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feat: update lc problems and solutions (#1055)
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‎.prettierignore

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/solution/0100-0199/0177.Nth Highest Salary/Solution.sql
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/solution/1000-1099/1076.Project Employees II/Solution.sql
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/solution/1000-1099/1082.Sales Analysis I/Solution.sql
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/solution/1100-1199/1173.Immediate Food Delivery I/Solution.sql
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/solution/1400-1499/1454.Active Users/Solution.sql
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/solution/1400-1499/1484.Group Sold Products By The Date/Solution.sql
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/solution/1500-1599/1511.Customer Order Frequency
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/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
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/solution/1600-1699/1613.Find the Missing IDs/Solution.sql
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/solution/1400-1499/1484.Group Sold Products By The Date/Solution.sql
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/solution/1100-1199/1173.Immediate Food Delivery I/Solution.sql
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/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql

‎solution/1000-1099/1034.Coloring A Border/README.md

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<p>给你一个大小为 <code>m x n</code> 的整数矩阵 <code>grid</code> ，表示一个网格。另给你三个整数&nbsp;<code>row</code>、<code>col</code> 和 <code>color</code> 。网格中的每个值表示该位置处的网格块的颜色。</p>
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<p>两个网格块属于同一 <strong>连通分量</strong> 需满足下述全部条件：</p>
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<p>如果两个方块在任意 4 个方向上相邻，则称它们&nbsp;<strong>相邻 </strong>。</p>
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<ul>
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<li>两个网格块颜色相同</li>
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<li>在上、下、左、右任意一个方向上相邻</li>
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</ul>
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<p>如果两个方块具有相同的颜色且相邻，它们则属于同一个 <strong>连通分量</strong> 。</p>
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<p><strong>连通分量的边界</strong><strong> </strong>是指连通分量中满足下述条件之一的所有网格块：</p>
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<li>在网格的边界上（第一行/列或最后一行/列）</li>
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</ul>
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<p>请你使用指定颜色&nbsp;<code>color</code> 为所有包含网格块&nbsp;<code>grid[row][col]</code> 的 <strong>连通分量的边界</strong> 进行着色，并返回最终的网格&nbsp;<code>grid</code> 。</p>
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<p>请你使用指定颜色&nbsp;<code>color</code> 为所有包含网格块&nbsp;<code>grid[row][col]</code> 的 <strong>连通分量的边界</strong> 进行着色。</p>
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<p>并返回最终的网格&nbsp;<code>grid</code> 。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<p><strong class="example">示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>grid = [[1,1],[1,2]], row = 0, col = 0, color = 3
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<strong>输出：</strong>[[3,3],[3,2]]</pre>
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<p><strong>示例 2：</strong></p>
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<p><strong class="example">示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>grid = [[1,2,2],[2,3,2]], row = 0, col = 1, color = 3
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<strong>输出：</strong>[[1,3,3],[2,3,3]]</pre>
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<p><strong>示例 3：</strong></p>
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<p><strong class="example">示例 3：</strong></p>
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<pre>
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<strong>输入：</strong>grid = [[1,1,1],[1,1,1],[1,1,1]], row = 1, col = 1, color = 2
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<li><code>0 &lt;= col &lt; n</code></li>
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</ul>
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<p>&nbsp;</p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->

‎solution/1200-1299/1211.Queries Quality and Percentage/README.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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query_name,
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ROUND(AVG(rating / position), 2) AS quality,
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ROUND(100 * AVG(rating < 3), 2) AS poor_query_percentage
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FROM Queries
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GROUP BY query_name;
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```
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<!-- tabs:end -->

‎solution/1200-1299/1211.Queries Quality and Percentage/README_EN.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT
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query_name,
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ROUND(AVG(rating / position), 2) AS quality,
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ROUND(100 * AVG(rating < 3), 2) AS poor_query_percentage
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FROM Queries
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GROUP BY query_name;
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```
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<!-- tabs:end -->

‎solution/1200-1299/1211.Queries Quality and Percentage/Solution.sql

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# Write your MySQL query statement below
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SELECT
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query_name,
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ROUND(AVG(rating / position), 2) AS quality,
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ROUND(100 * AVG(rating < 3), 2) AS poor_query_percentage
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FROM Queries
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GROUP BY query_name;

‎solution/1200-1299/1232.Check If It Is a Straight Line/README.md

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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<p><strong class="example">示例 1：</strong></p>
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<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1232.Check%20If%20It%20Is%20a%20Straight%20Line/images/untitled-diagram-2.jpg" /></p>
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<strong>输出：</strong>true
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</pre>
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<p><strong>示例 2：</strong></p>
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<p><strong class="example">示例 2：</strong></p>
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<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1232.Check%20If%20It%20Is%20a%20Straight%20Line/images/untitled-diagram-1.jpg" /></strong></p>
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‎solution/1300-1399/1321.Restaurant Growth/README.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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t AS (
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SELECT
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visited_on,
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sum(amount) OVER (
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ORDER BY visited_on
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ROWS 6 PRECEDING
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) AS amount,
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rank() OVER (
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ORDER BY visited_on
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ROWS 6 PRECEDING
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) AS rk
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FROM
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(
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SELECT visited_on, sum(amount) AS amount
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FROM Customer
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GROUP BY visited_on
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) AS tt
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)
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SELECT visited_on, amount, round(amount / 7, 2) AS average_amount
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FROM t
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WHERE rk > 6;
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```
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<!-- tabs:end -->

‎solution/1300-1399/1321.Restaurant Growth/README_EN.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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t AS (
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SELECT
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visited_on,
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sum(amount) OVER (
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ORDER BY visited_on
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ROWS 6 PRECEDING
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) AS amount,
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rank() OVER (
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ORDER BY visited_on
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ROWS 6 PRECEDING
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) AS rk
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FROM
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(
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SELECT visited_on, sum(amount) AS amount
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FROM Customer
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GROUP BY visited_on
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) AS tt
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)
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SELECT visited_on, amount, round(amount / 7, 2) AS average_amount
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FROM t
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WHERE rk > 6;
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```
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<!-- tabs:end -->

‎solution/1300-1399/1321.Restaurant Growth/Solution.sql

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# Write your MySQL query statement below
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WITH
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t AS (
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SELECT
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visited_on,
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sum(amount) OVER (
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ORDER BY visited_on
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ROWS 6 PRECEDING
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) AS amount,
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rank() OVER (
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ORDER BY visited_on
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ROWS 6 PRECEDING
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) AS rk
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FROM
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(
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SELECT visited_on, sum(amount) AS amount
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FROM Customer
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GROUP BY visited_on
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) AS tt
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)
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SELECT visited_on, amount, round(amount / 7, 2) AS average_amount
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FROM t
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WHERE rk > 6;

‎solution/1400-1499/1454.Active Users/README.md

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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH t AS
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(SELECT *,
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sum(id) over(partition by id
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ORDER BY login_date range interval 4 day preceding)/id cnt
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FROM
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(SELECT DISTINCT *
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FROM Accounts
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JOIN Logins using(id) ) tt )
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SELECT DISTINCT id,
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name
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FROM t
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WHERE cnt=5;
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```
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<!-- tabs:end -->

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