Problem Description

This problem asks you to convert a given sentence into "Goat Latin", which is a made-up language with specific transformation rules.

Given a string sentence containing words separated by spaces (where each word consists only of lowercase and uppercase letters), you need to transform each word according to three rules:

Rule 1: Vowel vs Consonant

• If a word starts with a vowel ('a', 'e', 'i', 'o', or 'u' - case insensitive), append "ma" to the end of the word.
  • Example: "apple" → "applema"
• If a word starts with a consonant, remove the first letter, append it to the end, then add "ma".
  • Example: "goat" → "oatgma"

Rule 2: Position-based 'a' Addition

• Add the letter 'a' to the end of each word based on its position in the sentence (1-indexed).
  • The 1st word gets one 'a' added
  • The 2nd word gets two 'a's added
  • The 3rd word gets three 'a's added, and so on

Example Walkthrough:

For the sentence "I speak Goat Latin":

1. "I" (1st word, starts with vowel) → "I" + "ma" + "a" = "Imaa"
2. "speak" (2nd word, starts with consonant 's') → "peaks" + "ma" + "aa" = "peaksmaaa"
3. "Goat" (3rd word, starts with consonant 'G') → "oatG" + "ma" + "aaa" = "oatGmaaaa"
4. "Latin" (4th word, starts with consonant 'L') → "atinL" + "ma" + "aaaa" = "atinLmaaaaa"

The final result would be: "Imaa peaksmaaa oatGmaaaa atinLmaaaaa"

The task is to return the complete transformed sentence with all words converted to Goat Latin, maintaining the space-separated format.

Intuition

The problem requires us to transform each word independently according to specific rules, then reassemble them into a sentence. This naturally suggests processing the sentence word by word.

The key insight is that we need to:

1. Split the sentence into individual words
2. Transform each word based on its starting letter and position
3. Join the transformed words back together

For each word transformation, we need to check if the first letter is a vowel. Since the problem states this check is case-insensitive (uppercase vowels like 'A', 'E' should also be treated as vowels), we can convert the first character to lowercase before checking against our vowel set.

The position-dependent suffix (adding 'a's) suggests we need to track which word we're processing. When iterating through words, we can use the index to determine how many 'a's to append - the first word (index 0) gets 1 'a', the second word (index 1) gets 2 'a's, and so on. This is simply (index + 1) repetitions of 'a'.

The transformation process for each word follows a clear sequence:

1. Check the first letter and apply the vowel/consonant rule
2. Add "ma"
3. Add the position-based 'a's

Since we're building a new sentence from transformed words, using a list to collect the transformed words and then joining them with spaces at the end is more efficient than string concatenation. This approach processes each word exactly once in a single pass through the sentence, giving us a linear time solution.

Solution Approach

The implementation follows a straightforward approach of processing each word individually:

Step 1: Split the sentence into words

sentence.split()

This breaks the input string into a list of words using spaces as delimiters.

Step 2: Process each word with enumeration

for i, word in enumerate(sentence.split()):

Using enumerate() gives us both the word and its index i (0-based), which we'll need for determining how many 'a's to append.

Step 3: Check if the word starts with a vowel

if word.lower()[0] not in ['a', 'e', 'i', 'o', 'u']:

• word.lower()[0] converts the first character to lowercase for case-insensitive comparison
• If the first letter is NOT a vowel (consonant case), we need to move it to the end

Step 4: Apply consonant transformation if needed

word = word[1:] + word[0]

For consonants, slice the word from index 1 to the end (word[1:]) and append the first character (word[0]).

Step 5: Add the required suffixes

word += 'ma'
word += 'a' * (i + 1)

• First append 'ma' to all words
• Then append 'a' repeated (i + 1) times
  • Since i is 0-based, (i + 1) gives us 1 for the first word, 2 for the second, etc.

Step 6: Collect and join the results

ans.append(word)
return ' '.join(ans)

Store each transformed word in a list ans, then join them with spaces to create the final sentence.

The algorithm processes each word exactly once, performing constant-time operations for each word (string slicing and concatenation for fixed-size operations). The time complexity is O(n) where n is the total number of characters in the sentence, and space complexity is also O(n) for storing the result.

Example Walkthrough

Let's trace through the solution with a simple example: "The quick fox"

Initial Setup:

• Input: "The quick fox"
• Create empty list ans = []
• Define vowels: ['a', 'e', 'i', 'o', 'u']

Processing Word 1: "The" (index = 0)

• First letter: 'T' → lowercase: 't'
• Is 't' a vowel? No (consonant)
• Apply consonant rule: "The" → "he" + "T" = "heT"
• Add "ma": "heT" + "ma" = "heTma"
• Add position-based 'a's: "heTma" + "a"*(0+1) = "heTmaa"
• Append to ans: ans = ["heTmaa"]

Processing Word 2: "quick" (index = 1)

• First letter: 'q' → lowercase: 'q'
• Is 'q' a vowel? No (consonant)
• Apply consonant rule: "quick" → "uick" + "q" = "uickq"
• Add "ma": "uickq" + "ma" = "uickqma"
• Add position-based 'a's: "uickqma" + "a"*(1+1) = "uickqmaaa"
• Append to ans: ans = ["heTmaa", "uickqmaaa"]

Processing Word 3: "fox" (index = 2)

• First letter: 'f' → lowercase: 'f'
• Is 'f' a vowel? No (consonant)
• Apply consonant rule: "fox" → "ox" + "f" = "oxf"
• Add "ma": "oxf" + "ma" = "oxfma"
• Add position-based 'a's: "oxfma" + "a"*(2+1) = "oxfmaaaa"
• Append to ans: ans = ["heTmaa", "uickqmaaa", "oxfmaaaa"]

Final Step:

• Join with spaces: " ".join(ans) = "heTmaa uickqmaaa oxfmaaaa"

Result: "heTmaa uickqmaaa oxfmaaaa"

Notice how each word gets progressively more 'a's at the end (1, 2, then 3), and consonant-starting words have their first letter moved to the end before adding "ma".

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm iterates through each word in the sentence once. Let n be the total number of characters in the input sentence.

• Splitting the sentence: O(n)
• For each word, we perform constant-time operations (checking vowel, string slicing, concatenation)
• String slicing word[1:] and word[0] takes O(w) where w is the length of the word
• String concatenation with 'ma' takes O(w)
• Adding 'a' * (i + 1) takes O(i) in the worst case, where i is the word index
• The join operation at the end takes O(n + m) where m is the total number of 'a's added

Since the sum of all word lengths equals n, and the maximum value of i is the number of words (which is at most n), the total time complexity is O(n).

Space Complexity: O(n)

• The ans list stores modified versions of all words
• Each modified word can be longer than the original due to the 'ma' suffix and additional 'a's
• In the worst case with k words, we add 2k + k(k+1)/2 extra characters (2 for 'ma' per word, plus triangular sum of 'a's)
• Since k is bounded by n, the total space used is O(n + n²/n) = O(n)
• The output string also requires O(n) space

Therefore, the overall space complexity is O(n).

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

Common Pitfalls

1. Case Sensitivity in Vowel Detection

Pitfall: A common mistake is only checking for lowercase vowels, forgetting that words can start with uppercase vowels (like "I" or "Apple").

Incorrect approach:

if word[0] not in ['a', 'e', 'i', 'o', 'u']:  # Misses uppercase vowels!
    word = word[1:] + word[0]

Solution: Either check both cases explicitly or convert to lowercase for comparison:

# Option 1: Include both cases in the set
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
if word[0] not in vowels:
    word = word[1:] + word[0]

# Option 2: Convert to lowercase for checking
if word[0].lower() not in ['a', 'e', 'i', 'o', 'u']:
    word = word[1:] + word[0]

2. Off-by-One Error with 'a' Suffix Count

Pitfall: Using the wrong index calculation for appending 'a's, typically forgetting that enumerate starts at 0 while the problem requires 1-based indexing.

Incorrect approach:

for i, word in enumerate(sentence.split()):
    # ... process word ...
    word += 'a' * i  # Wrong! First word gets 0 'a's instead of 1

Solution: Add 1 to the enumeration index:

for i, word in enumerate(sentence.split()):
    # ... process word ...
    word += 'a' * (i + 1)  # Correct: First word (i=0) gets 1 'a'

3. Modifying the Original Word Variable

Pitfall: Trying to build the result by repeatedly concatenating to the same string variable can lead to confusion about what transformations have been applied.

Potentially confusing approach:

for i, word in enumerate(sentence.split()):
    if word[0].lower() not in vowels:
        word = word[1:] + word[0]
    word = word + 'ma'
    word = word + 'a' * (i + 1)
    # Multiple reassignments can be error-prone

Solution: Use clear, sequential transformations or build a new string:

for i, word in enumerate(sentence.split()):
    # Build transformed word step by step with clear logic
    if word[0] not in vowels:
        transformed = word[1:] + word[0]
    else:
        transformed = word
  
    transformed += 'ma' + 'a' * (i + 1)
    result_words.append(transformed)

4. Empty String or Single Character Edge Cases

Pitfall: Not considering edge cases like empty strings or single-character words when slicing.

Potential issue:

word = word[1:] + word[0]  # What if word is empty or has only 1 character?

Solution: While the problem states words consist of letters (implying non-empty), it's good practice to handle edge cases:

if word and word[0] not in vowels:  # Check word is not empty
    if len(word) > 1:
        word = word[1:] + word[0]
    # Single consonant just gets 'ma' and 'a's appended