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Active Users - Problem

Database Medium

Active Users Analytics

You're building a user engagement analytics system for a social media platform. Your task is to identify active users - those who demonstrate consistent engagement by logging in for 5 or more consecutive days.

Given two database tables:

Accounts Table: Contains user account information (id and name)
Logins Table: Contains login records (account id and login date)

Your goal is to write an SQL query that finds all users who have logged in for at least 5 consecutive days. Return their id and name, ordered by id.

Note: A user may log in multiple times per day, but we only care about the distinct login dates for consecutive day calculations.

Input & Output

example_1.sql — Basic consecutive logins

$ Input: Accounts: [(1,'Winston'), (7,'Jonathan')] Logins: [(7,'2020-05-30'), (1,'2020-05-30'), (7,'2020-05-31'), (7,'2020-06-01'), (7,'2020-06-02'), (7,'2020-06-03'), (7,'2020-06-04'), (1,'2020-06-01')]

› Output: [(7,'Jonathan')]

💡 Note: User 7 (Jonathan) logged in for 5 consecutive days from 2020-05-30 to 2020-06-03, making him an active user. User 1 (Winston) only has 2 separate login days, not meeting the 5 consecutive days requirement.

example_2.sql — Multiple consecutive streaks

$ Input: Accounts: [(1,'Alice'), (2,'Bob'), (3,'Charlie')] Logins: [(1,'2023-01-01'), (1,'2023-01-02'), (1,'2023-01-03'), (1,'2023-01-04'), (1,'2023-01-05'), (2,'2023-01-01'), (2,'2023-01-03'), (2,'2023-01-05'), (3,'2023-01-10'), (3,'2023-01-11'), (3,'2023-01-12'), (3,'2023-01-13'), (3,'2023-01-14'), (3,'2023-01-15')]

› Output: [(1,'Alice'), (3,'Charlie')]

💡 Note: Alice has 5 consecutive days (Jan 1-5), Charlie has 6 consecutive days (Jan 10-15). Bob has non-consecutive login days, so he doesn't qualify as an active user.

example_3.sql — Edge case with duplicates

$ Input: Accounts: [(1,'John')] Logins: [(1,'2023-01-01'), (1,'2023-01-01'), (1,'2023-01-02'), (1,'2023-01-03'), (1,'2023-01-04'), (1,'2023-01-05'), (1,'2023-01-05')]

› Output: [(1,'John')]

💡 Note: Even with duplicate login entries for the same day, John still has 5 consecutive unique login days (Jan 1-5), making him an active user. Duplicates are handled by using DISTINCT in the query.

Constraints

1 ≤ Accounts table size ≤ 500
1 ≤ Logins table size ≤ 1000
1 ≤ id ≤ 500
All login_date values are valid dates
A user may have multiple login records for the same date

Visualization

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5 consecutive daysGroup B: 2 consecutive days✅ User 1 is ACTIVE (Group A ≥ 5 days)

Understanding the Visualization

Remove Duplicates

Get unique login dates per user (same person can't visit gym twice in one day)

Rank Dates

Apply DENSE_RANK to order login dates chronologically for each user

Group Consecutive

Calculate date groups by subtracting rank from date - consecutive dates get same group

Count & Filter

Count group sizes and keep users with 5+ consecutive login days

Key Takeaway

🎯 Key Insight: DENSE_RANK creates consistent group identifiers for consecutive date sequences, allowing efficient detection of active users in O(n log n) time

Asked in

f Meta 45 a Amazon 38 G Google 32 ⊞ Microsoft 25

The optimal solution uses DENSE_RANK() window function to efficiently detect consecutive login patterns. The key insight is that consecutive dates have the same date_group when we subtract the rank from the login date. This allows us to group consecutive sequences and count their lengths in O(n log n) time.

Common Approaches

Approach	Time	Space	Notes
✓ DENSE_RANK Window Function (Optimal)	O(n log n)	O(n)	Use DENSE_RANK to efficiently identify consecutive login streaks in one pass
Window Function with Row Number	O(n log n)	O(n)	Use ROW_NUMBER() to detect consecutive date patterns efficiently
Brute Force (Self-Join Approach)	O(n⁵)	O(n)	Use self-joins to check every possible 5-day consecutive sequence

DENSE_RANK Window Function (Optimal) — Algorithm Steps

Remove duplicate login dates using DISTINCT
Apply DENSE_RANK() to order dates for each user
Calculate date_group by subtracting rank from login_date
Group by user_id and date_group to find consecutive sequences
Filter groups having COUNT >= 5 (5 or more consecutive days)
Join with Accounts table to get user names and return results

Visualization

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Step-by-Step Walkthrough

Remove duplicates

Get distinct login dates for each user

Apply DENSE_RANK

Assign ranks to consecutive dates

Create date groups

Subtract rank from date to group consecutive sequences

Count and filter

Keep groups with 5+ consecutive days

Code -

solution.c — C

// Optimal SQL Query using DENSE_RANK Window Function
char* query = "\
SELECT a.id, a.name \
FROM Accounts a \
WHERE a.id IN ( \
    SELECT id \
    FROM ( \
        SELECT id, \
               DATE_SUB(login_date, INTERVAL DENSE_RANK() OVER (PARTITION BY id ORDER BY login_date) - 1 DAY) as date_group \
        FROM ( \
            SELECT DISTINCT id, login_date \
            FROM Logins \
        ) t \
    ) ranked \
    GROUP BY id, date_group \
    HAVING COUNT(*) >= 5 \
) \
ORDER BY a.id;";

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Single sort for DENSE_RANK operation, then linear grouping and filtering

⚡ Linearithmic

Space Complexity

O(n)

Space for intermediate results and window function processing

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

DENSE_RANK Window Function (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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