How do you solve Detect Capital on LeetCode?

Detect Capital (LeetCode #520) 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: Understanding the conditions for capital usage is crucial.

What is the brute force solution for Detect Capital?

The brute force approach for Detect Capital is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks each character in the string to determine if it meets the capital usage rules. This involves counting the number of uppercase letters and comparing it against the total length of the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Detect Capital?

Detect Capital uses the following concepts: String. The recommended patterns to study are: String Manipulation, Character Counting.

What are the interview tips for Detect Capital?

Clearly explain your thought process and approach. Consider edge cases, such as single-character words. Optimize your solution after the brute-force implementation.

What are common mistakes when solving Detect Capital?

Not considering the case where the first letter is lowercase. Overcomplicating the checks with unnecessary loops.

#520

Detect Capital

Easy

StringString ManipulationCharacter Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution directly checks the conditions for valid capital usage without counting characters multiple times. This approach is more efficient as it only requires a single pass through the string.

⚙️

Algorithm

2 steps

1Step 1: If the first character is lowercase, check if all characters are lowercase.
2Step 2: If the first character is uppercase, check if all characters are uppercase or if only the first character is uppercase.

solution.py4 lines

1def detectCapitalUse(word):
2    if word[0].islower():
3        return word.islower()
4    return word.isupper() or (word[1:].islower())

ℹ

Complexity note: The time complexity is O(n) because we only pass through the string once to check the conditions. The space complexity is O(1) since we use a constant amount of space.

1Understanding the conditions for capital usage is crucial.
2Single-pass checks are more efficient than counting characters.

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