Go Aaaaaaaaaaah

The problem is described in full here - Aaah!.

Input

The input consists of two lines. The first line is the “aaah”
Jon Marius is able to say that day. The second line is the “aah” the
doctor wants to hear. Only lowercase ’a’ and ’h’ will be used in the
input, and each line will contain between 0 and 999 ’a’s, inclusive,
followed by a single ’h’.
Output

Output “go” if Jon Marius can go to that doctor, and output “no”
otherwise.

Sample Input 1

aaah

aaaaah

Sample Output1

no

Sample Input 2

aaah
ah

Sample Output 2

go

My source code for the problem is below. I wrote one of the fast running implementations, but not the best. What changes can I make so as to improve the efficiency of the code?

#include
#include
#include
 
#define MAX_LEN 1000
#define MAX_INP 2000
 
//Checks whether the input is in the required format
int isValidInput(char*);
 
int main(int __argc, char* __argv[]) {
    char strJon[MAX_INP], strDoc[MAX_INP];
    int resJon, resDoc;
 
    gets(strJon);
    gets(strDoc);
 
    resJon = isValidInput(strJon);
 
    if(resJon  resJon) { 
            printf("no\n");
            return 0;
    }
 
    printf("go\n");
 
    return 0;
}
 
//Checks valid input & returns len
int isValidInput(char* strInp) {
    char *str = strInp;
    int len = 0;
 
    if(str == NULL) return -1;
 
    //Check if the string either starts with 'a' or 'h' 
    if(!(str[0] == 'h' || str[0] == 'H') && !(str[0] == 'a' || str[0] == 'A')) return -1; 
 
    len = strlen(str);
    if((str[0] == 'h' || str[0] == 'H') &&  len > 1) return -1;
 
    if(str[len - 1] != 'h' && str[len - 1] != 'H') return -1;
 
    if(len > MAX_LEN) return -1;
 
    while(*(str + 1) != NULL) {
            if(*str != 'a' && *str != 'A') return -1;
            ++str;
    }
 
    str = NULL;
 
    return len;
}

I don't see any requirements about actually valuating the input. This gives that the problem to be a simple length calculation of two strings. It's possible that you can get even fast behaviour if you can get the strings into two c-strings and do pointer arithmetic to calc i instead of the expensive i++ and i--.

#include
#include
int main(int argc, char ** argv)
{
        char c;
        int i = 0;
        while (getc(stdin) != '\n')
                i++;
        while (getc(stdin) != '\n')
                i--;
        if (i < 0)
                printf("no\n");
        else
                printf("go\n");
        return 0;
}

StackExchange Code Review Q#55426, answer score: 6