Table: Users
+----------------+---------+ | Column Name | Type | +----------------+---------+ | user_id | int | | name | varchar | +----------------+---------+ user_id is the primary key (column with unique values) for this table. This table contains the ID and the name of the user. The name consists of only lowercase and uppercase characters.
Write a solution to fix the names so that only the first character is uppercase and the rest are lowercase.
Return the result table ordered by user_id.
The result format is in the following example.
Example 1:
Input: Users table: +---------+-------+ | user_id | name | +---------+-------+ | 1 | aLice | | 2 | bOB | +---------+-------+ Output: +---------+-------+ | user_id | name | +---------+-------+ | 1 | Alice | | 2 | Bob | +---------+-------+
This approach involves iterating through each name in the list, transforming the name so that the first letter is uppercase and the rest are lowercase. This is a common string manipulation problem that can be solved using built-in string functions in most programming languages.
In C, we use the toupper and tolower functions from ctype.h to modify the casing of each character in the string. The first character is capitalized, while the rest are converted to lowercase.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n) for each name, where n is the length of the name.
Space Complexity: O(1) since we modify the names in place.
This approach applies regular expressions to locate patterns that match the criteria and replace them with the desired format. This leverages the pattern matching capability of regex to achieve the transformation in fewer lines of code.
In this Python solution, the regex (\w)(\w*) captures the first character and the remainder of the string separately. The lambda function then applies the uppercase and lowercase transformations in one substitution step.
JavaScript
Time Complexity: O(n) for each name due to regex matching and replacement, where n is the length of the name.
Space Complexity: O(n) for storing the modified strings.
| Approach | Complexity |
|---|---|
| Approach 1: Use Loop to Transform Each Name | Time Complexity: O(n) for each name, where n is the length of the name. |
| Approach 2: Use Regex Replacement Method | Time Complexity: O(n) for each name due to regex matching and replacement, where n is the length of the name. |
LeetCode 1667 Interview SQL Question with Detailed Explanation | Practice SQL • Everyday Data Science • 4,317 views views
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