How do you solve Fix Names in a Table on LeetCode?

Fix Names in a Table (LeetCode #1667) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using SQL string functions can optimize performance.

What is the brute force solution for Fix Names in a Table?

The brute force approach for Fix Names in a Table is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will iterate through each name in the Users table and manually adjust the case of each character. This is straightforward but inefficient, as it involves multiple passes over the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fix Names in a Table?

Fix Names in a Table uses the following concepts: Database. The recommended patterns to study are: String Manipulation, SQL Functions.

What are the interview tips for Fix Names in a Table?

Be clear about the SQL functions available for string manipulation. Practice writing SQL queries to manipulate data effectively. Always consider performance implications of your approach.

What are common mistakes when solving Fix Names in a Table?

Not using the correct SQL functions for string manipulation. Overcomplicating the problem with unnecessary loops.

#1667

Fix Names in a Table

Easy

DatabaseString ManipulationSQL Functions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Using SQL string functions efficiently allows us to transform the names in a single pass. This is much faster than the brute force method as we leverage built-in functions.

⚙️

Algorithm

3 steps

1Step 1: Use the SQL function CONCAT to combine the uppercase first character with the lowercase rest of the name.
2Step 2: Use the SUBSTRING function to extract the first character and the rest of the name.
3Step 3: Return the modified names ordered by user_id.

solution.py1 lines

1SELECT user_id, CONCAT(UPPER(SUBSTRING(name, 1, 1)), LOWER(SUBSTRING(name, 2))) AS name FROM Users ORDER BY user_id;

ℹ

Complexity note: The time complexity is O(n) because we are processing each name in a single pass using SQL string functions, which are optimized for performance.

1Using SQL string functions can optimize performance.
2Understanding string manipulation is crucial for data formatting.

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