How do you solve Count the Number of Special Characters I on LeetCode?

Count the Number of Special Characters I (LeetCode #3120) 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: Characters can be represented using their ASCII values for efficient indexing.

What is the brute force solution for Count the Number of Special Characters I?

The brute force approach for Count the Number of Special Characters I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each character in the string to see if it has both uppercase and lowercase versions. This is straightforward but inefficient as it involves checking pairs for each character. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Special Characters I?

Count the Number of Special Characters I uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count the Number of Special Characters I?

Clarify the problem requirements before coding. Think about edge cases, such as strings with only one case. Optimize your solution after implementing a brute force version.

What are common mistakes when solving Count the Number of Special Characters I?

Not considering both cases of a character when counting. Overlooking the need to check both uppercase and lowercase versions.

#3120

Count the Number of Special Characters I

Easy

Hash TableStringHash MapArray

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 approach uses a single pass to count occurrences of each character, allowing us to determine if a character is special in a more efficient manner. This reduces the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency array of size 52 to count occurrences of each letter (26 lowercase + 26 uppercase).
2Step 2: Traverse the word and update the frequency array based on the character's case.
3Step 3: Count how many characters have both lowercase and uppercase occurrences.

solution.py8 lines

1# Full working Python code
2word = 'aaAbcBC'
3frequency = [0] * 52
4for char in word:
5    index = ord(char) - ord('A') if char.isupper() else ord(char) - ord('a') + 26
6    frequency[index] += 1
7special_count = sum(1 for i in range(26) if frequency[i] > 0 and frequency[i + 26] > 0)
8print(special_count)

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once to count characters and then check the frequency array, which has a constant size of 52.

1Characters can be represented using their ASCII values for efficient indexing.
2Using a frequency array allows for quick checks of character presence.

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