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feat: update solutions to lc problems: No.1454,1455 (doocs#3356)
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‎.prettierignore

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@@ -16,7 +16,6 @@ node_modules/
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/solution/0100-0199/0177.Nth Highest Salary/Solution.sql
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/solution/0100-0199/0178.Rank Scores/Solution2.sql
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/solution/0500-0599/0586.Customer Placing the Largest Number of Orders/Solution2.sql
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/solution/1400-1499/1454.Active Users/Solution.sql
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/solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
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/solution/2100-2199/2118.Build the Equation/Solution.sql
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/solution/2100-2199/2175.The Change in Global Rankings/Solution.sql
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/solution/2200-2299/2252.Dynamic Pivoting of a Table/Solution.sql
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/solution/2200-2299/2253.Dynamic Unpivoting of a Table/Solution.sql
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/solution/3100-3199/3150.Invalid Tweets II/Solution.sql
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/solution/3100-3199/3198.Find Cities in Each State/Solution.sql
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/solution/3100-3199/3198.Find Cities in Each State/Solution.sql

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

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@@ -104,26 +104,44 @@ id = 7 的用户 Jonathon 在不同的 6 天内登录了 7 次, , 6 天中有 5
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<!-- solution:start -->
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### 方法一
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### 方法一: 使用窗口函数
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我们先将 `Logins` 表和 `Accounts` 表连接起来，并且去重，得到临时表 `T`。
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然后我们使用窗口函数 `ROW_NUMBER()`，计算出每个用户 `id` 的登录日期的基准日期 `g`，如果用户连续登录 5 天，那么他们的 `g` 值是相同的。
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最后，我们按照 `id` 和 `g` 进行分组，统计每个用户的登录次数，如果登录次数大于等于 5，那么这个用户就是活跃用户。
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<!-- tabs:start -->
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#### MySQL
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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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WITH
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T AS (
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SELECT DISTINCT *
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FROM
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Logins
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JOIN Accounts USING (id)
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),
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P AS (
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SELECT
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*,
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DATE_SUB(
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login_date,
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INTERVAL ROW_NUMBER() OVER (
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PARTITION BY id
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ORDER BY login_date
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) DAY
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) g
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FROM T
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)
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SELECT DISTINCT id, name
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FROM P
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GROUP BY id, g
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HAVING COUNT(*) >= 5
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ORDER BY 1;
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```
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<!-- tabs:end -->

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

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@@ -58,7 +58,7 @@ This table contains the account id of the user who logged in and the login date.
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong>
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<strong>Input:</strong>
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Accounts table:
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+----+----------+
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| id | name |
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| 1 | 2020-06-07 |
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| 7 | 2020-06-10 |
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+----+------------+
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<strong>Output:</strong>
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<strong>Output:</strong>
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+----+----------+
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| id | name |
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+----+----------+
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| 7 | Jonathan |
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+----+----------+
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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User Winston with id = 1 logged in 2 times only in 2 different days, so, Winston is not an active user.
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User Jonathan with id = 7 logged in 7 times in 6 different days, five of them were consecutive days, so, Jonathan is an active user.
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</pre>
@@ -100,26 +100,44 @@ User Jonathan with id = 7 logged in 7 times in 6 different days, five of them we
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Using Window Functions
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First, we join the `Logins` table and the `Accounts` table, and remove duplicates to get the temporary table `T`.
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Then, we use the window function `ROW_NUMBER()` to calculate the base login date `g` for each user `id`. If a user logs in for 5 consecutive days, their `g` values are the same.
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Finally, we group by `id` and `g` to count the number of logins for each user. If the number of logins is greater than or equal to 5, then the user is considered active.
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<!-- tabs:start -->
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#### MySQL
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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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WITH
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T AS (
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SELECT DISTINCT *
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FROM
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Logins
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JOIN Accounts USING (id)
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),
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P AS (
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SELECT
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*,
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DATE_SUB(
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login_date,
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INTERVAL ROW_NUMBER() OVER (
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PARTITION BY id
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ORDER BY login_date
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) DAY
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) g
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FROM T
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)
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SELECT DISTINCT id, name
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FROM P
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GROUP BY id, g
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HAVING COUNT(*) >= 5
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/1400-1499/1454.Active Users/Solution.sql

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@@ -1,13 +1,25 @@
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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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WITH
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T AS (
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SELECT DISTINCT *
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FROM
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Logins
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JOIN Accounts USING (id)
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),
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P AS (
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SELECT
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*,
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DATE_SUB(
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login_date,
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INTERVAL ROW_NUMBER() OVER (
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PARTITION BY id
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ORDER BY login_date
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) DAY
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) g
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FROM T
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)
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SELECT DISTINCT id, name
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FROM P
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GROUP BY id, g
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HAVING COUNT(*) >= 5
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ORDER BY 1;

‎solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README.md

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@@ -71,9 +71,9 @@ tags:
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### 方法一：字符串分割
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将 $sentence$ 按空格分割为 $words$，然后遍历 $words$，检查 $words[i]$ 是否是 $searchWord$ 的前缀，是则返回 $i+1$。若遍历结束，所有单词都不满足，返回 $-1$。
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我们将 $\textit{sentence}$ 按空格分割为 $\textit{words}$，然后遍历 $\textit{words}$，检查 $\textit{words}[i]$ 是否是 $\textit{searchWord}$ 的前缀，是则返回 $i+1$。若遍历结束，所有单词都不满足，返回 $-1$。
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时间复杂度 $O(mn)$。其中 $m$ 是 $sentence$ 的长度，而 $n$ 是 $searchWord$ 的长度。
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时间复杂度 $O(m \times n)$，空间复杂度 $O(m)$。其中 $m$ 和 $n$ 分别是 $\textit{sentence}$ 和 $\textit{searchWord}$ 的长度。
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<!-- tabs:start -->
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* @return Integer
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*/
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function isPrefixOfWord($sentence, $searchWord) {
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$arr = explode(' ', $sentence);
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for ($i = 0; $i < count($arr); $i++) {
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if (strpos($arr[$i], $searchWord) === 0) {
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$words = explode(' ', $sentence);
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for ($i = 0; $i < count($words); ++$i) {
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if (strpos($words[$i], $searchWord) === 0) {
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return $i + 1;
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}
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}

‎solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/README_EN.md

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@@ -67,7 +67,11 @@ tags:
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<!-- solution:start -->
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### Solution 1
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### Solution 1: String Splitting
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We split $\textit{sentence}$ by spaces into $\textit{words}$, then iterate through $\textit{words}$ to check if $\textit{words}[i]$ is a prefix of $\textit{searchWord}$. If it is, we return $i+1$. If the iteration completes and no words satisfy the condition, we return $-1$.
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The time complexity is $O(m \times n)$, and the space complexity is $O(m)$. Here, $m$ and $n$ are the lengths of $\textit{sentence}$ and $\textit{searchWord}$, respectively.
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<!-- tabs:start -->
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* @return Integer
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*/
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function isPrefixOfWord($sentence, $searchWord) {
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$arr = explode(' ', $sentence);
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for ($i = 0; $i < count($arr); $i++) {
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if (strpos($arr[$i], $searchWord) === 0) {
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$words = explode(' ', $sentence);
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for ($i = 0; $i < count($words); ++$i) {
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if (strpos($words[$i], $searchWord) === 0) {
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return $i + 1;
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}
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}

‎solution/1400-1499/1455.Check If a Word Occurs As a Prefix of Any Word in a Sentence/Solution.php

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@@ -5,9 +5,9 @@ class Solution {
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* @return Integer
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*/
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function isPrefixOfWord($sentence, $searchWord) {
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$arr = explode(' ', $sentence);
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for ($i = 0; $i < count($arr); $i++) {
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if (strpos($arr[$i], $searchWord) === 0) {
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$words = explode(' ', $sentence);
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for ($i = 0; $i < count($words); ++$i) {
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if (strpos($words[$i], $searchWord) === 0) {
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return $i + 1;
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}
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}

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