Rank Scores — LeetCode #178 (Medium)
Tags: Database
Related patterns: Hash Map, Array
Brute Force approach
Time complexity: O(n²). Space complexity: O(1).
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 time complexity is O(n²) because for each score, we compare it with every other score, leading to a quadratic number of comparisons.
- Step 1: For each score, initialize a rank variable to 1.
- Step 2: Compare the current score with every other score in the list. If another score is higher, increment the rank.
- Step 3: Store the score and its rank in the result table.
- Step 4: Repeat for all scores.
- Step 5: Return the result table ordered by score in descending order.
1. Score = 4.00, Rank = 1 (compared with 3.85, 3.65, 3.50)
2. Score = 4.00, Rank = 1 (same as above)
3. Score = 3.85, Rank = 2 (compared with 4.00, 3.65, 3.50)
4. Score = 3.65, Rank = 3 (compared with 4.00, 3.85, 3.50)
5. Score = 3.65, Rank = 3 (same as above)
6. Score = 3.50, Rank = 4 (compared with 4.00, 3.85, 3.65)
Optimal Solution approach
Time complexity: O(n log n). Space complexity: 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.
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.
- Step 1: Use the DENSE_RANK() function to assign ranks based on scores.
- Step 2: Order the scores in descending order to ensure higher scores get lower rank numbers.
- Step 3: Select the score and its corresponding rank from the result.
1. Scores: [4.00, 4.00, 3.85, 3.65, 3.65, 3.50]
2. DENSE_RANK() assigns ranks: [1, 1, 2, 3, 3, 4]
3. Final output: [(4.00, 1), (4.00, 1), (3.85, 2), (3.65, 3), (3.65, 3), (3.50, 4)]
Key Insights
- Using ranking functions like DENSE_RANK() simplifies the problem significantly.
- Understanding how to handle ties in ranking is crucial.
Common Mistakes
- Not considering how to handle ties correctly.
- Overcomplicating the ranking logic instead of using built-in functions.
Interview Tips
- 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.
