How do you solve Daily Leads and Partners on LeetCode?

Daily Leads and Partners (LeetCode #1693) 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 GROUP BY can significantly reduce the complexity of aggregation tasks.

What is the brute force solution for Daily Leads and Partners?

The brute force approach for Daily Leads and Partners is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves iterating through the entire dataset for each unique combination of date_id and make_name to count distinct lead_id and partner_id. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Daily Leads and Partners?

Daily Leads and Partners uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation, Group By.

What are the interview tips for Daily Leads and Partners?

Always clarify the requirements before diving into coding. Think about the efficiency of your approach and discuss it with the interviewer. Be prepared to explain your SQL queries step by step.

What are common mistakes when solving Daily Leads and Partners?

Not using DISTINCT when counting leads and partners. Forgetting to group by both date_id and make_name.

#1693

Daily Leads and Partners

Easy

DatabaseHash MapAggregationGroup By

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)

The optimal solution leverages SQL's grouping and counting capabilities to efficiently aggregate the data in one pass, reducing the need for nested loops.

⚙️

Algorithm

3 steps

1Step 1: Use the SQL GROUP BY clause to group the data by date_id and make_name.
2Step 2: Use COUNT(DISTINCT lead_id) to count unique leads and COUNT(DISTINCT partner_id) for unique partners.
3Step 3: Select the grouped results directly.

solution.py3 lines

1SELECT date_id, make_name, COUNT(DISTINCT lead_id) AS unique_leads, COUNT(DISTINCT partner_id) AS unique_partners
2FROM DailySales
3GROUP BY date_id, make_name;

ℹ

Complexity note: The complexity is O(n) because we only need to scan the table once to group and count distinct values, making it much more efficient than the brute force approach.

1Using GROUP BY can significantly reduce the complexity of aggregation tasks.
2Counting distinct values is a common requirement in data analysis.

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