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Jul 15, 2025
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task: #3497 #69

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1 change: 1 addition & 0 deletions README.md
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Expand Up @@ -174,6 +174,7 @@ Useful for preparing for technical interviews and improving your SQL skills.
- [1907. Count Salary Categories](./leetcode/medium/1907.%20Count%20Salary%20Categories.sql)
- [1934. Confirmation Rate](./leetcode/medium/1934.%20Confirmation%20Rate.sql)
- [3475. DNA Pattern Recognition](./leetcode/medium/3475.%20DNA%20Pattern%20Recognition.sql)
- [3497. Analyze Subscription Conversion](./leetcode/medium/3497.%20Analyze%20Subscription%20Conversion.sql)
3. [Hard](./leetcode/hard/)
- [185. Department Top Three Salaries](./leetcode/hard/185.%20Department%20Top%20Three%20Salaries.sql)
- [3374. First Letter Capitalization II](./leetcode/hard/3374.%20First%20Letter%20Capitalization%20II.sql)
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56 changes: 56 additions & 0 deletions leetcode/medium/3497. Analyze Subscription Conversion.sql
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@@ -0,0 +1,56 @@
/*
Question 3497. Analyze Subscription Conversion
Link: https://leetcode.com/problems/analyze-subscription-conversion/description/?envType=problem-list-v2&envId=database

Table: UserActivity

+------------------+---------+
| Column Name | Type |
+------------------+---------+
| user_id | int |
| activity_date | date |
| activity_type | varchar |
| activity_duration| int |
+------------------+---------+
(user_id, activity_date, activity_type) is the unique key for this table.
activity_type is one of ('free_trial', 'paid', 'cancelled').
activity_duration is the number of minutes the user spent on the platform that day.
Each row represents a user's activity on a specific date.
A subscription service wants to analyze user behavior patterns. The company offers a 7-day free trial, after which users can subscribe to a paid plan or cancel. Write a solution to:

Find users who converted from free trial to paid subscription
Calculate each user's average daily activity duration during their free trial period (rounded to 2 decimal places)
Calculate each user's average daily activity duration during their paid subscription period (rounded to 2 decimal places)
Return the result table ordered by user_id in ascending order.
*/

WITH trial_durations AS (
SELECT
user_id,
AVG(activity_duration) AS trial_avg_duration
FROM UserActivity
WHERE activity_type = 'free_trial'
GROUP BY user_id
),

paid_duration AS (
SELECT
user_id,
AVG(activity_duration) AS paid_avg_duration
FROM UserActivity
WHERE activity_type = 'paid'
GROUP BY user_id
)

SELECT DISTINCT
u.user_id,
ROUND(t.trial_avg_duration, 2) AS trial_avg_duration,
ROUND(p.paid_avg_duration, 2) AS paid_avg_duration
FROM UserActivity AS u
INNER JOIN
trial_durations AS t
ON u.user_id = t.user_id
INNER JOIN
paid_duration AS p
ON u.user_id = p.user_id