LeetCode #824 — EASY

Goat Latin

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"I speak Goat Latin"

Example 2

"The quick brown fox jumped over the lazy dog"

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #824: Goat Latin
class Solution {
    public String toGoatLatin(String sentence) {
        List<String> ans = new ArrayList<>();
        Set<Character> vowels
            = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
        int i = 1;
        for (String word : sentence.split(" ")) {
            StringBuilder t = new StringBuilder();
            if (!vowels.contains(word.charAt(0))) {
                t.append(word.substring(1));
                t.append(word.charAt(0));
            } else {
                t.append(word);
            }
            t.append("ma");
            for (int j = 0; j < i; ++j) {
                t.append("a");
            }
            ++i;
            ans.add(t.toString());
        }
        return String.join(" ", ans);
    }
}

// Accepted solution for LeetCode #824: Goat Latin
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #824: Goat Latin
// class Solution {
//     public String toGoatLatin(String sentence) {
//         List<String> ans = new ArrayList<>();
//         Set<Character> vowels
//             = new HashSet<>(Arrays.asList('a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'));
//         int i = 1;
//         for (String word : sentence.split(" ")) {
//             StringBuilder t = new StringBuilder();
//             if (!vowels.contains(word.charAt(0))) {
//                 t.append(word.substring(1));
//                 t.append(word.charAt(0));
//             } else {
//                 t.append(word);
//             }
//             t.append("ma");
//             for (int j = 0; j < i; ++j) {
//                 t.append("a");
//             }
//             ++i;
//             ans.add(t.toString());
//         }
//         return String.join(" ", ans);
//     }
// }

# Accepted solution for LeetCode #824: Goat Latin
class Solution:
    def toGoatLatin(self, sentence: str) -> str:
        ans = []
        for i, word in enumerate(sentence.split()):
            if word.lower()[0] not in ['a', 'e', 'i', 'o', 'u']:
                word = word[1:] + word[0]
            word += 'ma'
            word += 'a' * (i + 1)
            ans.append(word)
        return ' '.join(ans)

// Accepted solution for LeetCode #824: Goat Latin
use std::collections::HashSet;
impl Solution {
    pub fn to_goat_latin(sentence: String) -> String {
        let set: HashSet<&char> = ['a', 'e', 'i', 'o', 'u'].into_iter().collect();
        sentence
            .split_whitespace()
            .enumerate()
            .map(|(i, s)| {
                let first = char::from(s.as_bytes()[0]);
                let mut res = if set.contains(&first.to_ascii_lowercase()) {
                    s.to_string()
                } else {
                    s[1..].to_string() + &first.to_string()
                };
                res.push_str("ma");
                res.push_str(&"a".repeat(i + 1));
                res
            })
            .into_iter()
            .collect::<Vec<String>>()
            .join(" ")
    }
}

// Accepted solution for LeetCode #824: Goat Latin
function toGoatLatin(sentence: string): string {
    return sentence
        .split(' ')
        .map((s, i) => {
            let startStr: string;
            if (/[aeiou]/i.test(s[0])) {
                startStr = s;
            } else {
                startStr = s.slice(1) + s[0];
            }
            return `${startStr}ma${'a'.repeat(i + 1)}`;
        })
        .join(' ');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.