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feat: add solutions for lc No.3860 (#5060)
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‎solution/3800-3899/3860.Unique Email Groups/README.md‎

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---
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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3860.Unique%20Email%20Groups/README.md
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---
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<!-- problem:start -->
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# [3860. Unique Email Groups 🔒](https://leetcode.cn/problems/unique-email-groups)
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[English Version](/solution/3800-3899/3860.Unique%20Email%20Groups/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 an array of strings <code>emails</code>, where each string is a valid email address.</p>
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<p>Two email addresses belong to the same group if <strong>both</strong> their <strong>normalized</strong> local names and <strong>normalized</strong> domain names are <strong>identical</strong>.</p>
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<p>The normalization rules are as follows:</p>
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<ul>
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<li>The local name is the part <strong>before</strong> the <code>&#39;@&#39;</code> symbol.
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<ul>
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<li>Ignore all dots <code>&#39;.&#39;</code>.</li>
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<li>Ignore everything after the first <code>&#39;+&#39;</code>, if present.</li>
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<li>Convert to lowercase.</li>
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</ul>
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</li>
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<li>The domain name is the part <strong>after</strong> the <code>&#39;@&#39;</code> symbol.
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<ul>
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<li>Convert to lowercase.</li>
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</ul>
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</li>
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</ul>
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<p>Return an integer denoting the number of <strong>unique</strong> email groups after normalization.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">emails = [&quot;test.email+alex@leetcode.com&quot;, &quot;test.e.mail+bob.cathy@leetcode.com&quot;, &quot;testemail+david@lee.tcode.com&quot;]</span></p>
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<p><strong>Output:</strong> <span class="example-io">2</span></p>
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<p><strong>Explanation:</strong></p>
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</div>
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">Email</th>
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<th style="border: 1px solid black;">Local</th>
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<th style="border: 1px solid black;">Normalized Local</th>
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<th style="border: 1px solid black;">Domain</th>
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<th style="border: 1px solid black;">Normalized Domain</th>
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<th style="border: 1px solid black;">Final Email</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">test.email+alex@leetcode.com</td>
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<td style="border: 1px solid black;">test.email+alex</td>
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<td style="border: 1px solid black;">testemail</td>
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<td style="border: 1px solid black;">leetcode.com</td>
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<td style="border: 1px solid black;">leetcode.com</td>
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<td style="border: 1px solid black;">testemail@leetcode.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">test.e.mail+bob.cathy@leetcode.com</td>
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<td style="border: 1px solid black;">test.e.mail+bob.cathy</td>
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<td style="border: 1px solid black;">testemail</td>
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<td style="border: 1px solid black;">leetcode.com</td>
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<td style="border: 1px solid black;">leetcode.com</td>
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<td style="border: 1px solid black;">testemail@leetcode.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">testemail+david@lee.tcode.com</td>
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<td style="border: 1px solid black;">testemail+david</td>
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<td style="border: 1px solid black;">testemail</td>
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<td style="border: 1px solid black;">lee.tcode.com</td>
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<td style="border: 1px solid black;">lee.tcode.com</td>
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<td style="border: 1px solid black;">testemail@lee.tcode.com</td>
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</tr>
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</tbody>
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</table>
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<p>Unique emails are [<code>&quot;testemail@leetcode.com&quot;</code>, <code>&quot;testemail@lee.tcode.com&quot;</code>]. Thus, the answer is 2.</p>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">emails = [&quot;A@B.com&quot;, &quot;a@b.com&quot;, &quot;ab+xy@b.com&quot;, &quot;a.b@b.com&quot;]</span></p>
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<p><strong>Output:</strong> <span class="example-io">2</span></p>
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<p><strong>Explanation:</strong></p>
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">Email</th>
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<th style="border: 1px solid black;">Local</th>
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<th style="border: 1px solid black;">Normalized Local</th>
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<th style="border: 1px solid black;">Domain</th>
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<th style="border: 1px solid black;">Normalized Domain</th>
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<th style="border: 1px solid black;">Final Email</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">A@B.com</td>
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<td style="border: 1px solid black;">A</td>
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<td style="border: 1px solid black;">a</td>
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<td style="border: 1px solid black;">B.com</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">a@b.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">a@b.com</td>
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<td style="border: 1px solid black;">a</td>
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<td style="border: 1px solid black;">a</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">a@b.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">ab+xy@b.com</td>
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<td style="border: 1px solid black;">ab+xy</td>
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<td style="border: 1px solid black;">ab</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">ab@b.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">a.b@b.com</td>
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<td style="border: 1px solid black;">a.b</td>
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<td style="border: 1px solid black;">ab</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">b.com</td>
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<td style="border: 1px solid black;">ab@b.com</td>
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</tr>
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</tbody>
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</table>
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<p>Unique emails are [<code>&quot;a@b.com&quot;</code>, <code>&quot;ab@b.com&quot;</code>]. Thus, the answer is 2.</p>
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</div>
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<p><strong class="example">Example 3:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">emails = [&quot;a.b+c.d+e@DoMain.com&quot;, &quot;ab+xyz@domain.com&quot;, &quot;ab@domain.com&quot;]</span></p>
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<p><strong>Output:</strong> <span class="example-io">1</span></p>
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<p><strong>Explanation:</strong></p>
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<table style="border: 1px solid black;">
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<thead>
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<tr>
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<th style="border: 1px solid black;">Email</th>
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<th style="border: 1px solid black;">Local</th>
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<th style="border: 1px solid black;">Normalized Local</th>
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<th style="border: 1px solid black;">Domain</th>
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<th style="border: 1px solid black;">Normalized Domain</th>
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<th style="border: 1px solid black;">Final Email</th>
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</tr>
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</thead>
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<tbody>
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<tr>
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<td style="border: 1px solid black;">a.b+c.d+e@DoMain.com</td>
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<td style="border: 1px solid black;">a.b+c.d+e</td>
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<td style="border: 1px solid black;">ab</td>
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<td style="border: 1px solid black;">DoMain.com</td>
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<td style="border: 1px solid black;">domain.com</td>
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<td style="border: 1px solid black;">ab@domain.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">ab+xyz@domain.com</td>
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<td style="border: 1px solid black;">ab+xyz</td>
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<td style="border: 1px solid black;">ab</td>
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<td style="border: 1px solid black;">domain.com</td>
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<td style="border: 1px solid black;">domain.com</td>
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<td style="border: 1px solid black;">ab@domain.com</td>
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</tr>
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<tr>
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<td style="border: 1px solid black;">ab@domain.com</td>
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<td style="border: 1px solid black;">ab</td>
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<td style="border: 1px solid black;">ab</td>
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<td style="border: 1px solid black;">domain.com</td>
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<td style="border: 1px solid black;">domain.com</td>
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<td style="border: 1px solid black;">ab@domain.com</td>
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</tr>
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</tbody>
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</table>
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<p>All emails normalize to <code>&quot;ab@domain.com&quot;</code>. Thus, the answer is 1.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= emails.length &lt;= 1000</code></li>
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<li><code>1 &lt;= emails[i].length &lt;= 100</code></li>
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<li><code>emails[i]</code> consists of lowercase and uppercase English letters, digits, and the characters <code>&#39;.&#39;</code>, <code>&#39;+&#39;</code>, and <code>&#39;@&#39;</code>.</li>
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<li>Each <code>emails[i]</code> contains <strong>exactly</strong> one <code>&#39;@&#39;</code> character.</li>
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<li>All local and domain names are non-empty; local names do not start with <code>&#39;+&#39;</code>.</li>
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<li>Domain names end with the <code>&quot;.com&quot;</code> suffix and contain at least one character before <code>&quot;.com&quot;</code>.</li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一：哈希表
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我们可以使用一个哈希表 $\textit{st}$ 来存储每个邮箱地址的规范化结果。对于每个邮箱地址，我们按照题目要求进行规范化处理：
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- 将邮箱地址分为本地名和域名两部分。
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- 对于本地名，去掉所有的点 `.`，并且如果存在加号 `+`，则去掉加号及其后面的部分。最后将本地名转换为小写。
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- 对于域名，将其转换为小写。
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- 将规范化后的本地名和域名拼接起来，得到规范化后的邮箱地址，并将其加入哈希表 $\textit{st}$ 中。
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最后，哈希表 $\textit{st}$ 中的元素个数即为唯一邮箱组的数量。
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时间复杂度 $O(n \cdot m)$，其中 $n$ 和 $m$ 分别是邮箱地址的数量和每个邮箱地址的平均长度。空间复杂度 $O(n \cdot m)$，最坏情况下所有邮箱地址都不同。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def uniqueEmailGroups(self, emails: list[str]) -> int:
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st = set()
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for email in emails:
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local, domain = email.split("@")
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local = local.split("+")[0].replace(".", "").lower()
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domain = domain.lower()
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normalized = local + domain
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st.add(normalized)
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return len(st)
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```
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#### Java
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```java
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class Solution {
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public int uniqueEmailGroups(String[] emails) {
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Set<String> st = new HashSet<>();
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for (String email : emails) {
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String[] parts = email.split("@");
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String local = parts[0];
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String domain = parts[1];
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int plusIndex = local.indexOf('+');
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if (plusIndex != -1) {
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local = local.substring(0, plusIndex);
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}
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local = local.replace(".", "").toLowerCase();
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domain = domain.toLowerCase();
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String normalized = local + domain;
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st.add(normalized);
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}
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return st.size();
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int uniqueEmailGroups(vector<string>& emails) {
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unordered_set<string> st;
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for (auto& email : emails) {
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int atPos = email.find('@');
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string local = email.substr(0, atPos);
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string domain = email.substr(atPos + 1);
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int plusPos = local.find('+');
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if (plusPos != string::npos) {
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local = local.substr(0, plusPos);
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}
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string cleaned;
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for (char c : local) {
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if (c != '.') {
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cleaned += tolower(c);
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}
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}
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for (char& c : domain) {
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c = tolower(c);
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}
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st.insert(cleaned + domain);
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}
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return st.size();
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}
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};
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```
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#### Go
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```go
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func uniqueEmailGroups(emails []string) int {
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st := make(map[string]struct{})
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for _, email := range emails {
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parts := strings.Split(email, "@")
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local := parts[0]
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domain := parts[1]
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if idx := strings.Index(local, "+"); idx != -1 {
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local = local[:idx]
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}
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local = strings.ReplaceAll(local, ".", "")
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local = strings.ToLower(local)
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domain = strings.ToLower(domain)
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normalized := local + domain
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st[normalized] = struct{}{}
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}
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return len(st)
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}
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```
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#### TypeScript
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```ts
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function uniqueEmailGroups(emails: string[]): number {
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const st = new Set<string>();
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for (const email of emails) {
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let [local, domain] = email.split('@');
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local = local.split('+')[0].replace(/\./g, '').toLowerCase();
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domain = domain.toLowerCase();
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const normalized = local + domain;
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st.add(normalized);
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}
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return st.size;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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