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feat: add new lc problems #4634

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feat: add new lc problems
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4 changes: 3 additions & 1 deletion solution/0000-0099/0032.Longest Valid Parentheses/README.md
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<!-- description:start -->

<p>给你一个只包含 <code>'('</code>&nbsp;和 <code>')'</code>&nbsp;的字符串，找出最长有效（格式正确且连续）括号<span data-keyword="substring">子串</span>的长度。</p>
<p>给你一个只包含 <code>'('</code> 和 <code>')'</code> 的字符串，找出最长有效（格式正确且连续）括号 <span data-keyword="substring">子串</span> 的长度。</p>

<p>左右括号匹配，即每个左括号都有对应的右括号将其闭合的字符串是格式正确的，比如&nbsp;<code>"(()())"</code>。</p>

<p>&nbsp;</p>

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2 changes: 1 addition & 1 deletion solution/0000-0099/0033.Search in Rotated Sorted Array/README.md
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<p>整数数组 <code>nums</code> 按升序排列，数组中的值 <strong>互不相同</strong> 。</p>

<p>在传递给函数之前，<code>nums</code> 在预先未知的某个下标 <code>k</code>（<code>0 &lt;= k &lt; nums.length</code>）上进行了 <strong>旋转</strong>，使数组变为 <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code>（下标 <strong>从 0 开始</strong> 计数）。例如， <code>[0,1,2,4,5,6,7]</code> 在下标 <code>3</code> 处经旋转后可能变为&nbsp;<code>[4,5,6,7,0,1,2]</code> 。</p>
<p>在传递给函数之前，<code>nums</code> 在预先未知的某个下标 <code>k</code>（<code>0 &lt;= k &lt; nums.length</code>）上进行了 <strong>旋转</strong>，使数组变为 <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code>（下标 <strong>从 0 开始</strong> 计数）。例如， <code>[0,1,2,4,5,6,7]</code> 向左旋转&nbsp;<code>3</code>&nbsp;次后可能变为&nbsp;<code>[4,5,6,7,0,1,2]</code> 。</p>

<p>给你 <strong>旋转后</strong> 的数组 <code>nums</code> 和一个整数 <code>target</code> ，如果 <code>nums</code> 中存在这个目标值 <code>target</code> ，则返回它的下标，否则返回&nbsp;<code>-1</code>&nbsp;。</p>

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2 changes: 1 addition & 1 deletion solution/0000-0099/0033.Search in Rotated Sorted Array/README_EN.md
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<p>There is an integer array <code>nums</code> sorted in ascending order (with <strong>distinct</strong> values).</p>

<p>Prior to being passed to your function, <code>nums</code> is <strong>possibly rotated</strong> at an unknown pivot index <code>k</code> (<code>1 &lt;= k &lt; nums.length</code>) such that the resulting array is <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code> (<strong>0-indexed</strong>). For example, <code>[0,1,2,4,5,6,7]</code> might be rotated at pivot index <code>3</code> and become <code>[4,5,6,7,0,1,2]</code>.</p>
<p>Prior to being passed to your function, <code>nums</code> is <strong>possibly left rotated</strong> at an unknown index <code>k</code> (<code>1 &lt;= k &lt; nums.length</code>) such that the resulting array is <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code> (<strong>0-indexed</strong>). For example, <code>[0,1,2,4,5,6,7]</code> might be left rotated by&nbsp;<code>3</code>&nbsp;indices and become <code>[4,5,6,7,0,1,2]</code>.</p>

<p>Given the array <code>nums</code> <strong>after</strong> the possible rotation and an integer <code>target</code>, return <em>the index of </em><code>target</code><em> if it is in </em><code>nums</code><em>, or </em><code>-1</code><em> if it is not in </em><code>nums</code>.</p>

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8 changes: 4 additions & 4 deletions solution/0000-0099/0045.Jump Game II/README.md
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<p>给定一个长度为 <code>n</code> 的 <strong>0 索引</strong>整数数组 <code>nums</code>。初始位置为 <code>nums[0]</code>。</p>

<p>每个元素 <code>nums[i]</code> 表示从索引 <code>i</code> 向后跳转的最大长度。换句话说，如果你在 <code>nums[i]</code> 处，你可以跳转到任意 <code>nums[i + j]</code> 处:</p>
<p>每个元素 <code>nums[i]</code> 表示从索引 <code>i</code> 向后跳转的最大长度。换句话说，如果你在索引&nbsp;<code>i</code>&nbsp;处，你可以跳转到任意 <code>(i + j)</code> 处：</p>

<ul>
<li><code>0 &lt;= j &lt;= nums[i]</code>&nbsp;</li>
<li><code>0 &lt;= j &lt;= nums[i]</code>&nbsp;且</li>
<li><code>i + j &lt; n</code></li>
</ul>

<p>返回到达&nbsp;<code>nums[n - 1]</code> 的最小跳跃次数。生成的测试用例可以到达 <code>nums[n - 1]</code>。</p>
<p>返回到达&nbsp;<code>n - 1</code>&nbsp;的最小跳跃次数。测试用例保证可以到达 <code>n - 1</code>。</p>

<p>&nbsp;</p>

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<ul>
<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
<li>题目保证可以到达&nbsp;<code>nums[n-1]</code></li>
<li>题目保证可以到达&nbsp;<code>n - 1</code></li>
</ul>

<!-- description:end -->
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6 changes: 3 additions & 3 deletions solution/0000-0099/0045.Jump Game II/README_EN.md
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<!-- description:start -->

<p>You are given a <strong>0-indexed</strong> array of integers <code>nums</code> of length <code>n</code>. You are initially positioned at <code>nums[0]</code>.</p>
<p>You are given a <strong>0-indexed</strong> array of integers <code>nums</code> of length <code>n</code>. You are initially positioned at&nbsp;index 0.</p>

<p>Each element <code>nums[i]</code> represents the maximum length of a forward jump from index <code>i</code>. In other words, if you are at <code>nums[i]</code>, you can jump to any <code>nums[i + j]</code> where:</p>
<p>Each element <code>nums[i]</code> represents the maximum length of a forward jump from index <code>i</code>. In other words, if you are at index <code>i</code>, you can jump to any index <code>(i + j)</code>&nbsp;where:</p>

<ul>
<li><code>0 &lt;= j &lt;= nums[i]</code> and</li>
<li><code>i + j &lt; n</code></li>
</ul>

<p>Return <em>the minimum number of jumps to reach </em><code>nums[n - 1]</code>. The test cases are generated such that you can reach <code>nums[n - 1]</code>.</p>
<p>Return <em>the minimum number of jumps to reach index </em><code>n - 1</code>. The test cases are generated such that you can reach index&nbsp;<code>n - 1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
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2 changes: 1 addition & 1 deletion solution/0100-0199/0190.Reverse Bits/README.md
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<ul>
<li>请注意，在某些语言（如 Java）中，没有无符号整数类型。在这种情况下，输入和输出都将被指定为有符号整数类型，并且不应影响您的实现，因为无论整数是有符号的还是无符号的，其内部的二进制表示形式都是相同的。</li>
<li>在 Java 中，编译器使用<a href="https://baike.baidu.com/item/二进制补码/5295284" target="_blank">二进制补码</a>记法来表示有符号整数。因此，在 <strong>示例 2</strong>&nbsp;中，输入表示有符号整数 <code>-3</code>，输出表示有符号整数 <code>-1073741825</code>。</li>
<li>在 Java 中，编译器使用<a href="https://baike.baidu.com/item/二进制补码/5295284" target="_blank">二进制补码</a>记法来表示有符号整数。</li>
</ul>

<p>&nbsp;</p>
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2 changes: 1 addition & 1 deletion solution/0100-0199/0190.Reverse Bits/README_EN.md
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<ul>
<li>Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer&#39;s internal binary representation is the same, whether it is signed or unsigned.</li>
<li>In Java, the compiler represents the signed integers using <a href="https://en.wikipedia.org/wiki/Two%27s_complement" target="_blank">2&#39;s complement notation</a>. Therefore, in <strong class="example">Example 2</strong>&nbsp;below, the input represents the signed integer <code>-3</code> and the output represents the signed integer <code>-1073741825</code>.</li>
<li>In Java, the compiler represents the signed integers using <a href="https://en.wikipedia.org/wiki/Two%27s_complement" target="_blank">2&#39;s complement notation</a>.</li>
</ul>

<p>&nbsp;</p>
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1 change: 1 addition & 0 deletions solution/0300-0399/0307.Range Sum Query - Mutable/README.md
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- 树状数组
- 线段树
- 数组
- 分治
---

<!-- problem:start -->
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1 change: 1 addition & 0 deletions solution/0300-0399/0307.Range Sum Query - Mutable/README_EN.md
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- Binary Indexed Tree
- Segment Tree
- Array
- Divide and Conquer
---

<!-- problem:start -->
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6 changes: 3 additions & 3 deletions solution/0300-0399/0308.Range Sum Query 2D - Mutable/README.md
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Expand Up @@ -62,13 +62,13 @@ numMatrix.sumRegion(2, 1, 4, 3); // 返回 10 (即，右侧红色矩形的和)
<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>-10<sup>5</sup> &lt;= matrix[i][j] &lt;= 10<sup>5</sup></code></li>
<li><code>-10<font size="1">00</font>&nbsp;&lt;= matrix[i][j] &lt;= 10<font size="1">00</font></code></li>
<li><code>0 &lt;= row &lt; m</code></li>
<li><code>0 &lt;= col &lt; n</code></li>
<li><code>-10<sup>5</sup> &lt;= val &lt;= 10<sup>5</sup></code></li>
<li><code>-10<font size="1">00</font>&nbsp;&lt;= val &lt;= 10<font size="1">00</font></code></li>
<li><code>0 &lt;= row1 &lt;= row2 &lt; m</code></li>
<li><code>0 &lt;= col1 &lt;= col2 &lt; n</code></li>
<li>最多调用<code>10<sup>4</sup></code> 次&nbsp;<code>sumRegion</code> 和 <code>update</code> 方法</li>
<li>最多调用<code>5000</code>&nbsp;次&nbsp;<code>sumRegion</code> 和 <code>update</code> 方法</li>
</ul>

<!-- description:end -->
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2 changes: 2 additions & 0 deletions solution/0300-0399/0394.Decode String/README.md
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<p>此外，你可以认为原始数据不包含数字，所有的数字只表示重复的次数 <code>k</code> ，例如不会出现像&nbsp;<code>3a</code>&nbsp;或&nbsp;<code>2[4]</code>&nbsp;的输入。</p>

<p>测试用例保证输出的长度不会超过&nbsp;<code>10<sup>5</sup></code>。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>
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2 changes: 1 addition & 1 deletion solution/0400-0499/0475.Heaters/README_EN.md
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<p>Given the positions of <code>houses</code> and <code>heaters</code> on a horizontal line, return <em>the minimum radius standard of heaters&nbsp;so that those heaters could cover all houses.</em></p>

<p><strong>Notice</strong> that&nbsp;all the <code>heaters</code> follow your radius standard, and the warm radius will the same.</p>
<p><strong>Notice</strong> that&nbsp;all the <code>heaters</code> follow your radius standard, and the warm radius will be the same.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
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24 changes: 13 additions & 11 deletions solution/0400-0499/0499.The Maze III/README.md
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<!-- description:start -->

<p>由空地和墙组成的迷宫中有一个<strong>球</strong>。球可以向<strong>上（u）下（d）左（l）右（r）</strong>四个方向滚动，但在遇到墙壁前不会停止滚动。当球停下时，可以选择下一个方向。迷宫中还有一个<strong>洞</strong>，当球运动经过洞时，就会掉进洞里。</p>
<p>由空地和墙组成的迷宫中有一个<strong>球</strong>。球可以向<strong>上（u）下（d）左（l）右（r）</strong>四个方向滚动，但在遇到墙壁前不会停止滚动。当球停下时，可以选择下一个方向（必须与上一个选择的方向不同）。迷宫中还有一个<strong>洞</strong>，当球运动经过洞时，就会掉进洞里。</p>

<p>给定球的<strong>起始位置，目的地</strong>和<strong>迷宫</strong>，找出让球以最短距离掉进洞里的路径。&nbsp;距离的定义是球从起始位置（不包括）到目的地（包括）经过的<strong>空地</strong>个数。通过&#39;u&#39;, &#39;d&#39;, &#39;l&#39; 和&nbsp;&#39;r&#39;输出球的移动<strong>方向</strong>。&nbsp;由于可能有多条最短路径，&nbsp;请输出<strong>字典序最小</strong>的路径<strong>。</strong>如果球无法进入洞，输出&quot;impossible&quot;。</p>
<p>给定球的<strong>起始位置，目的地</strong>和<strong>迷宫</strong>，找出让球以最短距离掉进洞里的路径。&nbsp;距离的定义是球从起始位置（不包括）到目的地（包括）经过的<strong>空地</strong>个数。通过'u', 'd', 'l' 和&nbsp;'r'输出球的移动<strong>方向</strong>。&nbsp;由于可能有多条最短路径，&nbsp;请输出<strong>字典序最小</strong>的路径<strong>。</strong>如果球无法进入洞，输出"impossible"。</p>

<p>迷宫由一个0和1的二维数组表示。 1表示墙壁，0表示空地。你可以假定迷宫的边缘都是墙壁。起始位置和目的地的坐标通过行号和列号给出。</p>

<p>&nbsp;</p>

<p><strong>示例1:</strong></p>

<pre><strong>输入 1:</strong> 迷宫由以下二维数组表示
<pre>
<strong>输入 1:</strong> 迷宫由以下二维数组表示

0 0 0 0 0
1 1 0 0 1
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<strong>输入 2:</strong> 球的初始位置 (rowBall, colBall) = (4, 3)
<strong>输入 3:</strong> 洞的位置 (rowHole, colHole) = (0, 1)

<strong>输出:</strong> &quot;lul&quot;
<strong>输出:</strong> "lul"

<strong>解析:</strong> 有两条让球进洞的最短路径。
第一条路径是 左 -&gt; 上 -&gt; 左, 记为 &quot;lul&quot;.
第二条路径是 上 -&gt; 左, 记为 &#39;ul&#39;.
两条路径都具有最短距离6, 但&#39;l&#39; &lt; &#39;u&#39;，故第一条路径字典序更小。因此输出&quot;lul&quot;。
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0499.The%20Maze%20III/images/maze_2_example_1.png" style="width: 100%;">
第一条路径是 左 -&gt; 上 -&gt; 左, 记为 "lul".
第二条路径是 上 -&gt; 左, 记为 'ul'.
两条路径都具有最短距离6, 但'l' &lt; 'u'，故第一条路径字典序更小。因此输出"lul"。
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0499.The%20Maze%20III/images/maze_2_example_1.png" style="width: 100%;" />
</pre>

<p><strong>示例&nbsp;2:</strong></p>

<pre><strong>输入 1:</strong> 迷宫由以下二维数组表示
<pre>
<strong>输入 1:</strong> 迷宫由以下二维数组表示

0 0 0 0 0
1 1 0 0 1
Expand All @@ -66,10 +68,10 @@ tags:
<strong>输入 2:</strong> 球的初始位置 (rowBall, colBall) = (4, 3)
<strong>输入 3:</strong> 洞的位置 (rowHole, colHole) = (3, 0)

<strong>输出:</strong> &quot;impossible&quot;
<strong>输出:</strong> "impossible"

<strong>示例:</strong> 球无法到达洞。
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0499.The%20Maze%20III/images/maze_2_example_2.png" style="width: 100%;">
<img src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0499.The%20Maze%20III/images/maze_2_example_2.png" style="width: 100%;" />
</pre>

<p>&nbsp;</p>
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2 changes: 1 addition & 1 deletion solution/0500-0599/0544.Output Contest Matches/README.md
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<p><strong>提示：</strong></p>

<ul>
<li><code>n == 2x</code>，并且 <code>x</code> 在范围 <code>[1,12]</code> 内。</li>
<li><code>n == 2<sup>x</sup></code>，并且 <code>x</code> 在范围 <code>[1, 12]</code> 内。</li>
</ul>

<!-- description:end -->
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41 changes: 27 additions & 14 deletions solution/0800-0899/0808.Soup Servings/README.md
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<!-- description:start -->

<p>有&nbsp;<strong>A&nbsp;和&nbsp;B 两种类型&nbsp;</strong>的汤。一开始每种类型的汤有&nbsp;<code>n</code>&nbsp;毫升。有四种分配操作：</p>
<p>你有两种汤，<strong>A</strong> 和 <strong>B</strong>，每种初始为 <code>n</code>&nbsp;毫升。在每一轮中，会随机选择以下四种操作中的一种，每种操作的概率为 <code>0.25</code>，且与之前的所有轮次 <strong>无关</strong>：</p>

<ol>
<li>提供 <code>100ml</code> 的 <strong>汤A</strong> 和 <code>0ml</code> 的 <strong>汤B</strong> 。</li>
<li>提供 <code>75ml</code> 的 <strong>汤A</strong> 和 <code>25ml</code> 的 <strong>汤B</strong> 。</li>
<li>提供 <code>50ml</code> 的 <strong>汤A</strong> 和 <code>50ml</code> 的 <strong>汤B</strong> 。</li>
<li>提供 <code>25ml</code> 的 <strong>汤A</strong> 和 <code>75ml</code> 的 <strong>汤B</strong> 。</li>
<li>从汤 A 取 100 毫升，从汤 B 取 0 毫升</li>
<li>从汤 A 取 75 毫升，从汤 B 取 25 毫升</li>
<li>从汤 A 取 50 毫升，从汤 B 取 50 毫升</li>
<li>从汤 A 取 25 毫升，从汤 B 取 75 毫升</li>
</ol>

<p>当我们把汤分配给某人之后，汤就没有了。每个回合，我们将从四种概率同为 <code>0.25</code> 的操作中进行分配选择。如果汤的剩余量不足以完成某次操作，我们将尽可能分配。当两种类型的汤都分配完时，停止操作。</p>
<p><strong>注意：</strong></p>

<p><strong>注意&nbsp;</strong>不存在先分配 <code>100</code> ml <strong>汤B</strong> 的操作。</p>
<ul>
<li>不存在从汤 A 取&nbsp;<code>0</code>&nbsp;ml 和从汤 B 取&nbsp;<code>100</code> ml 的操作。</li>
<li>汤 A 和 B 在每次操作中同时被取出。</li>
<li>如果一次操作要求你取出比剩余的汤更多的量，请取出该汤剩余的所有部分。</li>
</ul>

<p>操作过程在任何回合中任一汤被取完后立即停止。</p>

<p>需要返回的值：&nbsp;<strong>汤A&nbsp;</strong>先分配完的概率 +&nbsp;&nbsp;<strong>汤A和汤B&nbsp;</strong>同时分配完的概率 / 2。返回值在正确答案&nbsp;<code>10<sup>-5</sup></code>&nbsp;的范围内将被认为是正确的。</p>
<p>返回汤 A 在 B 前取完的概率，加上两种汤在 <strong>同一回合&nbsp;</strong>取完概率的一半。返回值在正确答案&nbsp;<code>10<sup>-5</sup></code>&nbsp;的范围内将被认为是正确的。</p>

<p>&nbsp;</p>

<p><strong>示例 1:</strong></p>

<pre>
<strong>输入:</strong> n = 50
<strong>输出:</strong> 0.62500
<strong>解释:</strong>如果我们选择前两个操作<strong>，</strong>A 首先将变为空。
<strong>输入：</strong>n = 50
<strong>输出：</strong>0.62500
<strong>解释：
</strong>如果我们选择前两个操作<strong>，</strong>A 首先将变为空。
对于第三个操作，A 和 B 会同时变为空。
对于第四个操作，B 首先将变为空。<strong>
</strong>所以 A 变为空的总概率加上 A 和 B 同时变为空的概率的一半是 0.25 *(1 + 1 + 0.5 + 0)= 0.625。
Expand All @@ -49,16 +56,22 @@ tags:
<p><strong>示例 2:</strong></p>

<pre>
<strong>输入:</strong> n = 100
<strong>输出:</strong> 0.71875
<strong>输入：</strong>n = 100
<strong>输出：</strong>0.71875
<strong>解释：</strong>
如果我们选择第一个操作，A 首先将变为空。
如果我们选择第二个操作，A 将在执行操作 [1, 2, 3] 时变为空，然后 A 和 B 在执行操作 4 时同时变空。
如果我们选择第三个操作，A 将在执行操作 [1, 2] 时变为空，然后 A 和 B 在执行操作 3 时同时变空。
如果我们选择第四个操作，A 将在执行操作 1 时变为空，然后 A 和 B 在执行操作 2 时同时变空。
所以 A 变为空的总概率加上 A 和 B 同时变为空的概率的一半是 0.71875。
</pre>

<p>&nbsp;</p>

<p><strong>提示:</strong></p>

<ul>
<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code>​​​​​​​</li>
<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

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