Fine-tuning CodeLlama-7B for Fortran Code Generation

Given a string text, you want to use the characters of text to form as many instances of the word "balloon " as possible.

You can use each character in text at most once. Return the maximum number of instances that can be formed.

Example 1:

Input: text = "nlaebolko " Output: 1

Example 2:

Input: text = "loonbalxballpoon " Output: 2

Example 3:

Input: text = "leetcode " Output: 0

Constraints:

1 <= text.length <= 104
text consists of lower case English letters only.

Note: This problem is from LeetCode.

✅ Compiled

❌ Executed

❌ Correct

program main
    implicit none

    integer :: i, j, k, n, m
    character(len=100) :: text
    character(len=7), parameter :: balloon = "balloon "
    integer, allocatable :: dp(:,:)
    integer :: max_instances

    ! read input
    read(*,'(A)') text

    ! allocate dynamic memory for dp array
    n = len(text)
    allocate(dp(n+1,n))

    ! initialize dp array
    do i = 1, n+1
        dp(i,1) = 1
    end do
    do i = 2, n
        do j = 2, n
            if (text(j-1:j-1) == balloon(i-1:i-1)) then
                dp(j,i) = dp(j-1,i-1) + 1
            else
                dp(j,i) = max(dp(j-1,i), dp(j,i-1))
            end if
        end do
    end do

    ! find maximum number of instances
    max_instances = 0
    do i = 1, n
        do j = 1, n
            if (dp(j,i) > max_instances) then
                max_instances = dp(j,i)
            end if
        end do
    end do

    ! print result
    write(*,*) max_instances

    ! deallocate dynamic memory
    deallocate(dp)

end program main

stderr

At line 11 of file temp.f95 (unit = 5, file = 'stdin')
Fortran runtime error: End of file

Error termination. Backtrace:
#0  0x7bce65508960 in ???
#1  0x7bce655094d9 in ???
#2  0x7bce6575d17b in ???
#3  0x7bce6575d752 in ???
#4  0x7bce65759f1b in ???
#5  0x7bce6575ee3c in ???
#6  0x7bce6575fe55 in ???
#7  0x5d398932f2a5 in MAIN__
#8  0x5d398932f7ee in main

✅ Compiled

❌ Executed

❌ Correct

! Declare variables
integer :: i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z
character(len=100) :: text
integer :: max_instances

! Read input
read(*,*) text

! Initialize variables
max_instances = 0

! Loop through each character in the text
do i = 1, len_trim(text)
    ! Check if the current character is 'b'
    if (text(i:i) == 'b') then
        ! Loop through each character in the text starting from the next character
        do j = i + 1, len_trim(text)
            ! Check if the current character is 'a'
            if (text(j:j) == 'a') then
                ! Loop through each character in the text starting from the next character
                do k = j + 1, len_trim(text)
                    ! Check if the current character is 'l'
                    if (text(k:k) == 'l') then
                        ! Loop through each character in the text starting from the next character
                        do l = k + 1, len_trim(text)
                            ! Check if the current character is 'o'
                            if (text(l:l) == 'o') then
                                ! Loop through each character in the text starting from the next character
                                do m = l + 1, len_trim(text)
                                    ! Check if the current character is 'n'
                                    if (text(m:m) == 'n') then
                                        ! Loop through each character in the text starting from the next character
                                        do n = m + 1, len_trim(text)
                                            ! Check if the current character is ' '
                                            if (text(n:n) == ' ') then
                                                ! Increment the number of instances
                                                max_instances = max_instances + 1
                                            end if
                                        end do
                                    end if
                                end do
                            end if
                        end do
                    end if
                end do
            end if
        end do
    end if
end do

! Output the maximum number of instances
write(*,*) max_instances

end

stderr

At line 7 of file temp.f95 (unit = 5, file = 'stdin')
Fortran runtime error: End of file

Error termination. Backtrace:
#0  0x795d6ee1c960 in ???
#1  0x795d6ee1d4d9 in ???
#2  0x795d6f07117b in ???
#3  0x795d6f06a684 in ???
#4  0x795d6f06b2aa in ???
#5  0x59591ae02225 in MAIN__
#6  0x59591ae024c9 in main

🌐 Data from online sources

def max_number_of_balloons(text: str) -> int:
    letter_count = [0] * 5
    for c in text:
        if c == 'b': letter_count[0] += 1
        if c == 'a': letter_count[1] += 1
        if c == 'l': letter_count[2] += 1
        if c == 'o': letter_count[3] += 1
        if c == 'n': letter_count[4] += 1
    letter_count[2] //= 2
    letter_count[3] //= 2
    return min(letter_count)

First, we initialize an array letter_count with a length of 5 to represent each character in the word "balloon".
We then loop through each character in the input string text.
For each character, we increment the corresponding index in the letter_count array if it matches one of the characters in the word "balloon".
After the loop, we divide the count of 'l' and 'o' by 2 since they appear twice in the word "balloon".
Finally, we return the minimum value in the letter_count array, which represents the maximum number of instances of the word "balloon" that can be formed.

🌐 Data from online sources

int maxNumberOfBalloons(string text) {
    int letter_count[5] = {0};
    for (char c : text) {
        if (c == 'b') letter_count[0]++;
        if (c == 'a') letter_count[1]++;
        if (c == 'l') letter_count[2]++;
        if (c == 'o') letter_count[3]++;
        if (c == 'n') letter_count[4]++;
    }
    letter_count[2] /= 2;
    letter_count[3] /= 2;
    return *min_element(letter_count, letter_count + 5);
}

First, we initialize an array letter_count with a length of 5 to represent each character in the word "balloon".
We then loop through each character in the input string text.
For each character, we increment the corresponding index in the letter_count array if it matches one of the characters in the word "balloon".
After the loop, we divide the count of 'l' and 'o' by 2 since they appear twice in the word "balloon".
Finally, we return the minimum value in the letter_count array, which represents the maximum number of instances of the word "balloon" that can be formed.

Maximum Number of Balloons