Skip to content

3120. Count the Number of Special Characters I

Description

You are given a string word. A letter is called special if it appears both in lowercase and uppercase in word.

Return the number of special letters in word.

 

Example 1:

Input: word = "aaAbcBC"

Output: 3

Explanation:

The special characters in word are 'a', 'b', and 'c'.

Example 2:

Input: word = "abc"

Output: 0

Explanation:

No character in word appears in uppercase.

Example 3:

Input: word = "abBCab"

Output: 1

Explanation:

The only special character in word is 'b'.

 

Constraints:

  • 1 <= word.length <= 50
  • word consists of only lowercase and uppercase English letters.

Solutions

Solution 1: Hash Table or Array

Thinking

A letter is special when both cases appear. Scanning the string once per letter repeats \(O(n|\Sigma|)\) work.

Membership of each character is enough, and a set built in one pass answers all \(26\) pairs.

Insert \(word\) into a set, then count letters whose lower and upper forms both occur.

We use a hash table or array \(s\) to record the characters that appear in the string \(word\). Then we traverse the 26 letters. If both the lowercase and uppercase letters appear in \(s\), the count of special characters is incremented by one.

Finally, return the count of special characters.

The time complexity is \(O(n + |\Sigma|)\), and the space complexity is \(O(|\Sigma|)\). Where \(n\) is the length of the string \(word\), and \(|\Sigma|\) is the size of the character set. In this problem, \(|\Sigma| \leq 128\).

1
2
3
4
class Solution:
    def numberOfSpecialChars(self, word: str) -> int:
        s = set(word)
        return sum(a in s and b in s for a, b in zip(ascii_lowercase, ascii_uppercase))

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution {
    public int numberOfSpecialChars(String word) {
        boolean[] s = new boolean['z' + 1];
        for (int i = 0; i < word.length(); ++i) {
            s[word.charAt(i)] = true;
        }
        int ans = 0;
        for (int i = 0; i < 26; ++i) {
            if (s['a' + i] && s['A' + i]) {
                ++ans;
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
public:
    int numberOfSpecialChars(string word) {
        vector<bool> s('z' + 1);
        for (char& c : word) {
            s[c] = true;
        }
        int ans = 0;
        for (int i = 0; i < 26; ++i) {
            ans += s['a' + i] && s['A' + i];
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
func numberOfSpecialChars(word string) (ans int) {
    s := make([]bool, 'z'+1)
    for _, c := range word {
        s[c] = true
    }
    for i := 0; i < 26; i++ {
        if s['a'+i] && s['A'+i] {
            ans++
        }
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
function numberOfSpecialChars(word: string): number {
    const s: boolean[] = Array.from({ length: 'z'.charCodeAt(0) + 1 }, () => false);
    for (let i = 0; i < word.length; ++i) {
        s[word.charCodeAt(i)] = true;
    }
    let ans: number = 0;
    for (let i = 0; i < 26; ++i) {
        if (s['a'.charCodeAt(0) + i] && s['A'.charCodeAt(0) + i]) {
            ++ans;
        }
    }
    return ans;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
impl Solution {
    pub fn number_of_special_chars(word: String) -> i32 {
        let mut s = [false; 128];
        for ch in word.chars() {
            s[ch as u8 as usize] = true;
        }
        let mut ans = 0;
        for i in 0..26 {
            if s[(b'a' + i) as usize] && s[(b'A' + i) as usize] {
                ans += 1;
            }
        }
        ans
    }
}

Comments