How do you solve Rank Scores on LeetCode?

Rank Scores (LeetCode #178) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using ranking functions like DENSE_RANK() simplifies the problem significantly.

What is the brute force solution for Rank Scores?

The brute force approach for Rank Scores is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we calculate the rank for each score by comparing it with every other score in the list. This method is straightforward but inefficient, as it involves a lot of repeated comparisons. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Rank Scores?

Rank Scores uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Rank Scores?

Familiarize yourself with SQL ranking functions before the interview. Practice writing SQL queries that involve grouping and ordering. Think about performance and efficiency when writing queries.

What are common mistakes when solving Rank Scores?

Not considering how to handle ties correctly. Overcomplicating the ranking logic instead of using built-in functions.

#178

Rank Scores

Medium

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

This approach uses SQL's ranking functions to efficiently assign ranks based on scores without needing to compare each score individually. It leverages built-in functions to handle ties and ranking seamlessly.

⚙️

Algorithm

3 steps

1Step 1: Use the DENSE_RANK() function to assign ranks based on scores.
2Step 2: Order the scores in descending order to ensure higher scores get lower rank numbers.
3Step 3: Select the score and its corresponding rank from the result.

solution.py2 lines

1SELECT score, DENSE_RANK() OVER (ORDER BY score DESC) AS rank
2FROM Scores;

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step required for ranking, while space complexity is O(n) for storing the ranks.

1Using ranking functions like DENSE_RANK() simplifies the problem significantly.
2Understanding how to handle ties in ranking is crucial.

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