找到字符串中未使用的所有字母

Question

I want to be able to do something likes this:

$str="abc";
echo findNot($str); //will echo "defghijklomnopqrstuvwxyz"
$str2="happy birthday";
echo findNot($str2); //will echo "cfgjklmnoqsuvwxz"

Basically, it would find all letters not represented in the string and return them in an array or string.

I could do this easily with a foreach and arrays of characters, but I was wondering if anyone had a more elegant solution.

Answer 1

Here's what I came up with.

function findNot($str){
    return array_diff(range('a','z'), array_unique(str_split(strtolower($str))));
}