Filter an array of words by an array of single letters it should ONLY contain

Question

I currently have two PHP arrays:

1. array('a','c','r','r')
2. array('carr','car','arc','ra','c','abc','do','aa','rr')

My desired result is:

array('carr','car','arc','ra','c','rr')

i.e. filtering out 'abc','do','aa' because I wish to filter out words that:

1. don't contain a, c or r
2. contain more/multiple a, c, or r than I have in array1.

I have tried array_filter(), but I don't seem to be able to make it work.

Answer 1

One way to do this:

Count how many times each letter occurs in your first array, using array_count_values.

Then in your array_filter callback function, split each word into individual letters, and also count how many times each of them occurs. Then loop over those counted letters. If the current letter does not occur in your letter-count-array, or its count is greater than that in your letter-count-array, return false.

$letters = ['a','c','r','r'];
$words = ['carr','car','arc','ra','c','abc','do','aa','rr'];

$letterCounts = array_count_values($letters);

$filtered = array_filter($words, function($word) use ($letterCounts) {
    $wordLetterCounts = array_count_values(mb_str_split($word));
    foreach($wordLetterCounts as $wordLetter => $count) {
        if(!isset($letterCounts[$wordLetter]) || $letterCounts[$wordLetter] < $count) {
            return false;
        }
    }
    return true;
});

var_dump($filtered);