doocs · yanglbme · Mar 28, 2025 · Mar 28, 2025
diff --git a/solution/2000-2099/2033.Minimum Operations to Make a Uni-Value Grid/README.md b/solution/2000-2099/2033.Minimum Operations to Make a Uni-Value Grid/README.md
@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>输入：</strong>grid = [[2,4],[6,8]], x = 2
 <strong>输出：</strong>4
-<strong>解释：</strong>可以执行下述操作使所有元素都等于 4 ：
+<strong>解释：</strong>可以执行下述操作使所有元素都等于 4 ： 
 - 2 加 x 一次。
 - 6 减 x 一次。
 - 8 减 x 两次。

diff --git a/solution/2000-2099/2033.Minimum Operations to Make a Uni-Value Grid/README_EN.md b/solution/2000-2099/2033.Minimum Operations to Make a Uni-Value Grid/README_EN.md
@@ -33,7 +33,7 @@ tags:
 <pre>
 <strong>Input:</strong> grid = [[2,4],[6,8]], x = 2
 <strong>Output:</strong> 4
-<strong>Explanation:</strong> We can make every element equal to 4 by doing the following:
+<strong>Explanation:</strong> We can make every element equal to 4 by doing the following: 
 - Add x to 2 once.
 - Subtract x from 6 once.
 - Subtract x from 8 twice.

diff --git a/solution/3400-3499/3497.Analyze Subscription Conversion/README.md b/solution/3400-3499/3497.Analyze Subscription Conversion/README.md
@@ -0,0 +1,220 @@
+---
+comments: true
+difficulty: 中等
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3497.Analyze%20Subscription%20Conversion/README.md
+tags:
+    - 数据库
+---
+
+<!-- problem:start -->
+
+# [3497. 分析订阅转化](https://leetcode.cn/problems/analyze-subscription-conversion)
+
+[English Version](/solution/3400-3499/3497.Analyze%20Subscription%20Conversion/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>表：<code>UserActivity</code></p>
+
+<pre>
++------------------+---------+
+| Column Name      | Type    | 
++------------------+---------+
+| user_id          | int     |
+| activity_date    | date    |
+| activity_type    | varchar |
+| activity_duration| int     |
++------------------+---------+
+(user_id, activity_date, activity_type) 是这张表的唯一主键。
+activity_type 是('free_trial', 'paid', 'cancelled')中的一个。
+activity_duration 是用户当天在平台上花费的分钟数。
+每一行表示一个用户在特定日期的活动。
+</pre>
+
+<p>订阅服务想要分析用户行为模式。公司提供7天免费试用，试用结束后，用户可以选择订阅 <strong>付费计划</strong> 或 <strong>取消</strong>。编写解决方案：</p>
+
+<ol>
+	<li>查找从免费试用转为付费订阅的用户</li>
+	<li>计算每位用户在 <strong>免费试用</strong> 期间的 <strong>平均每日活动时长</strong>（四舍五入至小数点后 <code>2</code> 位）</li>
+	<li>计算每位用户在 <strong>付费</strong> 订阅期间的 <strong>平均每日活动时长</strong>（四舍五入到小数点后&nbsp;<code>2</code> 位）</li>
+</ol>
+
+<p>返回结果表以<em>&nbsp;</em><code>user_id</code><em> </em><strong>升序&nbsp;</strong>排序。</p>
+
+<p>结果格式如下所示。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong></p>
+
+<p>UserActivity 表：</p>
+
+<pre class="example-io">
++---------+---------------+---------------+-------------------+
+| user_id | activity_date | activity_type | activity_duration |
++---------+---------------+---------------+-------------------+
+| 1       | 2023-01-01    | free_trial    | 45                |
+| 1       | 2023-01-02    | free_trial    | 30                |
+| 1       | 2023-01-05    | free_trial    | 60                |
+| 1       | 2023-01-10    | paid          | 75                |
+| 1       | 2023-01-12    | paid          | 90                |
+| 1       | 2023-01-15    | paid          | 65                |
+| 2       | 2023-02-01    | free_trial    | 55                |
+| 2       | 2023-02-03    | free_trial    | 25                |
+| 2       | 2023-02-07    | free_trial    | 50                |
+| 2       | 2023-02-10    | cancelled     | 0                 |
+| 3       | 2023-03-05    | free_trial    | 70                |
+| 3       | 2023-03-06    | free_trial    | 60                |
+| 3       | 2023-03-08    | free_trial    | 80                |
+| 3       | 2023-03-12    | paid          | 50                |
+| 3       | 2023-03-15    | paid          | 55                |
+| 3       | 2023-03-20    | paid          | 85                |
+| 4       | 2023-04-01    | free_trial    | 40                |
+| 4       | 2023-04-03    | free_trial    | 35                |
+| 4       | 2023-04-05    | paid          | 45                |
+| 4       | 2023-04-07    | cancelled     | 0                 |
++---------+---------------+---------------+-------------------+
+</pre>
+
+<p><strong>输出：</strong></p>
+
+<pre class="example-io">
++---------+--------------------+-------------------+
+| user_id | trial_avg_duration | paid_avg_duration |
++---------+--------------------+-------------------+
+| 1       | 45.00              | 76.67             |
+| 3       | 70.00              | 63.33             |
+| 4       | 37.50              | 45.00             |
++---------+--------------------+-------------------+
+</pre>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><strong>用户 1:</strong>
+
+    <ul>
+    	<li>体验了 3 天免费试用，时长分别为 45，30 和 60 分钟。</li>
+    	<li>平均试用时长：(45 + 30 + 60) / 3 = 45.00 分钟。</li>
+    	<li>拥有 3 天付费订阅，时长分别为 75，90 和 65分钟。</li>
+    	<li>平均花费市场：(75 + 90 + 65) / 3 = 76.67 分钟。</li>
+    </ul>
+    </li>
+    <li><strong>用户 2:</strong>
+    <ul>
+    	<li>体验了 3 天免费试用，时长分别为 55，25 和 50 分钟。</li>
+    	<li>平均试用时长：(55 + 25 + 50) / 3 = 43.33 分钟。</li>
+    	<li>没有转为付费订阅（只有&nbsp;free_trial 和 cancelled 活动）。</li>
+    	<li>未包含在输出中，因为他未转换为付费用户。</li>
+    </ul>
+    </li>
+    <li><strong>用户 3:</strong>
+    <ul>
+    	<li>体验了 3 天免费试用，时长分别为 70，60 和 80 分钟。</li>
+    	<li>平均试用时长：(70 + 60 + 80) / 3 = 70.00 分钟。</li>
+    	<li>拥有 3 天付费订阅，时长分别为 50，55 和 85 分钟。</li>
+    	<li>平均花费时长：(50 + 55 + 85) / 3 = 63.33 分钟。</li>
+    </ul>
+    </li>
+    <li><strong>用户 4:</strong>
+    <ul>
+    	<li>体验了 2 天免费试用，时长分别为 40 和 35 分钟。</li>
+    	<li>平均试用时长：(40 + 35) / 2 = 37.50 分钟。</li>
+    	<li>在取消前有 1 天的付费订阅，时长为45分钟。</li>
+    	<li>平均花费时长：45.00 分钟。</li>
+    </ul>
+    </li>
+
+</ul>
+
+<p>结果表仅包括从免费试用转为付费订阅的用户（用户 1，3 和 4），并且以&nbsp;user_id 升序排序。</p>
+</div>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：分组 + 条件筛选 + 等值连接
+
+我们首先将表中的数据进行筛选，找出所有 `activity_type` 不等于 `cancelled` 的数据，将数据按照 `user_id` 和 `activity_type` 进行分组，求得每组的时长 `duration`，记录在表 `T` 中。
+
+接下来，我们从表 `T` 中筛选出 `activity_type` 为 `free_trial` 和 `paid` 的记录，分别记录在表 `F` 和 `P` 中，最后将这两张表按照 `user_id` 进行等值连接，并按照题目要求筛选出对应的字段并排序，得到最终结果。
+
+<!-- tabs:start -->
+
+#### MySQL
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT user_id, activity_type, ROUND(SUM(activity_duration) / COUNT(1), 2) duration
+        FROM UserActivity
+        WHERE activity_type != 'cancelled'
+        GROUP BY user_id, activity_type
+    ),
+    F AS (
+        SELECT user_id, duration trial_avg_duration
+        FROM T
+        WHERE activity_type = 'free_trial'
+    ),
+    P AS (
+        SELECT user_id, duration paid_avg_duration
+        FROM T
+        WHERE activity_type = 'paid'
+    )
+SELECT user_id, trial_avg_duration, paid_avg_duration
+FROM
+    F
+    JOIN P USING (user_id)
+ORDER BY 1;
+```
+
+#### Pandas
+
+```python
+import pandas as pd
+
+
+def analyze_subscription_conversion(user_activity: pd.DataFrame) -> pd.DataFrame:
+    df = user_activity[user_activity["activity_type"] != "cancelled"]
+
+    df_grouped = (
+        df.groupby(["user_id", "activity_type"])["activity_duration"]
+        .mean()
+        .add(0.0001)
+        .round(2)
+        .reset_index()
+    )
+
+    df_free_trial = (
+        df_grouped[df_grouped["activity_type"] == "free_trial"]
+        .rename(columns={"activity_duration": "trial_avg_duration"})
+        .drop(columns=["activity_type"])
+    )
+
+    df_paid = (
+        df_grouped[df_grouped["activity_type"] == "paid"]
+        .rename(columns={"activity_duration": "paid_avg_duration"})
+        .drop(columns=["activity_type"])
+    )
+
+    result = df_free_trial.merge(df_paid, on="user_id", how="inner").sort_values(
+        "user_id"
+    )
+
+    return result
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->