="example-io">[&quot;matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;]<br />
+<p><strong>输入：</strong><br />
+<span class="example-io">["Matrix3D", "setCell", "largestMatrix", "setCell", "largestMatrix", "setCell", "largestMatrix"]<br />
 [[3], [0, 0, 0], [], [1, 1, 2], [], [0, 0, 1], []]</span></p>
 
-<p><strong>Output:</strong><br />
+<p><strong>输出：</strong><br />
 <span class="example-io">[null, null, 0, null, 1, null, 0] </span></p>
 
-<p><strong>Explanation</strong></p>
-matrix3D matrix3D = new matrix3D(3); // Initializes a <code>3 x 3 x 3</code> 3D array <code>matrix</code>, filled with all 0&#39;s.<br />
-matrix3D.setCell(0, 0, 0); // Sets <code>matrix[0][0][0]</code> to 1.<br />
-matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1&#39;s.<br />
-matrix3D.setCell(1, 1, 2); // Sets <code>matrix[1][1][2]</code> to 1.<br />
-matrix3D.largestMatrix(); // Returns 1. <code>matrix[0]</code> and <code>matrix[1]</code> tie with the most number of 1&#39;s, but index 1 is bigger.<br />
-matrix3D.setCell(0, 0, 1); // Sets <code>matrix[0][0][1]</code> to 1.<br />
-matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1&#39;s.</div>
+<p><strong>解释：</strong></p>
+Matrix3D matrix3D = new Matrix3D(3); // 初始化一个&nbsp;<code>3 x 3 x 3</code>&nbsp;的三维数组&nbsp;<code>matrix</code>，用全 0 填充。<br />
+matrix3D.setCell(0, 0, 0); // 将&nbsp;<code>matrix[0][0][0]</code> 设为 1。<br />
+matrix3D.largestMatrix(); // 返回 0。<code>matrix[0]</code>&nbsp;1 的数量最多。<br />
+matrix3D.setCell(1, 1, 2); // 将 <code>matrix[1][1][2]</code> 设为 1。<br />
+matrix3D.largestMatrix(); // 返回 1。<code>matrix[0]</code> 和&nbsp;<code>matrix[1]</code>&nbsp;1 的数量一样多，但下标 1 更大。<br />
+matrix3D.setCell(0, 0, 1); // 将 <code>matrix[0][0][1]</code> 设为 1。<br />
+matrix3D.largestMatrix(); // 返回 0。<code>matrix[0]</code>&nbsp;1 的数量最多。</div>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong class="example">示例 2：</strong></p>
 
 <div class="example-block">
-<p><strong>Input:</strong><br />
-<span class="example-io">[&quot;matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;unsetCell&quot;, &quot;largestMatrix&quot;]<br />
+<p><strong>输入：</strong><br />
+<span class="example-io">["Matrix3D", "setCell", "largestMatrix", "unsetCell", "largestMatrix"]<br />
 [[4], [2, 1, 1], [], [2, 1, 1], []]</span></p>
 
-<p><strong>Output:</strong><br />
+<p><strong>输出：</strong><br />
 <span class="example-io">[null, null, 2, null, 3] </span></p>
 
-<p><strong>Explanation</strong></p>
-matrix3D matrix3D = new matrix3D(4); // Initializes a <code>4 x 4 x 4</code> 3D array <code>matrix</code>, filled with all 0&#39;s.<br />
-matrix3D.setCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 1.<br />
-matrix3D.largestMatrix(); // Returns 2. <code>matrix[2]</code> has the most number of 1&#39;s.<br />
-matrix3D.unsetCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 0.<br />
-matrix3D.largestMatrix(); // Returns 3. All indices from 0 to 3 tie with the same number of 1&#39;s, but index 3 is the biggest.</div>
+<p><strong>解释：</strong></p>
+Matrix3D matrix3D = new matrix3D(4); // 初始化一个&nbsp;<code>4 x 4 x 4</code>&nbsp;的三维数组&nbsp;<code>matrix</code>，用全 0 填充。<br />
+matrix3D.setCell(2, 1, 1); // 将&nbsp;<code>matrix[2][1][1]</code> 设为 1。<br />
+matrix3D.largestMatrix(); // 返回 2。<code>matrix[2]</code>&nbsp;1 的数量最多。<br />
+matrix3D.unsetCell(2, 1, 1); // 将 <code>matrix[2][1][1]</code> 设为 0。<br />
+matrix3D.largestMatrix(); // 返回 3。0 到 3 的对应值都有相同数量的 1，但下标 3 最大。</div>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>1 &lt;= n &lt;= 100</code></li>
 	<li><code>0 &lt;= x, y, z &lt; n</code></li>
-	<li>At most <code>10<sup>5</sup></code> calls are made in total to <code>setCell</code> and <code>unsetCell</code>.</li>
-	<li>At most <code>10<sup>4</sup></code> calls are made to <code>largestMatrix</code>.</li>
+	<li>最多总共调用&nbsp;<code>10<sup>5</sup></code>&nbsp;次&nbsp;<code>setCell</code> 和&nbsp;<code>unsetCell</code>。</li>
+	<li>最多调用&nbsp;<code>10<sup>4</sup></code> 次&nbsp;<code>largestMatrix</code>。</li>
 </ul>
 
 <!-- description:end -->

solution/3300-3399/3391.Design a 3D Binary Matrix with Efficient Layer Tracking/README_EN.md

@@ -16,10 +16,10 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3391.De
 
 <p>You are given a <code>n x n x n</code> <strong>binary</strong> 3D array <code>matrix</code>.</p>
 
-<p>Implement the <code>matrix3D</code> class:</p>
+<p>Implement the <code>Matrix3D</code> class:</p>
 
 <ul>
-	<li><code>matrix3D(int n)</code> Initializes the object with the 3D binary array <code>matrix</code>, where <strong>all</strong> elements are initially set to 0.</li>
+	<li><code>Matrix3D(int n)</code> Initializes the object with the 3D binary array <code>matrix</code>, where <strong>all</strong> elements are initially set to 0.</li>
 	<li><code>void setCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 1.</li>
 	<li><code>void unsetCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 0.</li>
 	<li><code>int largestMatrix()</code> Returns the index <code>x</code> where <code>matrix[x]</code> contains the most number of 1&#39;s. If there are multiple such indices, return the <strong>largest</strong> <code>x</code>.</li>
@@ -30,14 +30,14 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3391.De
 
 <div class="example-block">
 <p><strong>Input:</strong><br />
-<span class="example-io">[&quot;matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;]<br />
+<span class="example-io">[&quot;Matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;]<br />
 [[3], [0, 0, 0], [], [1, 1, 2], [], [0, 0, 1], []]</span></p>
 
 <p><strong>Output:</strong><br />
 <span class="example-io">[null, null, 0, null, 1, null, 0] </span></p>
 
 <p><strong>Explanation</strong></p>
-matrix3D matrix3D = new matrix3D(3); // Initializes a <code>3 x 3 x 3</code> 3D array <code>matrix</code>, filled with all 0&#39;s.<br />
+Matrix3D matrix3D = new Matrix3D(3); // Initializes a <code>3 x 3 x 3</code> 3D array <code>matrix</code>, filled with all 0&#39;s.<br />
 matrix3D.setCell(0, 0, 0); // Sets <code>matrix[0][0][0]</code> to 1.<br />
 matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1&#39;s.<br />
 matrix3D.setCell(1, 1, 2); // Sets <code>matrix[1][1][2]</code> to 1.<br />
@@ -49,14 +49,14 @@ matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most numb
 
 <div class="example-block">
 <p><strong>Input:</strong><br />
-<span class="example-io">[&quot;matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;unsetCell&quot;, &quot;largestMatrix&quot;]<br />
+<span class="example-io">[&quot;Matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;unsetCell&quot;, &quot;largestMatrix&quot;]<br />
 [[4], [2, 1, 1], [], [2, 1, 1], []]</span></p>
 
 <p><strong>Output:</strong><br />
 <span class="example-io">[null, null, 2, null, 3] </span></p>
 
 <p><strong>Explanation</strong></p>
-matrix3D matrix3D = new matrix3D(4); // Initializes a <code>4 x 4 x 4</code> 3D array <code>matrix</code>, filled with all 0&#39;s.<br />
+Matrix3D matrix3D = new Matrix3D(4); // Initializes a <code>4 x 4 x 4</code> 3D array <code>matrix</code>, filled with all 0&#39;s.<br />
 matrix3D.setCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 1.<br />
 matrix3D.largestMatrix(); // Returns 2. <code>matrix[2]</code> has the most number of 1&#39;s.<br />
 matrix3D.unsetCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 0.<br />

solution/README.md

@@ -644,7 +644,7 @@
 |  0631  |  [设计 Excel 求和公式](/solution/0600-0699/0631.Design%20Excel%20Sum%20Formula/README.md)  |  `图`,`设计`,`拓扑排序`,`数组`,`矩阵`  |  困难  |  🔒  |
 |  0632  |  [最小区间](/solution/0600-0699/0632.Smallest%20Range%20Covering%20Elements%20from%20K%20Lists/README.md)  |  `贪心`,`数组`,`哈希表`,`排序`,`滑动窗口`,`堆（优先队列）`  |  困难  |    |
 |  0633  |  [平方数之和](/solution/0600-0699/0633.Sum%20of%20Square%20Numbers/README.md)  |  `数学`,`双指针`,`二分查找`  |  中等  |    |
-|  0634  |  [寻找数组的错位排列](/solution/0600-0699/0634.Find%20the%20Derangement%20of%20An%20Array/README.md)  |  `数学`,`动态规划`  |  中等  |  🔒  |
+|  0634  |  [寻找数组的错位排列](/solution/0600-0699/0634.Find%20the%20Derangement%20of%20An%20Array/README.md)  |  `数学`,`动态规划`,`组合数学`  |  中等  |  🔒  |
 |  0635  |  [设计日志存储系统](/solution/0600-0699/0635.Design%20Log%20Storage%20System/README.md)  |  `设计`,`哈希表`,`字符串`,`有序集合`  |  中等  |  🔒  |
 |  0636  |  [函数的独占时间](/solution/0600-0699/0636.Exclusive%20Time%20of%20Functions/README.md)  |  `栈`,`数组`  |  中等  |    |
 |  0637  |  [二叉树的层平均值](/solution/0600-0699/0637.Average%20of%20Levels%20in%20Binary%20Tree/README.md)  |  `树`,`深度优先搜索`,`广度优先搜索`,`二叉树`  |  简单  |    |
@@ -3401,7 +3401,7 @@
 |  3388  |  [统计数组中的美丽分割](/solution/3300-3399/3388.Count%20Beautiful%20Splits%20in%20an%20Array/README.md)  |  `数组`,`动态规划`  |  中等  |  第 428 场周赛  |
 |  3389  |  [使字符频率相等的最少操作次数](/solution/3300-3399/3389.Minimum%20Operations%20to%20Make%20Character%20Frequencies%20Equal/README.md)  |  `哈希表`,`字符串`,`动态规划`,`计数`,`枚举`  |  困难  |  第 428 场周赛  |
 |  3390  |  [Longest Team Pass Streak](/solution/3300-3399/3390.Longest%20Team%20Pass%20Streak/README.md)  |  `数据库`  |  困难  |  🔒  |
-|  3391  |  [Design a 3D Binary Matrix with Efficient Layer Tracking](/solution/3300-3399/3391.Design%20a%203D%20Binary%20Matrix%20with%20Efficient%20Layer%20Tracking/README.md)  |    |  中等  |  🔒  |
+|  3391  |  [设计一个高效的层跟踪三维二进制矩阵](/solution/3300-3399/3391.Design%20a%203D%20Binary%20Matrix%20with%20Efficient%20Layer%20Tracking/README.md)  |    |  中等  |  🔒  |
 |  3392  |  [统计符合条件长度为 3 的子数组数目](/solution/3300-3399/3392.Count%20Subarrays%20of%20Length%20Three%20With%20a%20Condition/README.md)  |    |  简单  |  第 146 场双周赛  |
 |  3393  |  [统计异或值为给定值的路径数目](/solution/3300-3399/3393.Count%20Paths%20With%20the%20Given%20XOR%20Value/README.md)  |    |  中等  |  第 146 场双周赛  |
 |  3394  |  [判断网格图能否被切割成块](/solution/3300-3399/3394.Check%20if%20Grid%20can%20be%20Cut%20into%20Sections/README.md)  |    |  中等  |  第 146 场双周赛  |