Goat Latin

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Strings

You are given a string sentence that consist of words separated by spaces. Each word consists of lowercase and uppercase letters only.

We would like to convert the sentence to "Goat Latin" (a made-up language similar to Pig Latin.) The rules of Goat Latin are as follows:

If a word begins with a vowel ('a', 'e', 'i', 'o', or 'u'), append "ma" to the end of the word.
- For example, the word "apple" becomes "applema".
If a word begins with a consonant (i.e., not a vowel), remove the first letter and append it to the end, then add "ma".
- For example, the word "goat" becomes "oatgma".
Add one letter 'a' to the end of each word per its word index in the sentence, starting with 1.
- For example, the first word gets "a" added to the end, the second word gets "aa" added to the end, and so on.

Return the final sentence representing the conversion from sentence to Goat Latin.

Example 1:

Input: sentence = "I speak Goat Latin"
Output: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"

Example 2:

Input: sentence = "The quick brown fox jumped over the lazy dog"
Output: "heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa"

Constraints:

1 <= sentence.length <= 150
sentence consists of English letters and spaces.
sentence has no leading or trailing spaces.
All the words in sentence are separated by a single space.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What characters are considered vowels for the purpose of this transformation? Specifically, are uppercase vowels included?
Can the input string be empty or null? What should I return in those cases?
Are there any punctuation marks or special characters in the input string besides letters and spaces?
Is the input string guaranteed to have valid words separated by single spaces, or could there be multiple spaces between words?
What is the expected behavior if a word starts with a character that is neither a vowel nor a consonant (e.g., a number or special character)? Should I still append 'ma' and the 'a' string?

Brute Force Solution

Approach

The brute force method for Goat Latin directly translates the instructions. We examine each word individually and transform it according to the given rules of moving letters and adding suffixes. Then, we combine the modified words back into a complete sentence.

Here's how the algorithm would work step-by-step:

Take the first word of the sentence.
Check if the first letter of this word is a vowel (a, e, i, o, u).
If it is a vowel, add 'ma' to the end of the word.
If it is not a vowel, move the first letter to the end of the word, and then add 'ma'.
After doing the above, add a suffix 'a' to the end of the word. The number of 'a's added depends on the word's position in the sentence.
Repeat the above process for each word in the sentence.
Join all the modified words back together with spaces in between to create the final "Goat Latin" sentence.

Code Implementation

def goat_latin(sentence):
    words = sentence.split()
    goat_latin_words = []

    for word_index, word in enumerate(words):
        # Check if the word starts with a vowel.
        if word[0].lower() in 'aeiou':
            modified_word = word + 'ma'

        else:
            # Move the first letter to the end for non-vowels.
            modified_word = word[1:] + word[0] + 'ma'

        # Append 'a's based on the word's index.
        modified_word += 'a' * (word_index + 1)
        goat_latin_words.append(modified_word)

    return ' '.join(goat_latin_words)

Big(O) Analysis

Time Complexity

O(n^2) – The primary operation is processing each word in the input sentence. The number of words in the sentence defines 'n'. For each word, we may need to move the first letter to the end, which takes O(k) time where k is the length of the word. We also append 'ma' and a variable number of 'a's based on the word's position. In the worst case, we iterate 'n' times and for each iteration we can also have up to 'n' operations related to the string modifications to append the suffix of a's. Therefore, the complexity is O(n*k + n^2). Assuming the maximum length of a word is bounded by a constant, say M, then O(n*M + n^2). If n > M, the overall time complexity is O(n^2).

Space Complexity

O(N) – The brute force Goat Latin algorithm described creates a modified word for each word in the input sentence. Since each word's modified version can be as long as the original word plus some constant suffix (ma, and a's), each requires its own memory allocation. Therefore, the memory used to store the transformed words grows linearly with the number of words (N) in the input sentence. Thus, auxiliary space is O(N), where N is the length of the input string.

Optimal Solution

Approach

The goal is to transform each word in a sentence according to specific rules. We handle vowels and consonants differently, adding 'ma' to the end of each word, and then adding an increasing number of 'a's to the end of each word based on its position in the sentence.

Here's how the algorithm would work step-by-step:

Split the input sentence into individual words.
For each word, check if it starts with a vowel (a, e, i, o, u, case-insensitive).
If a word starts with a vowel, add 'ma' to the end of the word.
If a word starts with a consonant, move the first letter to the end, and then add 'ma'.
Add a string of 'a's to the end of each word, where the number of 'a's is equal to the word's position in the sentence (starting from 1). For example, the first word gets one 'a', the second word gets two 'a's, and so on.
Join the transformed words back together with spaces to form the final Goat Latin sentence.

Code Implementation

def goat_latin(sentence):
    words = sentence.split()
    goat_latin_words = []

    for index, word in enumerate(words):
        # Handle vowels and consonants differently
        if word[0].lower() in 'aeiou':
            transformed_word = word + 'ma'

        else:
            transformed_word = word[1:] + word[0] + 'ma'

        # Add the appropriate number of 'a's
        transformed_word += 'a' * (index + 1)
        goat_latin_words.append(transformed_word)

    # Join the words back into a sentence
    return ' '.join(goat_latin_words)

Big(O) Analysis

Time Complexity

O(n*m) – Let n be the number of words in the input sentence and m be the maximum length of a word. Splitting the string into words takes O(n) time, which is dominated by later operations. Iterating through each of the n words involves checking if it starts with a vowel, which takes constant time O(1). Moving the first character to the end of a word takes O(m) time. Adding 'ma' takes constant time. Adding 'a's requires appending to a string, and in the worst case, this takes O(m) time as the number of 'a's grows with the word index. Therefore the dominant cost is from string manipulations inside the loop that iterates through all n words, where we perform O(m) operations per word, leading to a final time complexity of O(n*m).

Space Complexity

O(N) – The space complexity is primarily determined by the space needed to store the individual words after splitting the input sentence. In the worst-case scenario, the split sentence creates a list containing N words where N is the number of words in the input sentence. Additional variables such as a counter or temporary string for individual words contribute only constant space. Therefore, the dominant factor is the list of split words, leading to O(N) space complexity.

Edge Cases

Case	How to Handle
Empty string input	Return an empty string immediately as there are no words to translate.
String with only spaces	The string should be split by spaces; if the result is an empty array, return an empty string.
String with leading/trailing spaces	Trim the input string to remove leading and trailing spaces before processing.
String with multiple spaces between words	The splitting mechanism should handle multiple spaces correctly, treating them as a single delimiter.
Words starting with upper and lowercase vowels	The solution should correctly identify vowels regardless of their case and apply the appropriate Goat Latin transformation.
Very long string with many words	Ensure the algorithm has reasonable time complexity, avoiding quadratic or exponential behavior, and consider memory usage.
Words containing punctuation	The problem statement does not explicitly define treatment of punctuation, so assume they are to be included as part of the word.
String with non-ASCII characters	Confirm whether non-ASCII characters are supported or should be handled specially (e.g., ignored, replaced, or throw an error).

Empty string input

How to Handle:

Return an empty string immediately as there are no words to translate.

String with only spaces

How to Handle:

The string should be split by spaces; if the result is an empty array, return an empty string.

String with leading/trailing spaces

How to Handle:

Trim the input string to remove leading and trailing spaces before processing.

String with multiple spaces between words

How to Handle:

The splitting mechanism should handle multiple spaces correctly, treating them as a single delimiter.

Words starting with upper and lowercase vowels

How to Handle:

The solution should correctly identify vowels regardless of their case and apply the appropriate Goat Latin transformation.

Very long string with many words

How to Handle:

Ensure the algorithm has reasonable time complexity, avoiding quadratic or exponential behavior, and consider memory usage.

Words containing punctuation

How to Handle:

The problem statement does not explicitly define treatment of punctuation, so assume they are to be included as part of the word.

String with non-ASCII characters

How to Handle:

Confirm whether non-ASCII characters are supported or should be handled specially (e.g., ignored, replaced, or throw an error).