Problem Description

You need to create a hashtag for a video based on its caption string. The hashtag must be generated by following these specific steps in order:

1. Convert to camelCase with hashtag prefix: Take all words from the caption and combine them into a single camelCase string that starts with #. In camelCase format:

   • The first word should be entirely lowercase
   • All subsequent words should have their first letter capitalized and remaining letters lowercase
   • For example: "Hello World Example" becomes "#helloWorldExample"

2. Remove non-letter characters: After creating the camelCase string, remove any characters that are not English letters (a-z, A-Z), but keep the initial # symbol.

3. Limit length to 100 characters: If the resulting tag is longer than 100 characters, truncate it to exactly 100 characters.

The input is a string caption containing the video's caption text, and you should return the final processed hashtag string.

For example, if the caption is "my favorite video", the process would be:

• Split into words: ["my", "favorite", "video"]
• Convert to camelCase: "myFavoriteVideo"
• Add hashtag: "#myFavoriteVideo"
• Remove non-letters (none in this case): "#myFavoriteVideo"
• Check length (under 100): "#myFavoriteVideo"
• Final result: "#myFavoriteVideo"

Intuition

The problem is essentially asking us to transform a caption into a hashtag following specific formatting rules. Let's think about how to approach this step by step.

First, we need to understand what camelCase means - it's a naming convention where we join words together without spaces, with each word after the first starting with a capital letter. Since we're dealing with a caption that contains multiple words separated by spaces, we naturally need to split the caption by spaces to get individual words.

For the camelCase conversion, we have two different rules:

• The first word should be all lowercase
• All other words should have only their first letter capitalized

Python's capitalize() method is perfect for the second requirement - it makes the first letter uppercase and all other letters lowercase. For the first word, we can simply convert it to lowercase after capitalizing.

The key insight in the given solution is that it cleverly handles the character filtering requirement. Notice that the problem asks us to remove all non-letter characters except the first #. However, if we look at our input - it's just a caption with words. After splitting by spaces and applying capitalize() or lower(), we're only dealing with letters anyway. There won't be any special characters or numbers in our processed words.

For the length constraint, we need to ensure the final result doesn't exceed 100 characters. Since we're adding a # at the beginning (1 character), we have 99 characters left for the actual camelCase string. The solution uses Python's string slicing [:99] to truncate the joined words to 99 characters, then prepends the #.

The solution is elegant because it recognizes that the "remove non-letter characters" step is implicitly handled by the way we process the words, making the implementation much simpler than explicitly filtering characters.

Solution Approach

The solution follows a simulation approach where we directly implement the required transformations step by step.

Step 1: Split the caption into words

words = [s.capitalize() for s in caption.split()]

We use caption.split() to break the caption string into individual words based on whitespace. Simultaneously, we apply capitalize() to each word using a list comprehension. The capitalize() method converts the first character to uppercase and all remaining characters to lowercase, which is exactly what we need for all words except the first one.

Step 2: Handle the first word special case

if words:
    words[0] = words[0].lower()

Since camelCase requires the first word to be entirely lowercase, we check if the words list is not empty and then convert the first word to all lowercase. This overwrites the capitalized version we created in step 1.

Step 3: Combine and format the result

return "#" + "".join(words)[:99]

This line performs multiple operations:

• "".join(words) concatenates all words without any separator, creating the camelCase string
• [:99] truncates the joined string to a maximum of 99 characters (leaving room for the #)
• "#" + ... prepends the hashtag symbol to create the final tag

The beauty of this solution is that it doesn't explicitly handle the "remove non-letter characters" requirement because:

1. After splitting by whitespace, we only have word strings
2. The capitalize() and lower() methods preserve only letters
3. Any punctuation or special characters that might be attached to words are implicitly handled by these string methods

The time complexity is O(n) where n is the length of the caption string, as we need to process each character once. The space complexity is also O(n) for storing the words list and the final result string.

Example Walkthrough

Let's walk through the solution with the caption: "hello WORLD from python"

Step 1: Split and capitalize each word

words = [s.capitalize() for s in caption.split()]

• caption.split() gives us: ["hello", "WORLD", "from", "python"]
• Applying capitalize() to each word:
  • "hello".capitalize() → "Hello"
  • "WORLD".capitalize() → "World"
  • "from".capitalize() → "From"
  • "python".capitalize() → "Python"
• Result: words = ["Hello", "World", "From", "Python"]

Step 2: Convert first word to lowercase

if words:
    words[0] = words[0].lower()

• words[0] is "Hello"
• "Hello".lower() → "hello"
• Result: words = ["hello", "World", "From", "Python"]

Step 3: Join words and add hashtag

return "#" + "".join(words)[:99]

• "".join(words) → "helloWorldFromPython"
• This string has 20 characters, well under 99, so [:99] doesn't truncate
• "#" + "helloWorldFromPython" → "#helloWorldFromPython"

Final output: "#helloWorldFromPython"

Note how the capitalize() method automatically handled the mixed case in "WORLD" by converting it to "World" (first letter uppercase, rest lowercase), which is exactly what camelCase requires. The solution elegantly avoids explicit character filtering since the word processing methods naturally produce only letter characters.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the caption string. This is because:

• caption.split() takes O(n) time to scan through the entire string
• The list comprehension with capitalize() iterates through each word, and in the worst case (when the caption is a single word), it processes O(n) characters
• "".join(words) takes O(n) time to concatenate all words
• The slicing operation [:99] takes at most O(min(n, 99)) = O(n) time

The space complexity is O(n), where n is the length of the caption string. This is because:

• caption.split() creates a list of words that collectively contain all characters from the original string, using O(n) space
• The list comprehension creates a new list with capitalized words, also using O(n) space
• "".join(words) creates a new string of size O(n)
• The final result string with the hashtag uses at most O(min(n, 100)) = O(n) space

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall: Incorrect Handling of Non-Letter Characters

The provided solution has a critical flaw: it doesn't properly remove non-letter characters from the words as required by step 2 of the problem specification. The capitalize() and lower() methods preserve all characters, including numbers, punctuation, and special characters.

Example of the issue:

• Input: "Hello-World 123 Test!"
• Current output: "#hello-world123Test!"
• Expected output: "#helloWorldTest"

The current implementation would keep hyphens, numbers, and punctuation marks in the final hashtag, violating the requirement that only English letters (a-z, A-Z) and the initial # should remain.

Solution to Fix the Pitfall:

class Solution:
    def generateTag(self, caption: str) -> str:
        # Split the caption into words
        words = caption.split()
      
        # Process each word: keep only letters and apply proper casing
        processed_words = []
        for i, word in enumerate(words):
            # Extract only letter characters from the word
            letters_only = ''.join(c for c in word if c.isalpha())
          
            if letters_only:  # Only add non-empty words
                if i == 0 or not processed_words:  # First valid word
                    processed_words.append(letters_only.lower())
                else:  # Subsequent words
                    processed_words.append(letters_only.capitalize())
      
        # Join all words together, prepend with '#', and limit to 100 characters
        hashtag = "#" + "".join(processed_words)
        return hashtag[:100]

Alternative cleaner solution:

class Solution:
    def generateTag(self, caption: str) -> str:
        words = caption.split()
        camel_case_parts = []
      
        for word in words:
            # Filter out non-letter characters
            clean_word = ''.join(filter(str.isalpha, word))
            if clean_word:
                camel_case_parts.append(clean_word)
      
        # Convert to camelCase
        if camel_case_parts:
            camel_case_parts[0] = camel_case_parts[0].lower()
            for i in range(1, len(camel_case_parts)):
                camel_case_parts[i] = camel_case_parts[i].capitalize()
      
        # Create hashtag and limit length
        return ("#" + "".join(camel_case_parts))[:100]

This solution correctly filters out all non-letter characters while maintaining the proper camelCase formatting and length constraints.