# 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

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, $|\Sigma| \leq 128$.

• 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;
}
}

• 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;
}
};

• 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))


• 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
}

• 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;
}