Title Case Converter - Problem

Given a sentence as a string, convert it to title case where the first letter of each word is capitalized and the rest are lowercase.

However, certain articles and prepositions should remain lowercase unless they are the first word of the sentence. These words include: 'a', 'an', 'the', 'and', 'but', 'or', 'for', 'nor', 'on', 'at', 'to', 'from', 'by', 'of', 'in'.

Rules:

The first word is always capitalized
Words not in the exception list are capitalized
Exception words remain lowercase unless they're the first word

Input & Output

Example 1 — Basic Case

$ Input: sentence = "the lord of the rings"

› Output: "The Lord of the Rings"

💡 Note: First word 'the' becomes 'The', 'lord' becomes 'Lord', 'of' stays lowercase (exception), second 'the' stays lowercase (exception), 'rings' becomes 'Rings'

Example 2 — All Exceptions

$ Input: sentence = "a tale of two cities"

› Output: "A Tale of Two Cities"

💡 Note: First word 'a' becomes 'A', 'tale' becomes 'Tale', 'of' stays lowercase (exception), 'two' becomes 'Two', 'cities' becomes 'Cities'

Example 3 — No Exceptions

$ Input: sentence = "harry potter"

› Output: "Harry Potter"

💡 Note: No exception words present, so both 'harry' and 'potter' get capitalized to 'Harry Potter'

Constraints

1 ≤ sentence.length ≤ 1000
sentence consists of lowercase/uppercase English letters and spaces
There is at least one word in the sentence

Visualization

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Asked in

G Google 25 M Microsoft 18 🍎 Apple 15 a Amazon 12

The key insight is using a hash set to store exception words for O(1) lookup instead of checking against each exception word individually. The optimal approach processes each word once, checking if it's an exception word using constant-time hash set lookup. Time: O(n), Space: O(n + k) where k=15 exception words.

Common Approaches

✓ Brute Force - Check Every Exception Word

⏱️ Time: O(n × m) Space: O(n)

Split the sentence into words, then for each word check if it matches any of the exception words by comparing with each one individually. Apply capitalization rules based on position and exception status.

Hash Set Optimization

⏱️ Time: O(n) Space: O(n + k)

Create a hash set containing all exception words, then for each word in the sentence, check if it exists in the set using constant-time lookup. Apply capitalization rules based on position and exception status.

Brute Force - Check Every Exception Word — Algorithm Steps

Split sentence into individual words
For each word, compare against all 15 exception words one by one
If first word or not an exception, capitalize first letter and lowercase rest
If exception word and not first, keep all lowercase
Join words back into final sentence

Visualization

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Step-by-Step Walkthrough

Split Into Words

Break sentence into individual words

Check Each Exception

For each word, compare against all 15 exceptions

Apply Rules

Capitalize based on position and exception status

Code -

solution.c — C

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

int isException(const char* word) {
    const char* exceptions[] = {"a", "an", "the", "and", "but", "or", "for", "nor", "on", "at", "to", "from", "by", "of", "in"};
    int numExceptions = 15;
    
    for (int i = 0; i < numExceptions; i++) {
        if (strcmp(word, exceptions[i]) == 0) {
            return 1;
        }
    }
    return 0;
}

char* solution(char* sentence) {
    if (!sentence || strlen(sentence) == 0) {
        char* result = malloc(1);
        result[0] = '\0';
        return result;
    }
    
    int len = strlen(sentence);
    char* result = malloc(len + 1);
    strcpy(result, sentence);
    
    // Convert to lowercase
    for (int i = 0; result[i]; i++) {
        result[i] = tolower(result[i]);
    }
    
    char* word = strtok(result, " ");
    int wordIndex = 0;
    int pos = 0;
    
    while (word != NULL) {
        if (wordIndex == 0 || !isException(word)) {
            word[0] = toupper(word[0]);
        }
        wordIndex++;
        word = strtok(NULL, " ");
    }
    
    return result;
}

int main() {
    char sentence[1000];
    fgets(sentence, sizeof(sentence), stdin);
    sentence[strcspn(sentence, "\n")] = 0;
    
    char* result = solution(sentence);
    printf("%s\n", result);
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × m)

n words × m exception words - each word checks against all 15 exceptions

✓ Linear Growth

Space Complexity

O(n)

Array to store n words after splitting the sentence

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

486 Likes

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Title Case Converter - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Every Exception Word — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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