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Commit 05880c3

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feat: add sql solution to lc problem: No.2153 (doocs#1167)
No.2153.The Number of Passengers in Each Bus II
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‎.prettierignore‎

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/solution/1600-1699/1651.Hopper Company Queries III/Solution.sql
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/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
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/solution/2100-2199/2118.Build the Equation/Solution.sql
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/solution/2100-2199/2153.The Number of Passengers in Each Bus II/Solution.sql
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/solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql

‎solution/0100-0199/0167.Two Sum II - Input Array Is Sorted/README.md‎

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**方法一:二分查找**
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注意到数组按照非递减顺序排列,因此对于每个 `numbers[i]`,可以通过二分查找的方式找到 `target - numbers[i]` 的位置,如果存在,那么返回 $[i + 1, j + 1]$ 即可。
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我们注意到数组按照非递减顺序排列,因此对于每个 $numbers[i]$,可以通过二分查找的方式找到 $target - numbers[i]$ 的位置,如果存在,那么返回 $[i + 1, j + 1]$ 即可。
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时间复杂度 $O(n \times \log n),ドル其中 $n$ 为数组 `numbers` 的长度。空间复杂度 $O(1)$。
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时间复杂度 $O(n \times \log n),ドル其中 $n$ 为数组 $numbers$ 的长度。空间复杂度 $O(1)$。
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**方法二:双指针**
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我们定义两个指针 $i$ 和 $j,ドル分别指向数组的第一个元素和最后一个元素。每次计算 $numbers[i] + numbers[j],ドル如果和等于目标值,那么返回 $[i + 1, j + 1]$ 即可。如果和小于目标值,那么将 $i$ 右移一位,如果和大于目标值,那么将 $j$ 左移一位。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 `numbers` 的长度。空间复杂度 $O(1)$。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 $numbers$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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‎solution/2100-2199/2153.The Number of Passengers in Each Bus II/README.md‎

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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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# 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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*,
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sum(cnt) OVER (ORDER BY dt, bus_id) AS cur,
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if(@t > 0, @t := cnt, @t := @t + cnt) AS cur_sum
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FROM
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(
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SELECT bus_id, arrival_time AS dt, capacity AS cnt FROM Buses
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UNION ALL
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SELECT -1, arrival_time AS dt, -1 FROM Passengers
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) AS a JOIN (SELECT @t := 0 x) AS b
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)
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SELECT
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bus_id,
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if(cur_sum > 0, cnt - cur_sum, cnt) AS passengers_cnt
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FROM T
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WHERE bus_id > 0
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ORDER BY bus_id;
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```
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<!-- tabs:end -->

‎solution/2100-2199/2153.The Number of Passengers in Each Bus II/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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*,
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sum(cnt) OVER (ORDER BY dt, bus_id) AS cur,
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if(@t > 0, @t := cnt, @t := @t + cnt) AS cur_sum
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FROM
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(
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SELECT bus_id, arrival_time AS dt, capacity AS cnt FROM Buses
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UNION ALL
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SELECT -1, arrival_time AS dt, -1 FROM Passengers
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) AS a JOIN (SELECT @t := 0 x) AS b
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)
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SELECT
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bus_id,
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if(cur_sum > 0, cnt - cur_sum, cnt) AS passengers_cnt
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FROM T
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WHERE bus_id > 0
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ORDER BY bus_id;
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```
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<!-- tabs:end -->
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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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*,
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sum(cnt) OVER (ORDER BY dt, bus_id) AS cur,
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if(@t > 0, @t := cnt, @t := @t + cnt) AS cur_sum
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FROM
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(
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SELECT bus_id, arrival_time AS dt, capacity AS cnt FROM Buses
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UNION ALL
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SELECT -1, arrival_time AS dt, -1 FROM Passengers
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) AS a JOIN (SELECT @t := 0 x) AS b
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)
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SELECT
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bus_id,
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if(cur_sum > 0, cnt - cur_sum, cnt) AS passengers_cnt
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FROM T
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WHERE bus_id > 0
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ORDER BY bus_id;

‎solution/2600-2699/2609.Find the Longest Balanced Substring of a Binary String/README.md‎

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if j - i + 1 < ans {
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break;
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}
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if check(i, j) {
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ans = std::cmp::max(ans, j - i + 1);
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break;

‎solution/2600-2699/2609.Find the Longest Balanced Substring of a Binary String/README_EN.md‎

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if j - i + 1 < ans {
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break;
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}
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if check(i, j) {
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ans = std::cmp::max(ans, j - i + 1);
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break;

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