Check for whether a string contains any duplicate characters
Missing return type

This function declaration is missing the return type:

isUnique(char* s1)


Sure, the default return type is int in C,
but it would be better to just write it anyway for clarity.

Use boolean

C99 has a bool type with true and false values,
which you can get from this #include:

#include &lt;stdbool.h&gt;


It would be better to rewrite isUnique to return a bool.
Even though, this will mean giving up the special value -1 for the NULL input case. But that doesn't seem so bad. A name like "is*" is commonly expected to return a boolean. If you really need a function that can handle the NULL case, you could create a wrapper function:

/* returns -1 on NULL input, 0 if not unique, 1 if unique */
int checkUnique(char* s1) {
    if (s1 == NULL) return -1;
    return isUnique(s1);
}


Note that the bool return value of isUnique will be converted to int, so that false is 0 and true is 1.

Avoid repeated calculations

You call strlen(s1) twice:


for(i = 0; i &lt; (strlen(s1) -1); i++)
{
    holder = s1[i];
    for(j = i; j &lt; (strlen(s1)-1); j++)



It would be better to call it once, store in a variable and reuse.

Time complexity

As you suspected, the time complexity of your algorithm is \$O(n^2)\$:
all characters are compared with all other.

You could improve that to \$O(n)\$ by using extra storage, the size of the alphabet.
You could create an array of bool, representing if a character was already seen or not:


All values are initialized to false
For character in the input, check if it was seen


If it was seen, return false
If it was not seen, mark it now (set value in the bool array to true)

If the end of the string is reached, return true


Something like this:

bool isUnique(char * string) {
    int len = strlen(string);

    bool seen[256];
    memset(seen, false, sizeof(seen));

    int i;
    for (i = 0; i &lt; len; i++) {
        char c = string[i];
        if (seen[c]) {
            return false;
        }
        seen[c] = true;
    }
    return true;
}


Printing the result

This will print a 0 or 1:


int x;
char s1[] = "dog hi cobe!";
x = isUnique(s1);
printf("%d", x);



It would be more informative to print the strings that was tested, and "false"/"true" instead of 0/1:

char s1[] = "dog hi cobe!";
printf("%s -&gt; %s\n", s1, isUnique(s1) ? "true" : "false");


Usability

It would be more interesting if the program took the strings from the command line, instead of using hardcoded values, for example:

void printWithResult(char* string) {
    printf("%s -&gt; %s\n", string, isUnique(string) ? "true" : "false");
}

int main(int argc, char ** argv) {   
    int i;
    for (i = 1; i &lt; argc; i++) {
        printWithResult(argv[i]);
    }
}