What is the brute force solution for Actors and Directors Who Cooperated At Least Three Times?

The brute force approach for Actors and Directors Who Cooperated At Least Three Times is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible pair of actor and director to count their collaborations. This is straightforward but inefficient for larger datasets. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Actors and Directors Who Cooperated At Least Three Times?

Actors and Directors Who Cooperated At Least Three Times uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation, Group By.

What are the interview tips for Actors and Directors Who Cooperated At Least Three Times?

Always think about how to reduce the number of operations — aggregation is your friend. Be clear about the difference between WHERE and HAVING clauses in SQL. Practice writing SQL queries to become familiar with syntax and common functions.

What are common mistakes when solving Actors and Directors Who Cooperated At Least Three Times?

Not using GROUP BY correctly, which can lead to incorrect counts. Overlooking the HAVING clause, which is essential for filtering aggregated results.

#1050

Actors and Directors Who Cooperated At Least Three Times

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 uses aggregation to count collaborations efficiently. By grouping the data and filtering based on the count, we minimize the number of operations needed.

⚙️

Algorithm

4 steps

1Step 1: Use a SQL query to group the records by actor_id and director_id.
2Step 2: Count the number of collaborations for each pair.
3Step 3: Filter the results to include only those pairs with a count of 3 or more.
4Step 4: Return the filtered results.

solution.py5 lines

1# Full working Python code
2SELECT actor_id, director_id
3FROM ActorDirector
4GROUP BY actor_id, director_id
5HAVING COUNT(*) >= 3;

ℹ

Complexity note: The time complexity is O(n) because we are processing each record once to group and count. The space complexity is O(n) due to the storage of intermediate results during grouping.

1Using aggregation functions like COUNT can significantly reduce the complexity of the problem.
2Understanding how to group data effectively is crucial in database queries.

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