Fine-tuning CodeLlama-7B for Fortran Code Generation

Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.

Example 1:

Input: nums = [1,2,3,1] Output: true

Example 2:

Input: nums = [1,2,3,4] Output: false

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2] Output: true

Constraints:

1 <= nums.length <= 105
-109 <= nums[i] <= 109

Note: This problem is from LeetCode.

❌ Compiled

❌ Executed

❌ Correct

program main
    implicit none

    integer, parameter :: n = 10
    integer, dimension(n) :: nums
    logical :: result

    ! Example 1
    nums = [1, 2, 3, 1]
    result = repeated_element(nums)
    write (*, '(L1)') result

    ! Example 2
    nums = [1, 2, 3, 4]
    result = repeated_element(nums)
    write (*, '(L1)') result

    ! Example 3
    nums = [1, 1, 1, 3, 3, 4, 3, 2, 4, 2]
    result = repeated_element(nums)
    write (*, '(L1)') result

contains

    function repeated_element(nums) result(result_)
        implicit none
        integer, intent(in) :: nums(:)
        logical :: result_

        integer :: i, j

        do i = 1, size(nums) - 1
            do j = i + 1, size(nums)
                if (nums(i) == nums(j)) then
                    result_ = .true.
                    exit
                end if
            end do
            if (result_) exit
        end do

    end function repeated_element

end program main

stderr

temp.f95:9:4:

    9 |     nums = [1, 2, 3, 1]
      |    1
Error: Different shape for array assignment at (1) on dimension 1 (10 and 4)
temp.f95:14:4:

   14 |     nums = [1, 2, 3, 4]
      |    1
Error: Different shape for array assignment at (1) on dimension 1 (10 and 4)

❌ Compiled

❌ Executed

❌ Correct

```
program duplicate_elements
    implicit none
    integer, parameter :: int64 = selected_int_kind(13)
    integer, dimension(:), allocatable :: nums
    logical :: has_duplicates

    ! read input
    read (*, *) nums

    ! check for duplicates
    has_duplicates = .false.
    do i = 1, size(nums) - 1
        do j = i + 1, size(nums)
            if (nums(i) == nums(j)) then
                has_duplicates = .true.
                exit
            end if
        end do
        if (has_duplicates) exit
    end do

    ! output result
    if (has_duplicates) then
        write (*, *) "true"
    else
        write (*, *) "false"
    end if

contains

    pure function contains_duplicate(nums) result(has_duplicates)
        implicit none
        integer(int64), dimension(:), intent(in) :: nums
        logical :: has_duplicates
        integer :: i, j

        has_duplicates = .false.
        do i = 1, size(nums) - 1
            do j = i + 1, size(nums)
                if (nums(i) == nums(j)) then
                    has_duplicates = .true.
                    exit
                end if
            end do
            if (has_duplicates) exit
        end do
    end function contains_duplicate
end program duplicate_elements
```

stderr

temp.f95:1:2:

    1 | ```
      |  1
Error: Invalid character in name at (1)
temp.f95:50:2:

   50 | ```
      |  1
Error: Invalid character in name at (1)
temp.f95:13:8:

   13 |     do i = 1, size(nums) - 1
      |        1
Error: Symbol ‘i’ at (1) has no IMPLICIT type
temp.f95:14:12:

   14 |         do j = i + 1, size(nums)
      |            1
Error: Symbol ‘j’ at (1) has no IMPLICIT type

🌐 Data from online sources

def containsDuplicate(nums):
    return len(nums) != len(set(nums))

The algorithm starts by creating an empty set. Then it iterates through the input array. For each number in the array, it checks if the number is already in the set. If the number is in the set, it means there's a duplicate, so it returns true. If the number is not in the set, it adds the number to the set. If the function has not returned true after iterating through the entire array, it means that there are no duplicates, so it returns false.

In Python, we use the fact that sets don't allow duplicates, so we can directly compare the length of the original list with the length of the set created from the list. If they are not equal, it means there are duplicates in the list, and the function returns true. Otherwise, it returns false.

🌐 Data from online sources

bool containsDuplicate(vector<int>& nums) {
    unordered_set<int> s;
    for (int num : nums) {
        if (s.find(num) != s.end()) {
            return true;
        }
        s.insert(num);
    }
    return false;
}

The algorithm starts by creating an empty set. Then it iterates through the input array. For each number in the array, it checks if the number is already in the set. If the number is in the set, it means there's a duplicate, so it returns true. If the number is not in the set, it adds the number to the set. If the function has not returned true after iterating through the entire array, it means that there are no duplicates, so it returns false.

Contains Duplicate