Pset2 Substitution -- check50 "timed out while waiting for program to exit"

Question

I'm aware that a similar question regarding the same error has been asked before but my situation appears to be different. Here is the specific error:

In my code, I have a function called "is_correct" that checks if the key adheres to the format specified in Pset2 and I think that logically the problem must be coming from there, but I have no idea how. The code runs perfectly fine when I test it and I'm unable to replicate this scenario that check50 keeps encountering. Here is my code:

#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <ctype.h>

bool is_correct(int c, string k);
int char_position(char letter);
string encrypt(string p, string k);

int main(int argc, string argv[])
{
    if (is_correct(argc, argv[1]) == true)
    {
        string p = get_string("plaintext:  ");
        printf("ciphertext: %s\n", encrypt(p, argv[1]));
        return 0;
    }
    else
    {
        printf("Error: Incorrect input. Please limit your key to a string of alphabetical characters not exceeding 26 in total\n");
        return 1;
    }
}

// c == "count" && k == "key"
bool is_correct(int c, string k)
{
    bool isco = false;
    int ad_asci = 0;
    if (c == 2)
    {
        for (int i = 0, len = strlen(k); i < len; i++)
        {
            ad_asci += toupper(k[i]);
            if (toupper(k[i]) < 65 || toupper(k[i]) > 90)
            {
                return isco;
            }
        }
        // 2015 is the sum of ASCII values between 'A' to 'Z'
        // Therefore k must be 26 characters long, must not have duplicates, etc.
        if (ad_asci != 2015)
        {
            return isco;
        }
        else
        {
            isco = true;
            return isco;
        }
    }
    else
    {
        return isco;
    }
}
string encrypt(string p, string k)
{
    string c = p;
    for (int i = 0, len = strlen(p); i < len; i++)
    {
        if (p[i] > 64 && p[i] < 91)
        {
            c[i] = toupper(k[char_position(p[i])]);
        }
        else if (p[i] > 96 && p[i] < 123)
        {
            c[i] = tolower(k[char_position(p[i])]);
        }
        else
        {
            c[i] = p[i];
        }
    }
    return c;
}

// Func will locate char position in alphabet, where 0 is A and 25 is Z
int char_position(char letter)
{
    string a = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    string b = "abcdefghijklmnopqrstuvwxyz";
    int position = 0;
    if (letter > 64 && letter < 91)
    {
        for (int i = 0, len = strlen(a); i < len; i++)
        {
            if (letter == a[i])
            {
                position += i;
                break;
            }
        }
    }
    else
    {
        for (int i = 0, len = strlen(b); i < len; i++)
        {
            if (letter == b[i])
            {
                position += i;
                break;
            }
        }
    }
    return position;
}

Answer 1

There are other combinations of (upper-case) letters whose sum of ascii values is 2015.