How do you solve User Activity for the Past 30 Days I on LeetCode?

User Activity for the Past 30 Days I (LeetCode #1141) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap can significantly reduce the time complexity of counting distinct elements.

What is the time complexity of User Activity for the Past 30 Days I?

The optimal solution for User Activity for the Past 30 Days I runs in O(n) time and uses O(n) space. This complexity is due to a single pass through the activities (O(n)) and storing the results in a HashMap (O(n)) for distinct user counts.

What is the brute force solution for User Activity for the Past 30 Days I?

The brute force approach for User Activity for the Past 30 Days I is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each day in the 30-day period and count the distinct users who were active on that day. This is straightforward but inefficient as it involves checking each activity for each day. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve User Activity for the Past 30 Days I?

User Activity for the Past 30 Days I uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation.

What are the interview tips for User Activity for the Past 30 Days I?

Always clarify the requirements and constraints before jumping into coding. Think about edge cases, such as no activity in the date range. Practice writing SQL queries as they often come up in data-related interviews.

What are common mistakes when solving User Activity for the Past 30 Days I?

Not considering duplicate activities for the same user on the same day. Failing to filter the date range correctly.

#1141

User Activity for the Past 30 Days I

Easy

DatabaseHash MapAggregation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of checking each day against all activities, we can use a single pass through the activity table to build a count of distinct users for each day. This reduces the number of operations significantly.

⚙️

Algorithm

3 steps

1Step 1: Use a HashMap to store the count of distinct users for each activity date.
2Step 2: Iterate through the activity table and for each activity, update the count in the HashMap.
3Step 3: Extract the results from the HashMap for the 30-day period.

solution.py4 lines

1SELECT activity_date, COUNT(DISTINCT user_id) AS active_users
2FROM Activity
3WHERE activity_date BETWEEN DATE('2019-06-28') AND DATE('2019-07-27')
4GROUP BY activity_date;

ℹ

Complexity note: This complexity is due to a single pass through the activities (O(n)) and storing the results in a HashMap (O(n)) for distinct user counts.

1Using a HashMap can significantly reduce the time complexity of counting distinct elements.
2Understanding the problem requirements helps in choosing the right data structure.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.