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feat: add solutions to lc problem: No.3476 (#4124)
No.3476.Maximize Profit from Task Assignment
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‎solution/3400-3499/3470.Permutations IV/README.md

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<!-- problem:start -->
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# [3470. 排列 IV](https://leetcode.cn/problems/permutations-iv)
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# [3470. 全排列 IV](https://leetcode.cn/problems/permutations-iv)
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[English Version](/solution/3400-3499/3470.Permutations%20IV/README_EN.md)
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‎solution/3400-3499/3475.DNA Pattern Recognition/README.md

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<!-- problem:start -->
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# [3475. DNA Pattern Recognition](https://leetcode.cn/problems/dna-pattern-recognition)
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# [3475. DNA 模式识别](https://leetcode.cn/problems/dna-pattern-recognition)
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[English Version](/solution/3400-3499/3475.DNA%20Pattern%20Recognition/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>Table: <code>Samples</code></p>
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<p>表：<code>Samples</code></p>
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<pre>
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+----------------+---------+
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| dna_sequence | varchar |
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| species | varchar |
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+----------------+---------+
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sample_id is the unique key for this table.
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Each row contains a DNA sequence represented as a string of characters (A, T, G, C) and the species it was collected from.
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sample_id 是这张表的唯一主键。
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每一行包含一个 DNA 序列以一个字符（A，T，G，C）组成的字符串表示以及它所采集自的物种。
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</pre>
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<p>Biologists are studying basic patterns in DNA sequences. Write a solution to identify <code>sample_id</code> with the following patterns:</p>
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<p>生物学家正在研究 DNA 序列中的基本模式。编写一个解决方案以识别具有以下模式的&nbsp;<code>sample_id</code>：</p>
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<ul>
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<li>Sequences that <strong>start</strong> with <strong>ATG</strong>&nbsp;(a common <strong>start codon</strong>)</li>
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<li>Sequences that <strong>end</strong> with either <strong>TAA</strong>, <strong>TAG</strong>, or <strong>TGA</strong>&nbsp;(<strong>stop codons</strong>)</li>
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<li>Sequences containing the motif <strong>ATAT</strong>&nbsp;(a simple repeated pattern)</li>
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<li>Sequences that have <strong>at least</strong> <code>3</code> <strong>consecutive</strong> <strong>G</strong>&nbsp;(like <strong>GGG</strong>&nbsp;or <strong>GGGG</strong>)</li>
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<li>以&nbsp;<strong>ATG</strong> <strong>开头</strong>&nbsp;的序列（一个常见的 <strong>起始密码子</strong>）</li>
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<li>以 <strong>TAA</strong>，<strong>TAG</strong>&nbsp;或&nbsp;<strong>TGA</strong>&nbsp;<strong>结尾</strong>&nbsp;的序列（终止密码子）</li>
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<li>包含基序 <strong>ATAT</strong> 的序列（一个简单重复模式）</li>
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<li>有 <strong>至少</strong>&nbsp;<code>3</code>&nbsp;<strong>个连续</strong>&nbsp;<strong>G</strong>&nbsp;的序列（如&nbsp;<strong>GGG</strong>&nbsp;或&nbsp;<strong>GGGG</strong>）</li>
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</ul>
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<p>Return <em>the result table ordered by&nbsp;</em><em>sample_id in <strong>ascending</strong> order</em>.</p>
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<p>返回结果表以&nbsp;sample_id <strong>升序</strong>&nbsp;排序<em>。</em></p>
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<p>The result format is in the following example.</p>
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<p>结果格式如下所示。</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example:</strong></p>
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<p><strong class="example">示例：</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong></p>
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<p><strong>输入：</strong></p>
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<p>Samples table:</p>
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<p>Samples 表：</p>
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<pre class="example-io">
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+-----------+------------------+-----------+
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</pre>
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<p><strong>Output:</strong></p>
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<p><strong>输出：</strong></p>
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<pre class="example-io">
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+-----------+------------------+-------------+-------------+------------+------------+------------+
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+-----------+------------------+-------------+-------------+------------+------------+------------+
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</pre>
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<p><strong>Explanation:</strong></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>Sample 1 (ATGCTAGCTAGCTAA):
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<li>样本 1（ATGCTAGCTAGCTAA）：
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<ul>
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<li>Starts with ATG&nbsp;(has_start = 1)</li>
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<li>Ends with TAA&nbsp;(has_stop = 1)</li>
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<li>Does not contain ATAT&nbsp;(has_atat = 0)</li>
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<li>Does not contain at least 3 consecutive &#39;G&#39;s (has_ggg = 0)</li>
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<li>以 ATG 开头（has_start = 1）</li>
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<li>以 TAA 结尾（has_stop = 1）</li>
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<li>不包含 ATAT（has_atat = 0）</li>
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<li>不包含至少 3 个连续 ‘G’（has_ggg = 0）</li>
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</ul>
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<li>Sample 2 (GGGTCAATCATC):
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<li>样本 2（GGGTCAATCATC）：
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<li>Does not start with ATG&nbsp;(has_start = 0)</li>
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<li>Does not end with TAA, TAG, or TGA&nbsp;(has_stop = 0)</li>
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<li>Does not contain ATAT&nbsp;(has_atat = 0)</li>
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<li>Contains GGG&nbsp;(has_ggg = 1)</li>
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<li>不以 ATG 开头（has_start = 0）</li>
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<li>不以 TAA，TAG 或 TGA 结尾（has_stop = 0）</li>
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<li>不包含 ATAT（has_atat = 0）</li>
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<li>包含 GGG（has_ggg = 1）</li>
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</ul>
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</li>
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<li>Sample 3 (ATATATCGTAGCTA):
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<li>样本 3（ATATATCGTAGCTA）：
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<li>Does not start with ATG&nbsp;(has_start = 0)</li>
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<li>Does not end with TAA, TAG, or TGA&nbsp;(has_stop = 0)</li>
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<li>Contains ATAT&nbsp;(has_atat = 1)</li>
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<li>Does not contain at least 3 consecutive &#39;G&#39;s (has_ggg = 0)</li>
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<li>不以 ATG 开头（has_start = 0）</li>
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<li>不以 TAA，TAG 或 TGA 结尾（has_stop = 0）</li>
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<li>包含 ATAT（has_atat = 1）</li>
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<li>不包含至少 3 个连续 ‘G’（has_ggg = 0）</li>
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<li>Sample 4 (ATGGGGTCATCATAA):
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<li>样本 4（ATGGGGTCATCATAA）：
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<li>Starts with ATG&nbsp;(has_start = 1)</li>
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<li>Ends with TAA&nbsp;(has_stop = 1)</li>
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<li>Does not contain ATAT&nbsp;(has_atat = 0)</li>
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<li>Contains GGGG&nbsp;(has_ggg = 1)</li>
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<li>以 ATG 开头（has_start = 1）</li>
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<li>以 TAA 结尾（has_stop = 1）</li>
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<li>不包含 ATAT（has_atat = 0）</li>
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<li>包含 GGGG（has_ggg = 1）</li>
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</ul>
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<li>Sample 5 (TCAGTCAGTCAG):
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<li>样本 5（TCAGTCAGTCAG）：
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<li>Does not match any patterns (all fields = 0)</li>
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<li>不匹配任何模式（所有字段 = 0）</li>
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</ul>
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<li>Sample 6 (ATATCGCGCTAG):
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<li>样本 6（ATATCGCGCTAG）：
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<li>Does not start with ATG&nbsp;(has_start = 0)</li>
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<li>Ends with TAG&nbsp;(has_stop = 1)</li>
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<li>Starts with ATAT&nbsp;(has_atat = 1)</li>
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<li>Does not contain at least 3 consecutive &#39;G&#39;s (has_ggg = 0)</li>
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<li>不以 ATG 开头（has_start = 0）</li>
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<li>以 TAG 结尾（has_stop = 1）</li>
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<li>包含 ATAT（has_atat = 1）</li>
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<li>不包含至少 3 个连续 ‘G’（has_ggg = 0）</li>
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<li>Sample 7 (CGTATGCGTCGTA):
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<li>样本 7（CGTATGCGTCGTA）：
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<li>Does not start with ATG&nbsp;(has_start = 0)</li>
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<li>Does not end with TAA, &quot;TAG&quot;, or &quot;TGA&quot; (has_stop = 0)</li>
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<li>Does not contain ATAT&nbsp;(has_atat = 0)</li>
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<li>Does not contain at least 3 consecutive &#39;G&#39;s (has_ggg = 0)</li>
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<li>不以 ATG 开头（has_start = 0）</li>
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<li>不以 TAA，TAG 或 TGA 结尾（has_stop = 0）</li>
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<li>不包含 ATAT（has_atat = 0）</li>
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<li>不包含至少 3 个连续 ‘G’（has_ggg = 0）</li>
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<p><strong>Note:</strong></p>
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<p><strong>注意：</strong></p>
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<ul>
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<li>The result is ordered by sample_id in ascending order</li>
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<li>For each pattern, 1 indicates the pattern is present and 0 indicates it is not present</li>
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<li>结果以 sample_id 升序排序</li>
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<li>对于每个模式，1 表示该模式存在，0 表示不存在</li>
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</ul>
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</div>
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