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Merge pull request #721 from openset/develop
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‎problems/active-businesses/README.md

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## [1126. Active Businesses (Medium)](https://leetcode.com/problems/active-businesses "")
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<p>Table: <code>Events</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| business_id | int |
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| event_type | varchar |
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| occurences | int |
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+---------------+---------+
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(business_id, event_type) is the primary key of this table.
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Each row in the table logs the info that an event of some type occured at some business for a number of times.</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to find all <em>active businesses</em>.</p>
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<p>An active business is a business that has more than one event type&nbsp;with occurences greater than the average occurences of that event type&nbsp;among all businesses.</p>
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<p>The query result format is in the following example:</p>
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<pre>
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Events table:
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+-------------+------------+------------+
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| business_id | event_type | occurences |
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+-------------+------------+------------+
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| 1 | reviews | 7 |
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| 3 | reviews | 3 |
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| 1 | ads | 11 |
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| 2 | ads | 7 |
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| 3 | ads | 6 |
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| 1 | page views | 3 |
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| 2 | page views | 12 |
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+-------------+------------+------------+
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Result table:
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+-------------+
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| business_id |
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+-------------+
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| 1 |
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+-------------+
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Average for &#39;reviews&#39;, &#39;ads&#39; and &#39;page views&#39; are (7+3)/2=5, (11+7+6)/3=8, (3+12)/2=7.5 respectively.
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Business with id 1 has 7 &#39;reviews&#39; events (more than 5) and 11 &#39;ads&#39; events (more than 8) so it is an active business.</pre>

‎problems/actors-and-directors-who-cooperated-at-least-three-times/README.md

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## [1050. Actors and Directors Who Cooperated At Least Three Times (Easy)](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times "合作过至少三次的演员和导演")
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<p>Table: <code>ActorDirector</code></p>
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<pre>
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| actor_id | int |
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| director_id | int |
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| timestamp | int |
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+-------------+---------+
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timestamp is the primary key column for this table.
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</pre>
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<p>&nbsp;</p>
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<p>Write a SQL query for a report that provides the pairs <code>(actor_id, director_id)</code> where the actor have cooperated with the director at least 3 times.</p>
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<p><strong>Example:</strong></p>
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<pre>
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ActorDirector table:
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+-------------+-------------+-------------+
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| actor_id | director_id | timestamp |
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+-------------+-------------+-------------+
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| 1 | 1 | 0 |
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| 1 | 1 | 1 |
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| 1 | 1 | 2 |
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| 1 | 2 | 3 |
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| 1 | 2 | 4 |
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| 2 | 1 | 5 |
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| 2 | 1 | 6 |
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+-------------+-------------+-------------+
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Result table:
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+-------------+-------------+
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| actor_id | director_id |
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+-------------+-------------+
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| 1 | 1 |
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+-------------+-------------+
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The only pair is (1, 1) where they cooperated exactly 3 times.
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</pre>

‎problems/add-bold-tag-in-string/README.md

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## [616. Add Bold Tag in String (Medium)](https://leetcode.com/problems/add-bold-tag-in-string "给字符串添加加粗标签")
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Given a string <b>s</b> and a list of strings <b>dict</b>, you need to add a closed pair of bold tag <code>&lt;b&gt;</code> and <code>&lt;/b&gt;</code> to wrap the substrings in s that exist in dict. If two such substrings overlap, you need to wrap them together by only one pair of closed bold tag. Also, if two substrings wrapped by bold tags are consecutive, you need to combine them.
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<p><b>Example 1:</b><br />
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<pre>
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<b>Input:</b>
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s = "abcxyz123"
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dict = ["abc","123"]
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<b>Output:</b>
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"&lt;b&gt;abc&lt;/b&gt;xyz&lt;b&gt;123&lt;/b&gt;"
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</pre>
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</p>
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<p><b>Example 2:</b><br />
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<pre>
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<b>Input:</b>
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s = "aaabbcc"
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dict = ["aaa","aab","bc"]
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<b>Output:</b>
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"&lt;b&gt;aaabbc&lt;/b&gt;c"
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</pre>
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</p>
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<p><b>Note:</b><br>
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<ol>
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<li>The given dict won't contain duplicates, and its length won't exceed 100.</li>
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<li>All the strings in input have length in range [1, 1000]. </li>
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</ol>
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</p>
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### Related Topics
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[[String](https://github.com/openset/leetcode/tree/master/tag/string/README.md)]

‎problems/alien-dictionary/README.md

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## [269. Alien Dictionary (Hard)](https://leetcode.com/problems/alien-dictionary "火星词典")
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<p>There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of <b>non-empty</b> words from the dictionary, where <b>words are sorted lexicographically by the rules of this new language</b>. Derive the order of letters in this language.</p>
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<p><b>Example 1:</b></p>
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<pre>
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<strong>Input:</strong>
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[
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&quot;wrt&quot;,
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&quot;wrf&quot;,
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&quot;er&quot;,
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&quot;ett&quot;,
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&quot;rftt&quot;
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]
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<strong>Output: </strong><code>&quot;wertf&quot;</code>
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</pre>
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<p><b>Example 2:</b></p>
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<pre>
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<strong>Input:</strong>
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[
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&quot;z&quot;,
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&quot;x&quot;
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]
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<strong>Output: </strong><code>&quot;zx&quot;</code>
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</pre>
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<p><b>Example 3:</b></p>
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<pre>
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<strong>Input:</strong>
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[
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&quot;z&quot;,
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&quot;x&quot;,
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&quot;z&quot;
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]
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<strong>Output:</strong> <code>&quot;&quot;</code>&nbsp;
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<strong>Explanation:</strong> The order is invalid, so return <code>&quot;&quot;</code>.
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</pre>
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<p><b>Note:</b></p>
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<ol>
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<li>You may assume all letters are in lowercase.</li>
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<li>You may assume that if a is a prefix of b, then a must appear before b in the given dictionary.</li>
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<li>If the order is invalid, return an empty string.</li>
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<li>There may be multiple valid order of letters, return any one of them is fine.</li>
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</ol>
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### Related Topics
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[[Graph](https://github.com/openset/leetcode/tree/master/tag/graph/README.md)]

‎problems/analyze-user-website-visit-pattern/README.md

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## [1152. Analyze User Website Visit Pattern (Medium)](https://leetcode.com/problems/analyze-user-website-visit-pattern "用户网站访问行为分析")
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<p>We are given some website visits: the user with name&nbsp;<code>username[i]</code> visited the website&nbsp;<code>website[i]</code> at time <code>timestamp[i]</code>.</p>
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<p>A <em>3-sequence</em>&nbsp;is a list of&nbsp;websites of length 3 sorted in ascending order&nbsp;by the time of their visits.&nbsp; (The websites in a 3-sequence are not necessarily distinct.)</p>
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<p>Find the 3-sequence visited&nbsp;by the largest number of users. If there is more than one solution, return the lexicographically smallest such 3-sequence.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input: </strong>username = <span>[&quot;joe&quot;,&quot;joe&quot;,&quot;joe&quot;,&quot;james&quot;,&quot;james&quot;,&quot;james&quot;,&quot;james&quot;,&quot;mary&quot;,&quot;mary&quot;,&quot;mary&quot;]</span>, timestamp = <span id="example-input-1-2">[1,2,3,4,5,6,7,8,9,10]</span>, website = <span id="example-input-1-3">[&quot;home&quot;,&quot;about&quot;,&quot;career&quot;,&quot;home&quot;,&quot;cart&quot;,&quot;maps&quot;,&quot;home&quot;,&quot;home&quot;,&quot;about&quot;,&quot;career&quot;]</span>
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<strong>Output: </strong><span id="example-output-1">[&quot;home&quot;,&quot;about&quot;,&quot;career&quot;]</span>
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<strong>Explanation: </strong>
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The tuples in this example are:
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[&quot;joe&quot;, 1, &quot;home&quot;]
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[&quot;joe&quot;, 2, &quot;about&quot;]
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[&quot;joe&quot;, 3, &quot;career&quot;]
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[&quot;james&quot;, 4, &quot;home&quot;]
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[&quot;james&quot;, 5, &quot;cart&quot;]
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[&quot;james&quot;, 6, &quot;maps&quot;]
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[&quot;james&quot;, 7, &quot;home&quot;]
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[&quot;mary&quot;, 8, &quot;home&quot;]
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[&quot;mary&quot;, 9, &quot;about&quot;]
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[&quot;mary&quot;, 10, &quot;career&quot;]
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The 3-sequence (&quot;home&quot;, &quot;about&quot;, &quot;career&quot;) was visited at least once by <strong>2</strong> users.
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The 3-sequence (&quot;home&quot;, &quot;cart&quot;, &quot;maps&quot;) was visited at least once by 1 user.
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The 3-sequence (&quot;home&quot;, &quot;cart&quot;, &quot;home&quot;) was visited at least once by 1 user.
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The 3-sequence (&quot;home&quot;, &quot;maps&quot;, &quot;home&quot;) was visited at least once by 1 user.
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The 3-sequence (&quot;cart&quot;, &quot;maps&quot;, &quot;home&quot;) was visited at least once by 1 user.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Note:</strong></p>
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<ol>
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<li><code>3 &lt;= N = username.length = timestamp.length = website.length &lt;= 50</code></li>
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<li><code>1 &lt;= username[i].length &lt;= 10</code></li>
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<li><code>0 &lt;= timestamp[i] &lt;= 10^9</code></li>
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<li><code>1 &lt;= website[i].length &lt;= 10</code></li>
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<li>Both <code>username[i]</code> and <code>website[i]</code> contain only lowercase characters.</li>
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<li>It is guaranteed that there is at least one user who visited at least 3 websites.</li>
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<li>No user visits two websites at the same time.</li>
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</ol>
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### Related Topics
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[[Sort](https://github.com/openset/leetcode/tree/master/tag/sort/README.md)]

‎problems/armstrong-number/README.md

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## [1134. Armstrong Number (Easy)](https://leetcode.com/problems/armstrong-number "阿姆斯特朗数")
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<p>The k-digit number <code>N</code> is an Armstrong number if and only if the k-th power of each digit sums to N.</p>
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<p>Given a positive integer <code>N</code>, return true if and only if it is an Armstrong number.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input: </strong><span id="example-input-1-1">153</span>
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<strong>Output: </strong><span id="example-output-1">true</span>
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<strong>Explanation: </strong>
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153 is a 3-digit number, and 153 = 1^3 + 5^3 + 3^3.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input: </strong><span id="example-input-2-1">123</span>
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<strong>Output: </strong><span id="example-output-2">false</span>
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<strong>Explanation: </strong>
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123 is a 3-digit number, and 123 != 1^3 + 2^3 + 3^3 = 36.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Note:</strong></p>
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<ol>
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<li><code>1 &lt;= N &lt;= 10^8</code></li>
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</ol>
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### Related Topics
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[[Math](https://github.com/openset/leetcode/tree/master/tag/math/README.md)]

‎problems/array-transformation/README.md

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## [1243. Array Transformation (Easy)](https://leetcode.com/problems/array-transformation "数组变换")
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<p>Given an initial array <code>arr</code>, every day you produce a new array using the array of the previous day.</p>
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<p>On the <code>i</code>-th day, you do the following operations on the array of day&nbsp;<code>i-1</code>&nbsp;to produce the array of day <code>i</code>:</p>
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<ol>
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<li>If an element is smaller than both its left neighbor and its right neighbor, then this element is incremented.</li>
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<li>If an element is bigger than both its left neighbor and its right neighbor, then this element is decremented.</li>
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<li>The first&nbsp;and last elements never change.</li>
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</ol>
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<p>After some days, the array does not change. Return that final array.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [6,2,3,4]
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<strong>Output:</strong> [6,3,3,4]
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<strong>Explanation: </strong>
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On the first day, the array is changed from [6,2,3,4] to [6,3,3,4].
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No more operations can be done to this array.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> arr = [1,6,3,4,3,5]
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<strong>Output:</strong> [1,4,4,4,4,5]
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<strong>Explanation: </strong>
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On the first day, the array is changed from [1,6,3,4,3,5] to [1,5,4,3,4,5].
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On the second day, the array is changed from [1,5,4,3,4,5] to [1,4,4,4,4,5].
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No more operations can be done to this array.
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</pre>
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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;= arr.length &lt;= 100</code></li>
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<li><code>1 &lt;= arr[i] &lt;= 100</code></li>
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</ul>
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### Related Topics
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[[Array](https://github.com/openset/leetcode/tree/master/tag/array/README.md)]

‎problems/article-views-i/README.md

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## [1148. Article Views I (Easy)](https://leetcode.com/problems/article-views-i "文章浏览 I")
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<p>Table: <code>Views</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| article_id | int |
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| author_id | int |
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| viewer_id | int |
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| view_date | date |
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+---------------+---------+
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There is no primary key for this table, it may have duplicate rows.
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Each row of this table indicates that some viewer viewed an article (written by some author) on some date.
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Note that equal author_id and viewer_id indicate the same person.</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to find all the authors that viewed at least one of their own articles, sorted in ascending order by their id.</p>
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<p>The query result format is in the following example:</p>
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<pre>
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Views table:
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+------------+-----------+-----------+------------+
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| article_id | author_id | viewer_id | view_date |
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+------------+-----------+-----------+------------+
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| 1 | 3 | 5 | 2019年08月01日 |
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| 1 | 3 | 6 | 2019年08月02日 |
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| 2 | 7 | 7 | 2019年08月01日 |
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| 2 | 7 | 6 | 2019年08月02日 |
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| 4 | 7 | 1 | 2019年07月22日 |
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| 3 | 4 | 4 | 2019年07月21日 |
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| 3 | 4 | 4 | 2019年07月21日 |
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+------------+-----------+-----------+------------+
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Result table:
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+------+
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| id |
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+------+
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| 4 |
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| 7 |
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+------+
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</pre>

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