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Commit d0e8c21

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feat: add new lc problems
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‎solution/0400-0499/0424.Longest Repeating Character Replacement/README_EN.md‎

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<strong>Output:</strong> 4
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<strong>Explanation:</strong> Replace the one &#39;A&#39; in the middle with &#39;B&#39; and form &quot;AABBBBA&quot;.
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The substring &quot;BBBB&quot; has the longest repeating letters, which is 4.
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</pre>
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There may exists other ways to achive this answer too.</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>

‎solution/0400-0499/0436.Find Right Interval/README.md‎

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<p>给你一个区间数组 <code>intervals</code> ,其中&nbsp;<code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> ,且每个&nbsp;<code>start<sub>i</sub></code> 都 <strong>不同</strong> 。</p>
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<p>区间 <code>i</code> 的 <strong>右侧区间</strong> 可以记作区间 <code>j</code> ,并满足 <code>start<sub>j</sub></code><code>&nbsp;&gt;= end<sub>i</sub></code> ,且 <code>start<sub>j</sub></code> <strong>最小化 </strong>。</p>
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<p>区间 <code>i</code> 的 <strong>右侧区间</strong> 可以记作区间 <code>j</code> ,并满足 <code>start<sub>j</sub></code><code>&nbsp;&gt;= end<sub>i</sub></code> ,且 <code>start<sub>j</sub></code> <strong>最小化 </strong>。注意 <code>i</code> 可能等于 <code>j</code> 。</p>
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<p>返回一个由每个区间 <code>i</code> 的 <strong>右侧区间</strong> 在&nbsp;<code>intervals</code> 中对应下标组成的数组。如果某个区间 <code>i</code> 不存在对应的 <strong>右侧区间</strong> ,则下标 <code>i</code> 处的值设为 <code>-1</code> 。</p>
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&nbsp;

‎solution/0800-0899/0852.Peak Index in a Mountain Array/README.md‎

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符合下列属性的数组 <code>arr</code> 称为 <strong>山脉数组</strong> :
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<ul>
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<li><code>arr.length >= 3</code></li>
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<li>存在 <code>i</code>(<code>0 < i < arr.length - 1</code>)使得:
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<li><code>arr.length &gt;= 3</code></li>
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<li>存在 <code>i</code>(<code>0 &lt; i&nbsp;&lt; arr.length - 1</code>)使得:
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<ul>
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<li><code>arr[0] < arr[1] < ... arr[i-1] < arr[i] </code></li>
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<li><code>arr[i] > arr[i+1] > ... > arr[arr.length - 1]</code></li>
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<li><code>arr[0] &lt; arr[1] &lt; ... arr[i-1] &lt; arr[i] </code></li>
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<li><code>arr[i] &gt; arr[i+1] &gt; ... &gt; arr[arr.length - 1]</code></li>
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</ul>
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</li>
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</ul>
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<p>给你由整数组成的山脉数组 <code>arr</code> ,返回任何满足 <code>arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1]</code> 的下标 <code>i</code> 。</p>
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<p>给你由整数组成的山脉数组 <code>arr</code> ,返回满足 <code>arr[0] &lt; arr[1] &lt; ... arr[i - 1] &lt; arr[i] &gt; arr[i + 1] &gt; ... &gt; arr[arr.length - 1]</code> 的下标 <code>i</code> 。</p>
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<p> </p>
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<p>你必须设计并实现时间复杂度为 <code>O(log(n))</code> 的解决方案。</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>1
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</pre>
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<p><strong>示例 4:</strong></p>
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<pre>
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<strong>输入:</strong>arr = [3,4,5,1]
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<strong>输出:</strong>2
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</pre>
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<p><strong>示例 5:</strong></p>
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<pre>
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<strong>输入:</strong>arr = [24,69,100,99,79,78,67,36,26,19]
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<strong>输出:</strong>2
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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>3 <= arr.length <= 10<sup>4</sup></code></li>
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<li><code>0 <= arr[i] <= 10<sup>6</sup></code></li>
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<li><code>3 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= arr[i] &lt;= 10<sup>6</sup></code></li>
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<li>题目数据保证 <code>arr</code> 是一个山脉数组</li>
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</ul>
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<p> </p>
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<p><strong>进阶:</strong>很容易想到时间复杂度 <code>O(n)</code> 的解决方案,你可以设计一个 <code>O(log(n))</code> 的解决方案吗?</p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

‎solution/0800-0899/0877.Stone Game/README_EN.md‎

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if i > j:
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return 0
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return max(piles[i] - dfs(i + 1, j), piles[j] - dfs(i, j - 1))
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return dfs(0, len(piles) - 1) > 0
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```
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‎solution/1000-1099/1071.Greatest Common Divisor of Strings/README.md‎

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<p>对于字符串&nbsp;<code>s</code> 和&nbsp;<code>t</code>,只有在&nbsp;<code>s = t + ... + t</code>(<code>t</code> 自身连接 1 次或多次)时,我们才认定&nbsp;"<code>t</code> 能除尽 <code>s</code>"。</p>
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<p>给定两个字符串&nbsp;<code>str1</code>&nbsp;和&nbsp;<code>str2</code>&nbsp;。返回 <em>最长字符串&nbsp;<code>x</code>,要求满足&nbsp;<code>x</code> 能除尽 <code>str1</code> 且&nbsp;<code>X</code> 能除尽 <code>str2</code></em> 。</p>
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<p>给定两个字符串&nbsp;<code>str1</code>&nbsp;和&nbsp;<code>str2</code>&nbsp;。返回 <em>最长字符串&nbsp;<code>x</code>,要求满足&nbsp;<code>x</code> 能除尽 <code>str1</code> 且<code>x</code> 能除尽 <code>str2</code></em> 。</p>
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<p>&nbsp;</p>
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‎solution/1300-1399/1339.Maximum Product of Splitted Binary Tree/README_EN.md‎

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}
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return t;
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};
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dfs(root);
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return ans % mod;
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}

‎solution/1300-1399/1384.Total Sales Amount by Year/README.md‎

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product_id 是这张表的主键。
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period_start&nbsp;和 period_end&nbsp;是该产品销售期的起始日期和结束日期,且这两个日期包含在销售期内。
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average_daily_sales 列存储销售期内该产品的日平均销售额。
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销售日期范围为2018年到2020年。
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</pre>
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<p>&nbsp;</p>
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<p>编写一段 SQL 查询每个产品每年的总销售额,并包含 product_id, product_name 以及 report_year 等信息。</p>
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<p>编写一段 SQL 查询,查找出每个产品每年的总销售额,并包含 <code>product_id</code> , <code>product_name</code> ,&nbsp;<code>report_year</code> 以及 <code>total_amount</code>&nbsp;。</p>
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<p>销售年份的日期介于 2018 年到 2020 年之间。你返回的结果需要按&nbsp;product_id 和 report_year<strong> 排序</strong>。</p>
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<p>返回结果并按&nbsp;<code>product_id</code> 和 <code>report_year</code><strong> 排序</strong>。</p>
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<p>查询结果格式如下例所示。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1:</strong></p>
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<p><strongclass="example">示例 1:</strong></p>
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<pre>
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<code><strong>输入:</strong>

‎solution/1300-1399/1384.Total Sales Amount by Year/README_EN.md‎

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<p>&nbsp;</p>
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<p>Write an SQL query to report the total sales amount of each item for each year, with corresponding <code>product_name</code>, <code>product_id</code>, <code>product_name</code>, and <code>report_year</code>.</p>
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<p>Write an SQL query to report the total sales amount of each item for each year, with corresponding <code>product_name</code>, <code>product_id</code>, <code>report_year</code>, and <code>total_amount</code>.</p>
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<p>Return the result table <strong>ordered</strong> by <code>product_id</code> and <code>report_year</code>.</p>
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‎solution/1600-1699/1601.Maximum Number of Achievable Transfer Requests/README_EN.md‎

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public int maximumRequests(int n, int[][] requests) {
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m = requests.length;
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this.n = n;
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this.requests = requests;
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this.requests = requests;
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int ans = 0;
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for (int mask = 0; mask < 1 << m; ++mask) {
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int cnt = Integer.bitCount(mask);

‎solution/1700-1799/1751.Maximum Number of Events That Can Be Attended II/README.md‎

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j = bisect_right(events, ed, lo=i + 1, key=lambda x: x[0])
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ans = max(ans, dfs(j, k - 1) + val)
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return ans
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events.sort()
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return dfs(0, k)
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```

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