LeetCode 3121: Count the Number of Special Characters II Solution

Master LeetCode problem 3121 (Count the Number of Special Characters II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3121. Count the Number of Special Characters II

Medium

Hash Table String

Problem Explanation

Explanation:

To solve this problem, we can iterate through the input string and keep track of the lowercase and uppercase characters we have encountered so far. We can use a set to store the lowercase characters and another set to store the uppercase characters. Whenever we encounter a character that is special according to the given condition, we increment our count of special characters.

We iterate through the input string.
For each character:
- If it is lowercase, we add it to the set of lowercase characters.
- If it is uppercase, we check if its corresponding lowercase character is in the set of lowercase characters. If so, we increment the count of special characters and remove the lowercase character from the set.
Finally, we return the count of special characters.

Time Complexity:

The time complexity of this solution is O(n), where n is the length of the input string.

Space Complexity:

The space complexity of this solution is O(1) since the sets used to store lowercase and uppercase characters will have a constant number of elements (26 characters).

Solution Code

class Solution {
    public int countSpecialCharacters(String word) {
        Set<Character> lower = new HashSet<>();
        Set<Character> upper = new HashSet<>();
        int count = 0;
        
        for (char c : word.toCharArray()) {
            if (Character.isLowerCase(c)) {
                lower.add(c);
            } else {
                if (lower.contains(Character.toLowerCase(c))) {
                    count++;
                    lower.remove(Character.toLowerCase(c));
                }
            }
        }
        
        return count;
    }
}

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