Fine-tuning CodeLlama-7B for Fortran Code Generation

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the ith row has exactly i coins. The last row of the staircase may be incomplete.

Given the integer n, return the number of complete rows of the staircase you will build.

Example 1:

Input: n = 5 Output: 2 Explanation: Because the 3rd row is incomplete, we return 2.

Example 2:

Input: n = 8 Output: 3 Explanation: Because the 4th row is incomplete, we return 3.

Constraints:

1 <= n <= 231 - 1

Note: This problem is from LeetCode.

✅ Compiled

✅ Executed

❌ Correct

program main
    implicit none

    integer :: n, k

    n = 5
    k = solve(n)
    write (*, '(A, I0)') 'Complete rows: ', k

    n = 8
    k = solve(n)
    write (*, '(A, I0)') 'Complete rows: ', k

contains

    function solve(n) result(k)
        implicit none
        integer, value :: n
        integer :: i, j, k

        do i = 1, n
            if (mod(n, i) == 0) then
                j = n / i
                if (j >= i) then
                    k = k + 1
                end if
            end if
        end do
    end function solve
end program main

stdout

Complete rows: 1
Complete rows: 2

❌ Compiled

❌ Executed

❌ Correct

```
program staircase
    implicit none
    integer :: n, k, i

    read(*,*) n

    k = 0
    do i = 1, n
        if (i == n) then
            exit
        end if
        if (i == n - i + 1) then
            k = k + 1
        end if
    end do

    write(*,*) k

end program staircase
```

stderr

temp.f95:1:2:

    1 | ```
      |  1
Error: Invalid character in name at (1)
temp.f95:21:2:

   21 | ```
      |  1
Error: Invalid character in name at (1)

🌐 Data from online sources

def arrange_coins(n: int) -> int:
    k = 0
    while n > k:
        k += 1
        n -= k
    return k

The algorithm initializes a variable k, which represents the current row number, to 0. Then it enters a loop where it checks if n is greater than k. If so, it increments k by 1 and subtracts k from n. This process continues until n is no longer greater than k. Finally, the value of k is returned, representing the number of complete rows of the staircase. This approach has a time complexity of O(sqrt(n)) since the number of iterations increases linearly with the size of the staircase.

🌐 Data from online sources

int arrangeCoins(int n) {
    int k = 0;
    while (n > k) {
        k++;
        n -= k;
    }
    return k;
}

Arranging Coins