Goat Latin

🏠 ⬅️ ➡️

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.
Note: This problem is from LeetCode.
❌ Compiled
❌ Executed
❌ Correct 
program main
    implicit none

    character(len=:), allocatable :: sentence
    integer :: i

    ! Example 1
    sentence = "I speak Goat Latin"
    print '(A)', goat_latin(sentence)

    ! Example 2
    sentence = "The quick brown fox jumped over the lazy dog"
    print '(A)', goat_latin(sentence)

contains

    function goat_latin(sentence) result(result_sentence)
        implicit none
        character(len=*), intent(in) :: sentence
        character(len=:), allocatable :: result_sentence
        logical :: is_vowel
        integer :: i

        allocate(character(len=len(sentence)+3) :: result_sentence)

        do i = 1, size(sentence)
            is_vowel = any(sentence(i:i) == ['a', 'e', 'i', 'o', 'u'])
            if (is_vowel) then
                result_sentence = trim(result_sentence) // sentence(i:i) // "ma "
            else
                result_sentence = trim(result_sentence) // sentence(i+1:) // sentence(i:i) // "ma "
            end if
            if (i < size(sentence)) then
                result_sentence = trim(result_sentence) // "a "
            end if
        end do
    end function goat_latin
end program main
❌ Compiled
❌ Executed
❌ Correct 
! Declare variables
integer :: i, j, k, l, m
character(len=150) :: sentence, word, new_word

! Read input sentence
read(*,*) sentence

! Initialize variables
i = 1
j = 1
k = 1
l = 1
m = 1

! Loop through each word in the sentence
do while (sentence(i:i) /= ' ')
    ! Get the current word
    word = sentence(j:l)

    ! Check if the word starts with a vowel
    if (index('aeiou', word(1:1)) /= 0) then
        ! Append "ma " to the end of the word
        new_word = word // 'ma '
    else
        ! Remove the first letter and append it to the end
        new_word = word(2:) // word(1:1) // 'ma '
    end if

    ! Add the letter 'a' to the end of the word
    new_word = new_word // 'a'

    ! Increment variables
    i = i + l
    j = j + l
    k = k + l
    l = l + l
    m = m + 1

    ! Print the new word
    write(*,*) new_word

    ! Check if we are at the end of the sentence
    if (sentence(i:i) == ' ') then
        exit
    end if

    ! Increment variables
    i = i + 1
    j = j + 1
    k = k + 1
    l = l + 1
    m = m + 1
end do

end [FORTRAN]
🌐 Data from online sources 
def number_of_lines(widths, s):
    lines = 1
    current_pixel = 0
    for c in s:
        letter_pixels = widths[ord(c) - ord('a')]
        if current_pixel + letter_pixels > 100:
            lines += 1
            current_pixel = 0
        current_pixel += letter_pixels
    return [lines, current_pixel]

The algorithm starts by initializing two variables, `lines` and `currentPixel`. `lines` denotes the number of lines it has taken so far to type out the string `s`, and `currentPixel` stores the width of the characters typed on the current line.

The algorithm iterates through each character c in the input string s. For each character, it calculates its width in pixels by using its respective entry in the widths array. If the addition of the width of the character to the currentPixel would cause it go over 100 pixels, it increases the line count by 1, resets the currentPixel to 0, and continues processing the rest of the string s. Otherwise, add the character width to the currentPixel.

Finally, when the loop finishes iterating all characters in the string s, the algorithm returns the lines and currentPixel as a result array of length 2.

🌐 Data from online sources 
#include <vector>
#include <string>
std::vector<int> numberOfLines(std::vector<int>& widths, std::string s) {
    int lines = 1;
    int currentPixel = 0;
    for (char c : s) {
        int letterPixels = widths[c - 'a'];
        if (currentPixel + letterPixels > 100) {
            lines++;
            currentPixel = 0;
        }
        currentPixel += letterPixels;
    }
    return {lines, currentPixel};
}

The algorithm starts by initializing two variables, `lines` and `currentPixel`. `lines` denotes the number of lines it has taken so far to type out the string `s`, and `currentPixel` stores the width of the characters typed on the current line.

The algorithm iterates through each character c in the input string s. For each character, it calculates its width in pixels by using its respective entry in the widths array. If the addition of the width of the character to the currentPixel would cause it go over 100 pixels, it increases the line count by 1, resets the currentPixel to 0, and continues processing the rest of the string s. Otherwise, add the character width to the currentPixel.

Finally, when the loop finishes iterating all characters in the string s, the algorithm returns the lines and currentPixel as a result array of length 2.