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feat: add solutions to lc problem: No.1940. Longest Common Subsequence
Between Sorted Arrays
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‎solution/1900-1999/1939.Users That Actively Request Confirmation Messages/README.md

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# [1939. Users That Actively Request Confirmation Messages](https://leetcode-cn.com/problems/users-that-actively-request-confirmation-messages)
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[English Version](/solution/1900-1999/1939.Users%20That%20Actively%20Request%20Confirmation%20Messages/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Signups</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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| user_id | int |
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| time_stamp | datetime |
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+----------------+----------+
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user_id is the primary key for this table.
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Each row contains information about the signup time for the user with ID user_id.
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</pre>
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<p>&nbsp;</p>
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<p>Table: <code>Confirmations</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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| user_id | int |
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| time_stamp | datetime |
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| action | ENUM |
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+----------------+----------+
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(user_id, time_stamp) is the primary key for this table.
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user_id is a foreign key with a reference to the Signups table.
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action is an ENUM of the type (&#39;confirmed&#39;, &#39;timeout&#39;)
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Each row of this table indicates that the user with ID user_id requested a confirmation message at time_stamp and that confirmation message was either confirmed (&#39;confirmed&#39;) or expired without confirming (&#39;timeout&#39;).</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to find the IDs of the users that requested a confirmation message <strong>twice</strong> within a 24-hour window. Two messages exactly 24 hours apart are considered to be within the window. The <code>action</code> does not affect the answer, only the request time.</p>
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<p>Return the result table <strong>in any order</strong>.</p>
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<p>The query result format is in the following example:</p>
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<p>&nbsp;</p>
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<pre>
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Signups table:
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+---------+---------------------+
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| user_id | time_stamp |
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+---------+---------------------+
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| 3 | 2020-03-21 10:16:13 |
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| 7 | 2020-01-04 13:57:59 |
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| 2 | 2020-07-29 23:09:44 |
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| 6 | 2020-12-09 10:39:37 |
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+---------+---------------------+
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Confirmations table:
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+---------+---------------------+-----------+
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| user_id | time_stamp | action |
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+---------+---------------------+-----------+
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| 3 | 2021-01-06 03:30:46 | timeout |
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| 3 | 2021-01-06 03:37:45 | timeout |
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| 7 | 2021-06-12 11:57:29 | confirmed |
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| 7 | 2021-06-13 11:57:30 | confirmed |
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| 2 | 2021-01-22 00:00:00 | confirmed |
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| 2 | 2021-01-23 00:00:00 | timeout |
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| 6 | 2021-10-23 14:14:14 | confirmed |
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| 6 | 2021-10-24 14:14:13 | timeout |
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+---------+---------------------+-----------+
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Result table
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+---------+
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| user_id |
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+---------+
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| 2 |
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| 3 |
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| 6 |
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+---------+
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User 2 requested two messages within exactly 24 hours of each other, so we include them.
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User 3 requested two messages within 6 minutes and 59 seconds of each other, so we include them.
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User 6 requested two messages within 23 hours, 59 minutes, and 59 seconds of each other, so we include them.
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User 7 requested two messages within 24 hours and 1 second of each other, so we exclude them from the answer.
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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SELECT
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DISTINCT c1.user_id AS user_id
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FROM
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Confirmations c1
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INNER JOIN Confirmations c2
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ON c1.user_id = c2.user_id
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WHERE c1.time_stamp < c2.time_stamp
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AND TIMESTAMPDIFF(SECOND, c1.time_stamp, c2.time_stamp) <= 24 * 60 * 60;
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```
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<!-- tabs:end -->

‎solution/1900-1999/1939.Users That Actively Request Confirmation Messages/README_EN.md

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# [1939. Users That Actively Request Confirmation Messages](https://leetcode.com/problems/users-that-actively-request-confirmation-messages)
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[中文文档](/solution/1900-1999/1939.Users%20That%20Actively%20Request%20Confirmation%20Messages/README.md)
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## Description
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<p>Table: <code>Signups</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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| user_id | int |
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| time_stamp | datetime |
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+----------------+----------+
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user_id is the primary key for this table.
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Each row contains information about the signup time for the user with ID user_id.
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</pre>
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<p>&nbsp;</p>
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<p>Table: <code>Confirmations</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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| user_id | int |
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| time_stamp | datetime |
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| action | ENUM |
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+----------------+----------+
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(user_id, time_stamp) is the primary key for this table.
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user_id is a foreign key with a reference to the Signups table.
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action is an ENUM of the type (&#39;confirmed&#39;, &#39;timeout&#39;)
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Each row of this table indicates that the user with ID user_id requested a confirmation message at time_stamp and that confirmation message was either confirmed (&#39;confirmed&#39;) or expired without confirming (&#39;timeout&#39;).</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to find the IDs of the users that requested a confirmation message <strong>twice</strong> within a 24-hour window. Two messages exactly 24 hours apart are considered to be within the window. The <code>action</code> does not affect the answer, only the request time.</p>
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<p>Return the result table <strong>in any order</strong>.</p>
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<p>The query result format is in the following example:</p>
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<p>&nbsp;</p>
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<pre>
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Signups table:
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+---------+---------------------+
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| user_id | time_stamp |
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+---------+---------------------+
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| 3 | 2020-03-21 10:16:13 |
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| 7 | 2020-01-04 13:57:59 |
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| 2 | 2020-07-29 23:09:44 |
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| 6 | 2020-12-09 10:39:37 |
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+---------+---------------------+
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Confirmations table:
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+---------+---------------------+-----------+
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| user_id | time_stamp | action |
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+---------+---------------------+-----------+
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| 3 | 2021-01-06 03:30:46 | timeout |
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| 3 | 2021-01-06 03:37:45 | timeout |
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| 7 | 2021-06-12 11:57:29 | confirmed |
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| 7 | 2021-06-13 11:57:30 | confirmed |
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| 2 | 2021-01-22 00:00:00 | confirmed |
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| 2 | 2021-01-23 00:00:00 | timeout |
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| 6 | 2021-10-23 14:14:14 | confirmed |
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| 6 | 2021-10-24 14:14:13 | timeout |
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+---------+---------------------+-----------+
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Result table
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+---------+
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| user_id |
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+---------+
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| 2 |
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| 3 |
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| 6 |
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+---------+
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User 2 requested two messages within exactly 24 hours of each other, so we include them.
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User 3 requested two messages within 6 minutes and 59 seconds of each other, so we include them.
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User 6 requested two messages within 23 hours, 59 minutes, and 59 seconds of each other, so we include them.
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User 7 requested two messages within 24 hours and 1 second of each other, so we exclude them from the answer.
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</pre>
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## Solutions
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<!-- tabs:start -->
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### **SQL**
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```sql
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SELECT
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DISTINCT c1.user_id AS user_id
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FROM
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Confirmations c1
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INNER JOIN Confirmations c2
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ON c1.user_id = c2.user_id
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WHERE c1.time_stamp < c2.time_stamp
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AND TIMESTAMPDIFF(SECOND, c1.time_stamp, c2.time_stamp) <= 24 * 60 * 60;
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```
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<!-- tabs:end -->

‎solution/1900-1999/1939.Users That Actively Request Confirmation Messages/Solution.sql

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SELECT
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DISTINCT c1.user_id AS user_id
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FROM
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Confirmations c1
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INNER JOIN Confirmations c2
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ON c1.user_id = c2.user_id
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WHERE c1.time_stamp < c2.time_stamp
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AND TIMESTAMPDIFF(SECOND, c1.time_stamp, c2.time_stamp) <= 24 * 60 * 60;

‎solution/1900-1999/1940.Longest Common Subsequence Between Sorted Arrays/README.md

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# [1940. 排序数组之间的最长公共子序列](https://leetcode-cn.com/problems/longest-common-subsequence-between-sorted-arrays)
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[English Version](/solution/1900-1999/1940.Longest%20Common%20Subsequence%20Between%20Sorted%20Arrays/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给定一个由整数数组组成的数组<code>arrays</code>，其中<code>arrays[i]</code>是严格递增排序的，返回一个表示所有数组之间的最长公共子序列的整数数组。</p>
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<p>子序列是从另一个序列派生出来的序列，删除一些元素或不删除任何元素，而不改变其余元素的顺序。</p>
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<p><strong>示例1:</strong></p>
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<pre>
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<strong>输入:</strong> arrays = [[<strong><em>1</em></strong>,3,<strong><em>4</em></strong>],
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  [<strong><em>1</em></strong>,<strong><em>4</em></strong>,7,9]]
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<strong>输出:</strong> [1,4]
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<strong>解释:</strong> 这两个数组中的最长子序列是[1,4]。
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</pre>
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<p><strong>示例 2:</strong></p>
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<pre>
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<strong>输入:</strong> arrays = [[<strong><em>2</em></strong>,<strong><em>3</em></strong>,<strong><em>6</em></strong>,8],
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  [1,<strong><em>2</em></strong>,<strong><em>3</em></strong>,5,<strong><em>6</em></strong>,7,10],
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  [<strong><em>2</em></strong>,<strong><em>3</em></strong>,4,<em><strong>6</strong></em>,9]]
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<strong>输出:</strong> [2,3,6]
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<strong>解释:</strong> 这三个数组中的最长子序列是[2,3,6]。
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</pre>
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<p><strong>示例 3:</strong></p>
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<pre>
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<strong>输入:</strong> arrays = [[1,2,3,4,5],
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  [6,7,8]]
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<strong>输出:</strong> []
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<strong>解释:</strong> 这两个数组之间没有公共子序列。
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</pre>
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<p> </p>
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<p><strong>限制条件:</strong></p>
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<ul>
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<li><code>2 <= arrays.length <= 100</code></li>
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<li><code>1 <= arrays[i].length <= 100</code></li>
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<li><code>1 <= arrays[i][j] <= 100</code></li>
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<li><code>arrays[i]</code> 是严格递增排序.</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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计数器或者双指针实现。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def longestCommomSubsequence(self, arrays: List[List[int]]) -> List[int]:
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n = len(arrays)
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counter = collections.defaultdict(int)
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for array in arrays:
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for e in array:
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counter[e] += 1
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return [e for e, count in counter.items() if count == n]
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```
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```python
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class Solution:
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def longestCommomSubsequence(self, arrays: List[List[int]]) -> List[int]:
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def common(l1, l2):
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i, j, n1, n2 = 0, 0, len(l1), len(l2)
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res = []
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while i < n1 and j < n2:
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if l1[i] == l2[j]:
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res.append(l1[i])
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i += 1
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j += 1
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elif l1[i] > l2[j]:
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j += 1
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else:
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i += 1
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return res
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n = len(arrays)
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for i in range(1, n):
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arrays[i] = common(arrays[i - 1], arrays[i])
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return arrays[n - 1]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public List<Integer> longestCommomSubsequence(int[][] arrays) {
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Map<Integer, Integer> counter = new HashMap<>();
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for (int[] array : arrays) {
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for (int e : array) {
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counter.put(e, counter.getOrDefault(e, 0) + 1);
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}
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}
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int n = arrays.length;
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List<Integer> res = new ArrayList<>();
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for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
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if (entry.getValue() == n) {
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res.add(entry.getKey());
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}
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}
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return res;
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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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vector<int> longestCommomSubsequence(vector<vector<int>>& arrays) {
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unordered_map<int, int> counter;
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vector<int> res;
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int n = arrays.size();
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for (auto array : arrays) {
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for (auto e : array) {
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counter[e] += 1;
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if (counter[e] == n) {
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res.push_back(e);
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}
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}
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}
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return res;
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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 longestCommomSubsequence(arrays [][]int) []int {
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counter := make(map[int]int)
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n := len(arrays)
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var res []int
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for _, array := range arrays {
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for _, e := range array {
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counter[e]++
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if counter[e] == n {
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res = append(res, e)
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}
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}
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}
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return res
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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