Panagram solution for a given string
Your solution is actually quite decent, and the code style is good, etc. There are a couple of style changes I would recommend:


In a for loop, the variable i typically indicates an int, and people tend to use c to indicate a char. So, I would write for (char c = 'a'; c &lt;= 'z'; c++) {...
your isPanagram method should be public.


Now, about the algorithm, your code will scan the result as many as 26 times. In reality, this will happen fast, and there's no pressing need to change that, but it could be faster, but would require a little work to get it right.

If you reverse the logic, and just iterate through each characters in the String once, then you can reference it back to an index for the characters.... It would go something like:

boolean[] seen = new boolean[26]; // or a constant for the number of characters.
int stillNeed = seen.length;
for (int i = 0; i &lt; inputString.length(); i++) {
    char c = Character.toLowerCase(inputString.charAt(i));
    if (c &gt;= 'a' &amp;&amp; c &lt;= 'z') {
        c = c - 'a';
        if (!seen[c]) {
            seen[c] = true;
            stillNeed--;
            if (stillNeed == 0) {
                return true;
            }
        }
    }
}
return false;


Now, for long strings, the above will be more efficient than your solution. It goes through each character in the string just once, but it has to create a list of what's been seen, or not. If it is a new character, it has to count it as seen.

Your mileage may vary.