PHP look-and-say sequence - php

Question

I'm trying to code Conway look-and-say sequence in PHP.
Here is my code:
function look_and_say ($number) {
$arr = str_split($number . " ");
$target = $arr[0];
$count = 0;
$res = "";
foreach($arr as $num){
if($num == $target){
$count++;
}else{
$res .= $count . $target;
$count = 1;
$target = $num;
}
}
return $res;
}
As I run the function, look_and_say(9900) I am getting value I expected: 2920.
My question is for assigning $arr to be $arr = str_split($number) rather than $arr = str_split($number . " "), the result omits the very last element of the $arr and return 29.
Is it normal to add empty space at the end of the $arr foreach to examine the last element or is there any better way to practice this code - besides regex way.

Answer 1

There are 2 methods I was able to come up with.
1 is to add concatenate at the result after the loop too.
function look_and_say ($number) {
$arr = str_split($number);
$target = $arr[0];
$count = 0;
$res = "";
foreach($arr as $num){
if($num == $target){
$count++;
}else{
$res .= $count . $target;
$count = 1;
$target = $num;
}
}
$res .= $count . $target;
return $res;
}
And the 2nd one is to add another if clause inside the loop and determine the last iteration:
function look_and_say ($number) {
$arr = str_split($number);
$target = $arr[0];
$count = 0;
$res = "";
$i=0;
$total = count($arr);
foreach($arr as $num){
if($i == ($total-1))
{
$count++;
$res .= $count . $target;
}
elseif($num == $target){
$count++;
}else{
$res .= $count . $target;
$count = 1;
$target = $num;
}
$i++;
}
return $res;
}

Answer 2

I want to suggest you other way using two nested while loops:
<?php
function lookAndSay($number) {
$digits = str_split($number);
$result = '';
$i = 0;
while ($i < count($digits)) {
$lastDigit = $digits[$i];
$count = 0;
while ($i < count($digits) && $lastDigit === $digits[$i]) {
$i++;
$count++;
}
$result .= $count . $lastDigit;
}
return $result;
}