Filter array to keep values containing the search word using word boundaries

I know this type of question has been asked before and I also saw those working answers. However, it's not working when there are no space between the search string and the rest of the array value. Here is my code -

$example = array ( 'ext_03.jpg', 'int_01_headlight.jpg');
$searchword = 'int_';
$matches = array_filter($example, function($var) use ($searchword) {
    return preg_match("/\b$searchword\b/i", $var);
});
echo array_values($matches)[0];`

The last value in the $example array doesn't have any space and this code doesn't work. However, if I put space after int_ it works. But I need it to work even if there are no spaces (should work when there is space too). How can I achieve that?

Here is the solution :

$example = array ( 'ext_03.jpg', 'int_01_headlight.jpg');
$searchword = 'int_';
$matches = array_filter($example, function($var) use ($searchword) { return 
preg_match("/\b$searchword/i", $var); });
var_dump($matches);

Remove second \b : The \b in the pattern indicates a word boundary

Documentation : http://php.net/manual/en/function.preg-match.php

EDIT :

The better way is to use \A : Beginning of the string

$example = array ( 'ext_03.jpg', 'int_01_headlight.jpg', 'ext_ int_01_headlight.jpg');
$searchword = 'int_';

// Return 2 results (wrong way)
$matches = array_filter($example, function($var) use ($searchword) { return preg_match("/\b$searchword/i", $var); });
var_dump($matches);

// Return 1 result
$matches = array_filter($example, function($var) use ($searchword) { return preg_match("/\A$searchword/i", $var); });
var_dump($matches);