计算ASCII码为3的倍数的字符串中的字符数，而不使用循环或递归

Question

I have the following PHP code which uses foreach and gives the correct count. This is what I tried with loop.

$str = 'hello world';
$cnt = 0;

$arr = str_split($str);
foreach($arr as $val){
  if( ord($val)%3 == 0 ){
   $cnt++;
  }
}
echo 'total count- '.$cnt; //count is 6 here which is correct

Is there a way of doing the same thing in PHP without the use of loops or recursion?

Answer 1

You can use PHPs inbuilt array functions to do this, but internally they loop over the values themselves so is that valid? One way would be to use array_filter on the results of str_split and then getting the count of the number of values in the resultant array:

$str = 'hello world';
$cnt = count(array_filter(str_split($str), function ($v) { return ord($v) % 3 == 0; }));
echo $cnt;

Output

6

You can also do something similar with array_reduce (this is most similar to your existing code):

$cnt = array_reduce(str_split($str), function ($c, $v) { return $c + (ord($v) % 3 == 0 ? 1 : 0); }, 0);

Demo on 3v4l.org