Find all letters not used in a string - php

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))));
}

Answer 2

How about this
$str="abc";
var_dump(findNot($str));
function findNot($string)
{
$letters = range('a', 'z');
$presents = array_map(function($i) { return chr($i); }, array_keys(array_filter(count_chars($string))));
return array_diff($letters, $presents);
}
PS: implode the result if you need a string of chars, not array
PPS: not sure if it is a "more elegant" solution :-)
PPPS: another solution I could think of is
$str="abc";
var_dump(findNot($str));
function findNot($string)
{
$letters = range('a', 'z');
$presents = str_split(count_chars($string, 4));
return array_intersect($letters, $presents);
}

Answer 3

You could do something like this:
$text = 'abcdefghijklmnop';
$search = array('a','b','c');
$result = str_replace($search, '', $text);

Answer 4

Zerk's solution is nice. I came up with this similar example, not as elegant.
<?php
$alphabet = 'abcdefghijklmnopqrstuvwxyz';
$alphabet = preg_split('//', $alphabet, -1, PREG_SPLIT_NO_EMPTY);
$str="abc";
var_dump( findNot($str) ); //will echo "defghijklomnopqrstuvwxyz"
$str2="happy birthday";
var_dump( findNot($str2) ); //will echo "cfgjklmnoqsuvwxz"
function findNot($str)
{
global $alphabet;
$str = str_replace(' ', '', $str);
$chars = preg_split('//', $str, -1, PREG_SPLIT_NO_EMPTY);
$chars = array_unique($chars);
sort($chars);
$diff = array_diff($alphabet, $chars);
return $diff;
}