feat: add sql solution to lc problem: No.3118 (#2584)

No.3118.Friday Purchase III

Author: yanglbme

solution/1200-1299/1249.Minimum Remove to Make Valid Parentheses/README_EN.md

@@ -47,7 +47,7 @@
 
 <ul>
 	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
-	<li><code>s[i]</code> is either<code>&#39;(&#39;</code> , <code>&#39;)&#39;</code>, or lowercase English letter<code>.</code></li>
+	<li><code>s[i]</code> is either&nbsp;<code>&#39;(&#39;</code> , <code>&#39;)&#39;</code>, or lowercase English letter.</li>
 </ul>
 
 ## Solutions

solution/2700-2799/2796.Repeat String/README_EN.md

@@ -43,6 +43,9 @@
 	<li><code>1 &lt;=&nbsp;str.length &lt;= 1000</code></li>
 </ul>
 
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Let&#39;s assume, for the sake of simplifying analysis, that concatenating strings is a constant time operation <code>O(1)</code>. With this assumption in mind, can you write an algorithm with a runtime complexity of <code>O(log n)</code>?
+
 ## Solutions
 
 ### Solution 1

solution/2800-2899/2821.Delay the Resolution of Each Promise/README_EN.md

@@ -10,10 +10,12 @@
 
 <ul>
 	<li><code>functions</code>&nbsp;is an array of functions that return promises.</li>
-	<li><code>ms</code>&nbsp;represents the delay duration in milliseconds. It determines the amount of time to wait before resolving each promise in the new array.</li>
+	<li><code>ms</code>&nbsp;represents the delay duration in milliseconds. It determines the amount of time to wait before resolving or rejecting each promise in the new array.</li>
 </ul>
 
-<p>Each function in the new array should return a promise that resolves after a delay of <code>ms</code>&nbsp;milliseconds, preserving the order of the original <code>functions</code>&nbsp;array. The <code>delayAll</code> function should ensure&nbsp;that each promise from <code>functions</code> is executed with a delay, forming the new array of functions returning delayed promises.</p>
+<p>Each function in the new array should return a promise that resolves or rejects after an additional delay of <code>ms</code>&nbsp;milliseconds, preserving the order of the original <code>functions</code>&nbsp;array.</p>
+
+<p>The&nbsp;<code>delayAll</code>&nbsp;function should ensure&nbsp;that each promise from&nbsp;<code>functions</code>&nbsp;is executed with a delay, forming the new array of functions returning delayed promises.</p>
 
 <p>&nbsp;</p>
 <p><strong class="example">Example 1:</strong></p>
@@ -41,6 +43,18 @@ ms = 70
 <strong>Explanation:</strong> The promises from the array would have resolved after 50 ms and 80 ms, but they were delayed by 70 ms, thus 50 ms + 70 ms = 120 ms and 80 ms + 70 ms = 150 ms.
 </pre>
 
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+functions = [
+&nbsp;   () =&gt; new Promise((resolve, reject) =&gt; setTimeout(reject, 20)), 
+&nbsp;   () =&gt; new Promise((resolve, reject) =&gt; setTimeout(reject, 100))
+], 
+ms = 30
+<strong>Output: </strong>[50,130]
+</pre>
+
 <p>&nbsp;</p>
 <p><strong>Constraints:</strong></p>
 

solution/3000-3099/3096.Minimum Levels to Gain More Points/README_EN.md

@@ -8,13 +8,13 @@
 
 <p>You are given a binary array <code>possible</code> of length <code>n</code>.</p>
 
-<p>Danielchandg and Bob are playing a game that consists of <code>n</code> levels. Some of the levels in the game are <strong>impossible</strong> to clear while others can <strong>always</strong> be cleared. In particular, if <code>possible[i] == 0</code>, then the <code>i<sup>th</sup></code> level is <strong>impossible</strong> to clear for <strong>both</strong> the players. A player gains <code>1</code> point on clearing a level and loses <code>1</code> point if the player fails to clear it.</p>
+<p>Alice and Bob are playing a game that consists of <code>n</code> levels. Some of the levels in the game are <strong>impossible</strong> to clear while others can <strong>always</strong> be cleared. In particular, if <code>possible[i] == 0</code>, then the <code>i<sup>th</sup></code> level is <strong>impossible</strong> to clear for <strong>both</strong> the players. A player gains <code>1</code> point on clearing a level and loses <code>1</code> point if the player fails to clear it.</p>
 
-<p>At the start of the game, Danielchandg will play some levels in the <strong>given order</strong> starting from the <code>0<sup>th</sup></code> level, after which Bob will play for the rest of the levels.</p>
+<p>At the start of the game, Alice will play some levels in the <strong>given order</strong> starting from the <code>0<sup>th</sup></code> level, after which Bob will play for the rest of the levels.</p>
 
-<p>Danielchandg wants to know the <strong>minimum</strong> number of levels he should play to gain more points than Bob, if both players play optimally to <strong>maximize</strong> their points.</p>
+<p>Alice wants to know the <strong>minimum</strong> number of levels she should play to gain more points than Bob, if both players play optimally to <strong>maximize</strong> their points.</p>
 
-<p>Return <em>the <strong>minimum</strong> number of levels danielchandg should play to gain more points</em>. <em>If this is <strong>not</strong> possible, return</em> <code>-1</code>.</p>
+<p>Return <em>the <strong>minimum</strong> number of levels Alice should play to gain more points</em>. <em>If this is <strong>not</strong> possible, return</em> <code>-1</code>.</p>
 
 <p><strong>Note</strong> that each player must play at least <code>1</code> level.</p>
 
@@ -28,15 +28,15 @@
 
 <p><strong>Explanation:</strong></p>
 
-<p>Let&#39;s look at all the levels that Danielchandg can play up to:</p>
+<p>Let&#39;s look at all the levels that Alice can play up to:</p>
 
 <ul>
-	<li>If Danielchandg plays only level 0 and Bob plays the rest of the levels, Danielchandg has 1 point, while Bob has -1 + 1 - 1 = -1 point.</li>
-	<li>If Danielchandg plays till level 1 and Bob plays the rest of the levels, Danielchandg has 1 - 1 = 0 points, while Bob has 1 - 1 = 0 points.</li>
-	<li>If Danielchandg plays till level 2 and Bob plays the rest of the levels, Danielchandg has 1 - 1 + 1 = 1 point, while Bob has -1 point.</li>
+	<li>If Alice plays only level 0 and Bob plays the rest of the levels, Alice has 1 point, while Bob has -1 + 1 - 1 = -1 point.</li>
+	<li>If Alice plays till level 1 and Bob plays the rest of the levels, Alice has 1 - 1 = 0 points, while Bob has 1 - 1 = 0 points.</li>
+	<li>If Alice plays till level 2 and Bob plays the rest of the levels, Alice has 1 - 1 + 1 = 1 point, while Bob has -1 point.</li>
 </ul>
 
-<p>Danielchandg must play a minimum of 1 level to gain more points.</p>
+<p>Alice must play a minimum of 1 level to gain more points.</p>
 </div>
 
 <p><strong class="example">Example 2:</strong></p>
@@ -48,16 +48,16 @@
 
 <p><strong>Explanation:</strong></p>
 
-<p>Let&#39;s look at all the levels that Danielchandg can play up to:</p>
+<p>Let&#39;s look at all the levels that Alice can play up to:</p>
 
 <ul>
-	<li>If Danielchandg plays only level 0 and Bob plays the rest of the levels, Danielchandg has 1 point, while Bob has 4 points.</li>
-	<li>If Danielchandg plays till level 1 and Bob plays the rest of the levels, Danielchandg has 2 points, while Bob has 3 points.</li>
-	<li>If Danielchandg plays till level 2 and Bob plays the rest of the levels, Danielchandg has 3 points, while Bob has 2 points.</li>
-	<li>If Danielchandg plays till level 3 and Bob plays the rest of the levels, Danielchandg has 4 points, while Bob has 1 point.</li>
+	<li>If Alice plays only level 0 and Bob plays the rest of the levels, Alice has 1 point, while Bob has 4 points.</li>
+	<li>If Alice plays till level 1 and Bob plays the rest of the levels, Alice has 2 points, while Bob has 3 points.</li>
+	<li>If Alice plays till level 2 and Bob plays the rest of the levels, Alice has 3 points, while Bob has 2 points.</li>
+	<li>If Alice plays till level 3 and Bob plays the rest of the levels, Alice has 4 points, while Bob has 1 point.</li>
 </ul>
 
-<p>Danielchandg must play a minimum of 3 levels to gain more points.</p>
+<p>Alice must play a minimum of 3 levels to gain more points.</p>
 </div>
 
 <p><strong class="example">Example 3:</strong></p>
@@ -69,7 +69,7 @@
 
 <p><strong>Explanation:</strong></p>
 
-<p>The only possible way is for both players to play 1 level each. Danielchandg plays level 0 and loses 1 point. Bob plays level 1 and loses 1 point. As both players have equal points, Danielchandg can&#39;t gain more points than Bob.</p>
+<p>The only possible way is for both players to play 1 level each. Alice plays level 0 and loses 1 point. Bob plays level 1 and loses 1 point. As both players have equal points, Alice can&#39;t gain more points than Bob.</p>
 </div>
 
 <p>&nbsp;</p>

solution/3100-3199/3112.Minimum Time to Visit Disappearing Nodes/README_EN.md

@@ -71,7 +71,7 @@
 <p><strong>Constraints:</strong></p>
 
 <ul>
-	<li><code>2 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
 	<li><code>0 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>
 	<li><code>edges[i] == [u<sub>i</sub>, v<sub>i</sub>, length<sub>i</sub>]</code></li>
 	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</code></li>

solution/3100-3199/3118.Friday Purchase III/README.md

@@ -0,0 +1,158 @@
+# [3118. Friday Purchase III](https://leetcode.cn/problems/friday-purchase-iii)
+
+[English Version](/solution/3100-3199/3118.Friday%20Purchase%20III/README_EN.md)
+
+<!-- tags: -->
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>Table: <code>Purchases</code></p>
+
+<pre>
++---------------+------+
+| Column Name   | Type |
++---------------+------+
+| user_id       | int  |
+| purchase_date | date |
+| amount_spend  | int  |
++---------------+------+
+(user_id, purchase_date, amount_spend) is the primary key (combination of columns with unique values) for this table.
+purchase_date will range from November 1, 2023, to November 30, 2023, inclusive of both dates.
+Each row contains user_id, purchase_date, and amount_spend.
+</pre>
+
+<p>Table: <code>Users</code></p>
+
+<pre>
++-------------+------+
+| Column Name | Type |
++-------------+------+
+| user_id     | int  |
+| membership  | enum |
++-------------+------+
+user_id is the primary key for this table.
+membership is an ENUM (category) type of (&#39;Standard&#39;, &#39;Premium&#39;, &#39;VIP&#39;).
+Each row of this table indicates the user_id, membership type.
+</pre>
+
+<p>Write a solution to calculate the <strong>total spending</strong> by <code>Premium</code>&nbsp;and <code>VIP</code> members on <strong>each Friday of every week</strong> in November 2023.&nbsp; If there are <strong>no purchases</strong> on a <strong>particular Friday</strong> by <code>Premium</code> or <code>VIP</code> members, it should be considered as <code>0</code>.</p>
+
+<p>Return <em>the result table</em>&nbsp;<em>ordered by week of the month,&nbsp; and </em><code>membership</code><em> in <strong>ascending</strong> order</em>.</p>
+
+<p>The result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong></p>
+
+<p>Purchases table:</p>
+
+<pre class="example-io">
++---------+---------------+--------------+
+| user_id | purchase_date | amount_spend |
++---------+---------------+--------------+
+| 11      | 2023-11-03    | 1126         |
+| 15      | 2023-11-10    | 7473         |
+| 17      | 2023-11-17    | 2414         |
+| 12      | 2023-11-24    | 9692         |
+| 8       | 2023-11-24    | 5117         |
+| 1       | 2023-11-24    | 5241         |
+| 10      | 2023-11-22    | 8266         |
+| 13      | 2023-11-21    | 12000        |
++---------+---------------+--------------+
+</pre>
+
+<p>Users table:</p>
+
+<pre class="example-io">
++---------+------------+
+| user_id | membership |
++---------+------------+
+| 11      | Premium    |
+| 15      | VIP        |
+| 17      | Standard   |
+| 12      | VIP        |
+| 8       | Premium    |
+| 1       | VIP        |
+| 10      | Standard   |
+| 13      | Premium    |
++---------+------------+
+</pre>
+
+<p><strong>Output:</strong></p>
+
+<pre class="example-io">
++---------------+-------------+--------------+
+| week_of_month | membership  | total_amount |
++---------------+-------------+--------------+
+| 1             | Premium     | 1126         |
+| 1             | VIP         | 0            |
+| 2             | Premium     | 0            |
+| 2             | VIP         | 7473         |
+| 3             | Premium     | 0            |
+| 3             | VIP         | 0            |
+| 4             | Premium     | 5117         |
+| 4             | VIP         | 14933        |
++---------------+-------------+--------------+
+        </pre>
+
+<p><strong>Explanation:</strong></p>
+
+<ul>
+	<li>During the first week of November 2023, a transaction occurred on Friday, 2023-11-03, by a Premium member amounting to $1,126. No transactions were made by VIP members on this day, resulting in a value of 0.</li>
+	<li>For the second week of November 2023, there was a transaction on Friday, 2023-11-10, and it was made by a VIP member, amounting to $7,473. Since there were no purchases by Premium members that Friday, the output shows 0 for Premium members.</li>
+	<li>Similarly, during the third week of November 2023, no transactions by Premium or VIP members occurred on Friday, 2023-11-17, which shows 0 for both categories in this week.</li>
+	<li>In the fourth week of November 2023, transactions occurred on Friday, 2023-11-24, involving one Premium member purchase of $5,117 and VIP member purchases totaling $14,933 ($9,692 from one and $5,241 from another).</li>
+</ul>
+
+<p><strong>Note:</strong> The output table is ordered by week_of_month and membership in ascending order.</p>
+</div>
+
+## 解法
+
+### 方法一：递归 + 连接
+
+我们首先创建一个递归表 `T`，其中包含 `week_of_month` 列，表示月份的第几周。然后创建一个表 `M`，包含 `membership` 列，表示会员类型，取值为 `'Premium'` 和 `'VIP'`。
+
+接着创建一个表 `P`，包含 `week_of_month`、`membership` 和 `amount_spend` 列，筛选出每个会员在每个月的第几周的周五的消费金额。最后，我们将 `T` 和 `M` 表连接，再左连接 `P` 表，并且按照 `week_of_month` 和 `membership` 列进行分组，计算每周每种会员的总消费金额。
+
+<!-- tabs:start -->
+
+```sql
+# Write your MySQL query statement below
+WITH RECURSIVE
+    T AS (
+        SELECT 1 AS week_of_month
+        UNION
+        SELECT week_of_month + 1
+        FROM T
+        WHERE week_of_month < 4
+    ),
+    M AS (
+        SELECT 'Premium' AS membership
+        UNION
+        SELECT 'VIP'
+    ),
+    P AS (
+        SELECT CEIL(DAYOFMONTH(purchase_date) / 7) AS week_of_month, membership, amount_spend
+        FROM
+            Purchases
+            JOIN Users USING (user_id)
+        WHERE DAYOFWEEK(purchase_date) = 6
+    )
+SELECT week_of_month, membership, IFNULL(SUM(amount_spend), 0) AS total_amount
+FROM
+    T
+    JOIN M
+    LEFT JOIN P USING (week_of_month, membership)
+GROUP BY 1, 2
+ORDER BY 1, 2;
+```
+
+<!-- tabs:end -->
+
+<!-- end -->