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Commit 1316a46

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feat: update lc problems (#3881)
1 parent b26ca5e commit 1316a46

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13 files changed

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

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<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>
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<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>
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<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>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<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;" />
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<strong>输出:</strong>1
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</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li>树中节点的数目范围是<code>[0, 5 * 10<sup>4</sup>]</code></li>
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<li><code>0 <= Node.val <= 5 * 10<sup>4</sup></code></li>
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<li><code>0 &lt;= Node.val &lt;= 5 * 10<sup>4</sup></code></li>
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<li>题目数据保证输入的树是 <strong>完全二叉树</strong></li>
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</ul>
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<p></p>
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<p>&nbsp;</p>
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<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>
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‎solution/0300-0399/0321.Create Maximum Number/README_EN.md‎

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

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

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tags:
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- 数学
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- 动态规划
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- 组合数学
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---
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<!-- problem:start -->

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

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tags:
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- Math
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- Dynamic Programming
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- Combinatorics
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---
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<!-- problem:start -->

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

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<!-- description:start -->
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<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row.</p>
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<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>
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<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>
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‎solution/0900-0999/0977.Squares of a Sorted Array/README.md‎

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<ul>
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</ul>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<strong>输出:</strong>[4,9,9,49,121]
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</pre>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code><span>1 <= nums.length <= </span>10<sup>4</sup></code></li>
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<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>
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<li><code><span>1 &lt;= nums.length &lt;= </span>10<sup>4</sup></code></li>
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<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
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<li><code>nums</code> 已按 <strong>非递减顺序</strong> 排序</li>
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</ul>
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<p></p>
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<p>&nbsp;</p>
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<p><strong>进阶:</strong></p>
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<ul>
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<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>
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<li>请你设计时间复杂度为 <code>O(n)</code> 的算法解决本问题</li>
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</ul>
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<!-- description:end -->

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

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<li>切割成四条长度为 <code>1</code>&nbsp;的绳子。</li>
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</ul>
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<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>
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<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>
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<p>&nbsp;</p>
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‎solution/2500-2599/2558.Take Gifts From the Richest Pile/README_EN.md‎

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<ul>
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<li>Choose the pile with the maximum number of gifts.</li>
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<li>If there is more than one pile with the maximum number of gifts, choose any.</li>
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<li>Leave behind the floor of the square root of the number of gifts in the pile. Take the rest of the gifts.</li>
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<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>
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</ul>
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<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

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<!-- problem:start -->
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# [3391. Design a 3D Binary Matrix with Efficient Layer Tracking 🔒](https://leetcode.cn/problems/design-a-3d-binary-matrix-with-efficient-layer-tracking)
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# [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)
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[English Version](/solution/3300-3399/3391.Design%20a%203D%20Binary%20Matrix%20with%20Efficient%20Layer%20Tracking/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>You are given a <code>n x n x n</code> <strong>binary</strong> 3D array <code>matrix</code>.</p>
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<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>
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<p>Implement the <code>matrix3D</code> class:</p>
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<p>实现&nbsp;<code>Matrix3D</code>&nbsp;类:</p>
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<ul>
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<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>
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<li><code>void setCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 1.</li>
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<li><code>void unsetCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 0.</li>
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<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>
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<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>
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<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>
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<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>
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<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>
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</ul>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
32-
<p><strong>Input:</strong><br />
33-
<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 />
33+
<p><strong>输入:</strong><br />
34+
<span class="example-io">["Matrix3D", "setCell", "largestMatrix", "setCell", "largestMatrix", "setCell", "largestMatrix"]<br />
3435
[[3], [0, 0, 0], [], [1, 1, 2], [], [0, 0, 1], []]</span></p>
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<p><strong>Output:</strong><br />
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<p><strong>输出:</strong><br />
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<span class="example-io">[null, null, 0, null, 1, null, 0] </span></p>
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<p><strong>Explanation</strong></p>
40-
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 />
41-
matrix3D.setCell(0, 0, 0); // Sets <code>matrix[0][0][0]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1&#39;s.<br />
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matrix3D.setCell(1, 1, 2); // Sets <code>matrix[1][1][2]</code> to 1.<br />
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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 />
45-
matrix3D.setCell(0, 0, 1); // Sets <code>matrix[0][0][1]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1&#39;s.</div>
40+
<p><strong>解释:</strong></p>
41+
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 />
42+
matrix3D.setCell(0, 0, 0); // 将&nbsp;<code>matrix[0][0][0]</code> 设为 1。<br />
43+
matrix3D.largestMatrix(); // 返回 0。<code>matrix[0]</code>&nbsp;1 的数量最多。<br />
44+
matrix3D.setCell(1, 1, 2); // 将 <code>matrix[1][1][2]</code> 设为 1。<br />
45+
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 />
46+
matrix3D.setCell(0, 0, 1); // 将 <code>matrix[0][0][1]</code> 设为 1。<br />
47+
matrix3D.largestMatrix(); // 返回 0。<code>matrix[0]</code>&nbsp;1 的数量最多。</div>
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<p><strong class="example">Example 2:</strong></p>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong><br />
52-
<span class="example-io">[&quot;matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;unsetCell&quot;, &quot;largestMatrix&quot;]<br />
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<p><strong>输入:</strong><br />
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<span class="example-io">["Matrix3D", "setCell", "largestMatrix", "unsetCell", "largestMatrix"]<br />
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[[4], [2, 1, 1], [], [2, 1, 1], []]</span></p>
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<p><strong>Output:</strong><br />
56+
<p><strong>输出:</strong><br />
5657
<span class="example-io">[null, null, 2, null, 3] </span></p>
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<p><strong>Explanation</strong></p>
59-
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 />
60-
matrix3D.setCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 1.<br />
61-
matrix3D.largestMatrix(); // Returns 2. <code>matrix[2]</code> has the most number of 1&#39;s.<br />
62-
matrix3D.unsetCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 0.<br />
63-
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>
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<p><strong>解释:</strong></p>
60+
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 />
61+
matrix3D.setCell(2, 1, 1); // 将&nbsp;<code>matrix[2][1][1]</code> 设为 1。<br />
62+
matrix3D.largestMatrix(); // 返回 2。<code>matrix[2]</code>&nbsp;1 的数量最多。<br />
63+
matrix3D.unsetCell(2, 1, 1); // 将 <code>matrix[2][1][1]</code> 设为 0。<br />
64+
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>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= n &lt;= 100</code></li>
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<li><code>0 &lt;= x, y, z &lt; n</code></li>
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<li>At most <code>10<sup>5</sup></code> calls are made in total to <code>setCell</code> and <code>unsetCell</code>.</li>
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<li>At most <code>10<sup>4</sup></code> calls are made to <code>largestMatrix</code>.</li>
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<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>
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<li>最多调用&nbsp;<code>10<sup>4</sup></code> 次&nbsp;<code>largestMatrix</code>。</li>
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</ul>
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<!-- description:end -->

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

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<p>You are given a <code>n x n x n</code> <strong>binary</strong> 3D array <code>matrix</code>.</p>
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<p>Implement the <code>matrix3D</code> class:</p>
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<p>Implement the <code>Matrix3D</code> class:</p>
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<ul>
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<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>
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<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>
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<li><code>void setCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 1.</li>
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<li><code>void unsetCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 0.</li>
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<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>
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<div class="example-block">
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<p><strong>Input:</strong><br />
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<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 />
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<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 />
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[[3], [0, 0, 0], [], [1, 1, 2], [], [0, 0, 1], []]</span></p>
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<p><strong>Output:</strong><br />
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<span class="example-io">[null, null, 0, null, 1, null, 0] </span></p>
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<p><strong>Explanation</strong></p>
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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 />
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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 />
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matrix3D.setCell(0, 0, 0); // Sets <code>matrix[0][0][0]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1&#39;s.<br />
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matrix3D.setCell(1, 1, 2); // Sets <code>matrix[1][1][2]</code> to 1.<br />
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<div class="example-block">
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<p><strong>Input:</strong><br />
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<span class="example-io">[&quot;matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;unsetCell&quot;, &quot;largestMatrix&quot;]<br />
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<span class="example-io">[&quot;Matrix3D&quot;, &quot;setCell&quot;, &quot;largestMatrix&quot;, &quot;unsetCell&quot;, &quot;largestMatrix&quot;]<br />
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[[4], [2, 1, 1], [], [2, 1, 1], []]</span></p>
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<p><strong>Output:</strong><br />
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<span class="example-io">[null, null, 2, null, 3] </span></p>
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<p><strong>Explanation</strong></p>
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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 />
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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 />
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matrix3D.setCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 2. <code>matrix[2]</code> has the most number of 1&#39;s.<br />
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matrix3D.unsetCell(2, 1, 1); // Sets <code>matrix[2][1][1]</code> to 0.<br />

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