Count the Number of Special Characters I - Problem

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.

Input & Output

Example 1 — Mixed Case Letters

$ Input: word = "aaAbcBC"

› Output: 3

💡 Note: Letters 'a', 'b', and 'c' each appear in both lowercase and uppercase forms

Example 2 — Only Lowercase

$ Input: word = "abc"

› Output: 0

💡 Note: All letters are lowercase only, no uppercase versions present

Example 3 — Partial Match

$ Input: word = "abBcab"

› Output: 1

💡 Note: Only letter 'b' appears in both cases (lowercase 'b' and uppercase 'B')

Constraints

1 ≤ word.length ≤ 1000
word consists of only lowercase and uppercase English letters.

Visualization

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Asked in

G Google 15 M Microsoft 12

The key insight is to track which letters appear in both lowercase and uppercase forms. Best approach uses two sets in a single pass to categorize characters, then counts the intersection. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check Each Letter Twice

⏱️ Time: O(n²) Space: O(1)

For each unique letter in the string, scan the entire string to check if both lowercase and uppercase versions exist. Count letters that have both forms.

Single Pass with Character Tracking

⏱️ Time: O(n) Space: O(1)

Use two sets to track which letters we've seen in lowercase and uppercase. In a single pass through the string, record each character in the appropriate set, then count letters present in both sets.

Brute Force - Check Each Letter Twice — Algorithm Steps

Find all unique letters in the string
For each letter, search string for lowercase version
For same letter, search string for uppercase version
Count letters found in both cases

Visualization

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Step-by-Step Walkthrough

Find unique letters

Identify each distinct letter

Search both cases

For each letter, search for lowercase and uppercase

Count matches

Count letters that exist in both forms

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>

int solution(char* word) {
    int specialCount = 0;
    bool checked[26] = {false};
    int len = strlen(word);
    
    for (int i = 0; i < len; i++) {
        char letter = tolower(word[i]);
        int index = letter - 'a';
        
        if (!checked[index]) {
            bool hasLower = false, hasUpper = false;
            
            for (int j = 0; j < len; j++) {
                if (word[j] == letter) hasLower = true;
                if (word[j] == toupper(letter)) hasUpper = true;
            }
            
            if (hasLower && hasUpper) {
                specialCount++;
            }
            
            checked[index] = true;
        }
    }
    
    return specialCount;
}

int main() {
    char word[1001];
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    
    int result = solution(word);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each unique letter, we scan the entire string twice

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables, no extra data structures

✓ Linear Space

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Count the Number of Special Characters I - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Letter Twice — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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