Fine-tuning CodeLlama-7B for Fortran Code Generation

You are given an integer num. You know that Danny Mittal will sneakily remap one of the 10 possible digits (0 to 9) to another digit.

Return the difference between the maximum and minimum values Danny can make by remapping exactly one digit in num.

Notes:

When Danny remaps a digit d1 to another digit d2, Danny replaces all occurrences of d1 in num with d2.
Danny can remap a digit to itself, in which case num does not change.
Danny can remap different digits for obtaining minimum and maximum values respectively.
The resulting number after remapping can contain leading zeroes.
We mentioned "Danny Mittal " to congratulate him on being in the top 10 in Weekly Contest 326.

Example 1:

Input: num = 11891 Output: 99009 Explanation: To achieve the maximum value, Danny can remap the digit 1 to the digit 9 to yield 99899. To achieve the minimum value, Danny can remap the digit 1 to the digit 0, yielding 890. The difference between these two numbers is 99009.

Example 2:

Input: num = 90 Output: 99 Explanation: The maximum value that can be returned by the function is 99 (if 0 is replaced by 9) and the minimum value that can be returned by the function is 0 (if 9 is replaced by 0). Thus, we return 99.

Constraints:

1 <= num <= 108

Note: This problem is from LeetCode.

✅ Compiled

✅ Executed

❌ Correct

program main
    implicit none

    integer :: num
    integer :: result

    num = 11891
    result = maxMinDiff(num)
    print '(I0)', result

    num = 90
    result = maxMinDiff(num)
    print '(I0)', result

contains

function maxMinDiff(num) result(diff)
    implicit none
    integer, intent(in) :: num
    integer :: diff
    integer :: i, j, k

    do i = 0, 9
        do j = 0, 9
            if (i /= j) then
                k = num
                call replaceDigit(k, i, j)
                if (k > diff) then
                    diff = k
                end if
            end if
        end do
    end do
end function maxMinDiff

subroutine replaceDigit(num, oldDigit, newDigit)
    implicit none
    integer, intent(inout) :: num
    integer, intent(in) :: oldDigit, newDigit
    integer :: digit

    digit = mod(num, 10)
    if (digit == oldDigit) then
        num = num / 10
        num = num * 10 + newDigit
    else
        num = num / 10
        num = num * 10 + digit
    end if
end subroutine replaceDigit

end program main

stdout

11899
23017

❌ Compiled

❌ Executed

❌ Correct

! This script solves the problem of Danny Mittal remaping a number to obtain the maximum and minimum values.

! The script takes an integer input num and returns the difference between the maximum and minimum values Danny can make by remapping exactly one digit in num.

! The script assumes that Danny can remap a digit to itself, in which case num does not change.

! The script assumes that Danny can remap different digits for obtaining minimum and maximum values respectively.

! The script assumes that the resulting number after remapping can contain leading zeroes.

! The script assumes that Danny Mittal is congratulated on being in the top 10 in Weekly Contest 326.

! The script assumes that the input num is a valid integer.

! The script assumes that the output is a valid integer.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to 108.

! The script assumes that the input and output are in the range of 1 to

stderr

/usr/bin/ld: /usr/lib/gcc/x86_64-linux-gnu/11/../../../x86_64-linux-gnu/Scrt1.o: in function `_start':
(.text+0x1b): undefined reference to `main'
collect2: error: ld returned 1 exit status

🌐 Data from online sources

def diffMaxMin(num):
    max_remap = 1
    min_remap = 9
    num_str = str(num)

    for c in num_str:
        digit = int(c)
        max_remap = max(max_remap, 9 - digit)
        if digit != 0:
            min_remap = min(min_remap, digit)

    max_value = num + max_remap * (len(num_str) - num_str.count(str(9 - max_remap)))
    min_value = num - min_remap * num_str.count(str(min_remap))
    return max_value - min_value

Initialize max_remap to 1 and min_remap to 9, as well as convert num to a string num_str.
Iterate through each character c in num_str and perform the following steps: a. Compute the integer digit from the character c. b. Update max_remap to be the maximum of its current value and 9 - digit. c. If digit is not 0, update min_remap to be the minimum of its current value and digit.
Calculate max_value by adding max_remap multiplied by the number of occurrences where the digit 9 - max_remap appears in the string num_str.
Calculate min_value by subtracting min_remap multiplied by the number of occurrences where the digit min_remap appears in the string num_str.
Return the difference between max_value and min_value.

🌐 Data from online sources

int diffMaxMin(int num) {
    int maxRemap = 1, minRemap = 9;
    string numStr = to_string(num);
    for (char c : numStr) {
        int digit = c - '0';
        maxRemap = max(maxRemap, 9 - digit);
        if (digit != 0) {
            minRemap = min(minRemap, digit);
        }
    }
    int maxValue = num + maxRemap * (numStr.length() - count(numStr.begin(), numStr.end(), '9' - maxRemap));
    int minValue = num - minRemap * count(numStr.begin(), numStr.end(), minRemap + '0');
    return maxValue - minValue;
}

Initialize max_remap to 1 and min_remap to 9, as well as convert num to a string num_str.
Iterate through each character c in num_str and perform the following steps: a. Compute the integer digit from the character c. b. Update max_remap to be the maximum of its current value and 9 - digit. c. If digit is not 0, update min_remap to be the minimum of its current value and digit.
Calculate max_value by adding max_remap multiplied by the number of occurrences where the digit 9 - max_remap appears in the string num_str.
Calculate min_value by subtracting min_remap multiplied by the number of occurrences where the digit min_remap appears in the string num_str.
Return the difference between max_value and min_value.

Maximum Difference by Remapping a Digit