feat: update solutions to lc problems: No.1454,1455

.prettierignore

@@ -16,7 +16,6 @@ node_modules/
 /solution/0100-0199/0177.Nth Highest Salary/Solution.sql
 /solution/0100-0199/0178.Rank Scores/Solution2.sql
 /solution/0500-0599/0586.Customer Placing the Largest Number of Orders/Solution2.sql
-/solution/1400-1499/1454.Active Users/Solution.sql
 /solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
 /solution/2100-2199/2118.Build the Equation/Solution.sql
 /solution/2100-2199/2175.The Change in Global Rankings/Solution.sql
@@ -25,4 +24,4 @@ node_modules/
 /solution/2200-2299/2252.Dynamic Pivoting of a Table/Solution.sql
 /solution/2200-2299/2253.Dynamic Unpivoting of a Table/Solution.sql
 /solution/3100-3199/3150.Invalid Tweets II/Solution.sql
-/solution/3100-3199/3198.Find Cities in Each State/Solution.sql
+/solution/3100-3199/3198.Find Cities in Each State/Solution.sql

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

@@ -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
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一: 使用窗口函数
+
+我们先将 `Logins` 表和 `Accounts` 表连接起来，并且去重，得到临时表 `T`。
+
+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。
+
+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。
 
 <!-- tabs:start -->
 
 #### MySQL
 
 ```sql
 # Write your MySQL query statement below
-WITH t AS
-    (SELECT *,
-		 SUM(id) over(partition by id
-    ORDER BY  login_date range interval 4 day preceding)/id cnt
-    FROM
-        (SELECT DISTINCT *
-        FROM Accounts
-        JOIN Logins using(id) ) tt )
-    SELECT DISTINCT id,
-		 name
-FROM t
-WHERE cnt=5;
+WITH
+    T AS (
+        SELECT DISTINCT *
+        FROM
+            Logins
+            JOIN Accounts USING (id)
+    ),
+    P AS (
+        SELECT
+            *,
+            DATE_SUB(
+                login_date,
+                INTERVAL ROW_NUMBER() OVER (
+                    PARTITION BY id
+                    ORDER BY login_date
+                ) DAY
+            ) g
+        FROM T
+    )
+SELECT DISTINCT id, name
+FROM P
+GROUP BY id, g
+HAVING COUNT(*) >= 5
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -58,7 +58,7 @@ This table contains the account id of the user who logged in and the login date.
 <p><strong class="example">Example 1:</strong></p>
 
 <pre>
-<strong>Input:</strong> 
+<strong>Input:</strong>
 Accounts table:
 +----+----------+
 | id | name     |
@@ -80,13 +80,13 @@ Logins table:
 | 1  | 2020-06-07 |
 | 7  | 2020-06-10 |
 +----+------------+
-<strong>Output:</strong> 
+<strong>Output:</strong>
 +----+----------+
 | id | name     |
 +----+----------+
 | 7  | Jonathan |
 +----+----------+
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 User Winston with id = 1 logged in 2 times only in 2 different days, so, Winston is not an active user.
 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.
 </pre>
@@ -100,26 +100,44 @@ User Jonathan with id = 7 logged in 7 times in 6 different days, five of them we
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Using Window Functions
+
+First, we join the `Logins` table and the `Accounts` table, and remove duplicates to get the temporary table `T`.
+
+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.
+
+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.
 
 <!-- tabs:start -->
 
 #### MySQL
 
 ```sql
 # Write your MySQL query statement below
-WITH t AS
-    (SELECT *,
-		 SUM(id) over(partition by id
-    ORDER BY  login_date range interval 4 day preceding)/id cnt
-    FROM
-        (SELECT DISTINCT *
-        FROM Accounts
-        JOIN Logins using(id) ) tt )
-    SELECT DISTINCT id,
-		 name
-FROM t
-WHERE cnt=5;
+WITH
+    T AS (
+        SELECT DISTINCT *
+        FROM
+            Logins
+            JOIN Accounts USING (id)
+    ),
+    P AS (
+        SELECT
+            *,
+            DATE_SUB(
+                login_date,
+                INTERVAL ROW_NUMBER() OVER (
+                    PARTITION BY id
+                    ORDER BY login_date
+                ) DAY
+            ) g
+        FROM T
+    )
+SELECT DISTINCT id, name
+FROM P
+GROUP BY id, g
+HAVING COUNT(*) >= 5
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -1,13 +1,25 @@
 # Write your MySQL query statement below
-WITH t AS
-    (SELECT *,
-		 SUM(id) over(partition by id
-    ORDER BY  login_date range interval 4 day preceding)/id cnt
-    FROM
-        (SELECT DISTINCT *
-        FROM Accounts
-        JOIN Logins using(id) ) tt )
-    SELECT DISTINCT id,
-		 name
-FROM t
-WHERE cnt=5;
+WITH
+    T AS (
+        SELECT DISTINCT *
+        FROM
+            Logins
+            JOIN Accounts USING (id)
+    ),
+    P AS (
+        SELECT
+            *,
+            DATE_SUB(
+                login_date,
+                INTERVAL ROW_NUMBER() OVER (
+                    PARTITION BY id
+                    ORDER BY login_date
+                ) DAY
+            ) g
+        FROM T
+    )
+SELECT DISTINCT id, name
+FROM P
+GROUP BY id, g
+HAVING COUNT(*) >= 5
+ORDER BY 1;

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

@@ -71,9 +71,9 @@ tags:
 
 ### 方法一：字符串分割
 
-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 $-1$。
+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 $-1$。
 
-时间复杂度 $O(mn)$。其中 $m$ 是 $sentence$ 的长度，而 $n$ 是 $searchWord$ 的长度。
+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。
 
 <!-- tabs:start -->
 
@@ -176,9 +176,9 @@ class Solution {
      * @return Integer
      */
     function isPrefixOfWord($sentence, $searchWord) {
-        $arr = explode(' ', $sentence);
-        for ($i = 0; $i < count($arr); $i++) {
-            if (strpos($arr[$i], $searchWord) === 0) {
+        $words = explode(' ', $sentence);
+        for ($i = 0; $i < count($words); ++$i) {
+            if (strpos($words[$i], $searchWord) === 0) {
                 return $i + 1;
             }
         }

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

@@ -67,7 +67,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: String Splitting
+
+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 $-1$.
+
+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.
 
 <!-- tabs:start -->
 
@@ -170,9 +174,9 @@ class Solution {
      * @return Integer
      */
     function isPrefixOfWord($sentence, $searchWord) {
-        $arr = explode(' ', $sentence);
-        for ($i = 0; $i < count($arr); $i++) {
-            if (strpos($arr[$i], $searchWord) === 0) {
+        $words = explode(' ', $sentence);
+        for ($i = 0; $i < count($words); ++$i) {
+            if (strpos($words[$i], $searchWord) === 0) {
                 return $i + 1;
             }
         }

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

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