LeetCode #3497 — MEDIUM

Analyze Subscription Conversion

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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.

The result format is in the following example.

Example:

Input:

UserActivity table:

+---------+---------------+---------------+-------------------+
| 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                 |
+---------+---------------+---------------+-------------------+

Output:

+---------+--------------------+-------------------+
| user_id | trial_avg_duration | paid_avg_duration |
+---------+--------------------+-------------------+
| 1       | 45.00              | 76.67             |
| 3       | 70.00              | 63.33             |
| 4       | 37.50              | 45.00             |
+---------+--------------------+-------------------+

Explanation:

User 1:
- Had 3 days of free trial with durations of 45, 30, and 60 minutes.
- Average trial duration: (45 + 30 + 60) / 3 = 45.00 minutes.
- Had 3 days of paid subscription with durations of 75, 90, and 65 minutes.
- Average paid duration: (75 + 90 + 65) / 3 = 76.67 minutes.
User 2:
- Had 3 days of free trial with durations of 55, 25, and 50 minutes.
- Average trial duration: (55 + 25 + 50) / 3 = 43.33 minutes.
- Did not convert to a paid subscription (only had free_trial and cancelled activities).
- Not included in the output because they didn't convert to paid.
User 3:
- Had 3 days of free trial with durations of 70, 60, and 80 minutes.
- Average trial duration: (70 + 60 + 80) / 3 = 70.00 minutes.
- Had 3 days of paid subscription with durations of 50, 55, and 85 minutes.
- Average paid duration: (50 + 55 + 85) / 3 = 63.33 minutes.
User 4:
- Had 2 days of free trial with durations of 40 and 35 minutes.
- Average trial duration: (40 + 35) / 2 = 37.50 minutes.
- Had 1 day of paid subscription with duration of 45 minutes before cancelling.
- Average paid duration: 45.00 minutes.

The result table only includes users who converted from free trial to paid subscription (users 1, 3, and 4), and is ordered by user_id in ascending order.

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers":{"UserActivity":["user_id","activity_date","activity_type","activity_duration"]},"rows":{"UserActivity":[[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]]}}

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// 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

// Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// 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

# Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
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

// Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (py):
// # Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// 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

// Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #3497: Analyze Subscription Conversion 
// 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

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.