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

Count the Number of Special Characters II (LeetCode #3121) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Special characters must have both lowercase and uppercase forms present.

What is the time complexity of Count the Number of Special Characters II?

The optimal solution for Count the Number of Special Characters II runs in O(n) time and uses O(n) space. This complexity is linear because we only make a single pass through the string and use additional space for sets and maps to track occurrences.

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

The brute force approach for Count the Number of Special Characters II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every character in the string and verifies if it meets the criteria of being special. This is straightforward but inefficient, as it requires multiple passes over the string. 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 II?

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

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

Always clarify the problem requirements and constraints before jumping into coding. Consider edge cases, such as strings with only uppercase or only lowercase letters. Discuss your thought process and approach clearly with the interviewer.

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

Not checking the order of occurrences properly. Using nested loops unnecessarily, leading to time inefficiency.

#3121

Count the Number of Special Characters II

Medium

Hash TableStringHash MapSet

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a single pass through the string to track the first occurrences of uppercase and lowercase letters. This reduces the need for nested loops and searches, making it efficient.

⚙️

Algorithm

6 steps

1Step 1: Create two sets to track lowercase and uppercase characters.
2Step 2: Create a map to track the first occurrence of each character's uppercase and lowercase versions.
3Step 3: Iterate through the string and populate the sets and map.
4Step 4: For each character, check if it is in both sets and if the lowercase occurs before the uppercase in the map.
5Step 5: Count valid special characters based on the conditions.
6Step 6: Return the count of special characters.

solution.py25 lines

1# Full working Python code
2
3def countSpecialCharacters(word):
4    lower_seen = set()
5    upper_seen = set()
6    position = {}
7    special_count = 0
8
9    for i, c in enumerate(word):
10        if c.islower():
11            lower_seen.add(c)
12            position[c] = (i, 'lower')
13        else:
14            upper_seen.add(c)
15            position[c.lower()] = (i, 'upper')
16
17    for c in lower_seen:
18        if c.upper() in upper_seen:
19            if position[c][0] < position[c.upper()][0]:
20                special_count += 1
21
22    return special_count
23
24# Example usage
25print(countSpecialCharacters('aaAbcBC'))  # Output: 3

ℹ

Complexity note: This complexity is linear because we only make a single pass through the string and use additional space for sets and maps to track occurrences.

1Special characters must have both lowercase and uppercase forms present.
2The order of occurrences is crucial: lowercase must appear before uppercase.

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