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"

Goat Latin Conversion

This problem requires converting a sentence into Goat Latin according to the specified rules. We need to handle words starting with vowels and consonants differently and append 'ma' and a series of 'a's based on the word index.

Naive Solution

A straightforward approach is to split the sentence into words, iterate through each word, and apply the Goat Latin rules directly. This involves checking if the word starts with a vowel, performing the necessary transformations, and appending the correct number of 'a's.

def to_goat_latin_naive(sentence):
    vowels = 'aeiouAEIOU'
    words = sentence.split()
    result = []
    for i, word in enumerate(words):
        if word[0] in vowels:
            new_word = word + 'ma'
        else:
            new_word = word[1:] + word[0] + 'ma'
        new_word += 'a' * (i + 1)
        result.append(new_word)
    return ' '.join(result)

# Example Usage
sentence = "I speak Goat Latin"
print(to_goat_latin_naive(sentence))
# Output: Imaa peaksmaaa oatGmaaaa atinLmaaaaa

sentence = "The quick brown fox jumped over the lazy dog"
print(to_goat_latin_naive(sentence))
# Output: heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa

Optimal Solution

The optimal solution is essentially the same as the naive solution because the problem constraints don't allow for significant optimizations. However, we can refactor the code to be more readable and slightly more efficient.

def to_goat_latin_optimal(sentence):
    vowels = set('aeiouAEIOU')
    words = sentence.split()
    goat_latin_words = []

    for index, word in enumerate(words, 1):
        first_char = word[0]
        if first_char in vowels:
            goat_word = word + 'ma' + 'a' * index
        else:
            goat_word = word[1:] + first_char + 'ma' + 'a' * index
        goat_latin_words.append(goat_word)

    return ' '.join(goat_latin_words)

# Example Usage
sentence = "I speak Goat Latin"
print(to_goat_latin_optimal(sentence))
# Output: Imaa peaksmaaa oatGmaaaa atinLmaaaaa

sentence = "The quick brown fox jumped over the lazy dog"
print(to_goat_latin_optimal(sentence))
# Output: heTmaa uickqmaaa rownbmaaaa oxfmaaaaa umpedjmaaaaaa overmaaaaaaa hetmaaaaaaaa azylmaaaaaaaaa ogdmaaaaaaaaaa

Big(O) Run-time Analysis

The time complexity of both the naive and optimal solutions is O(N*M), where N is the number of words in the sentence, and M is the average length of a word. This is because we iterate through each word, and for each word, we perform string operations (checking if the first letter is a vowel, appending characters) that take O(M) time.

Big(O) Space Usage Analysis

The space complexity is O(N*M), where N is the number of words in the sentence and M is the average length of a word. This is because we store the modified words in a list, which can grow proportionally to the size of the input sentence.

Edge Cases

1. Empty Sentence: If the input sentence is empty, the function should return an empty string. Handled by default.
2. Sentence with only spaces: If the sentence consists only of spaces, the function should return an empty string. Handled by split() which will return an empty list.
3. Words with mixed-case letters: The code should handle words with both lowercase and uppercase letters correctly. Addressed because the vowel check includes uppercase and lowercase vowels.
4. Very long sentence: For very long sentences, the space complexity might become a concern. However, given the problem constraints (sentence length <= 150), this is not a significant issue.