Remove zeros that follow a 1 in an array - php

Question

I am getting the values of an array of checkboxes (created dynamically). There were several issues, but I narrowed it to a certain point. Now the abstract of the problem is this (forget for a second these are checkboxes and just imagine we are working with an array of numbers).
I have an array store[20] which contains values of 0s or 1s.
I need to get rid of every 0 that follows a one.
Ex. lets assume my array is store={0, 1, 0, 1, 0, 0}.
I need to create a new array that will be like this check={0, 1, 1, 0}.
The third and fifth terms dont have to be inserted in the new array because they follow the second and fourth term (which are ones).
Also, in case you wonder, there are no occurrences of two consecutive ones, so no need to worry about that.
Here is my php code (but implementations in other languages are the same since its just for loops and arrays):
$j=0; $count=0;
for($j=0;$j<20; $j++) {
$check[$j] = $store[$j + $count];
if($store[$j]) {
$count = $count + 1;
}
}
The code is not working correctly. Maybe I am dead wrong in the way im implementing it. But Im a bit tired and started seeing things. I really can't see what im doing wrong.
PS: I didnt include the array declarations and other preceding code because i already tested it and I know they are ok. The problem is in the logic of this part of the code.

Answer 1

In PHP you could just use str_split, a little regex and an implode without having to worry about a for loop. Adding in array_map, the results will be returned as integers instead of strings.
$check = array_map('intval',str_split('',preg_replace('/01/','',implode('',$store))));
Demo:
<?php
$store = array(0, 1, 0, 1, 0, 0);
$check = str_split(preg_replace('/10/','1',implode('',$store)));
var_dump($check);
$check = array_map('intval',$check)
var_dump($check);
?>

Answer 2

This is the output I get from running your code:
Array
(
[0] => 0
[1] => 1
[2] => 1
[3] => 0
[4] =>
[5] =>
[6] =>
[7] =>
[8] =>
[9] =>
[10] =>
[11] =>
[12] =>
[13] =>
[14] =>
[15] =>
[16] =>
[17] =>
[18] =>
[19] =>
)
So basically it's working, except for the presence of null values at the end of your array since you're reading values outside of the bounds of the $store array. Simply check whether you're outside of the bounds of the array in your for loop. Notice I added +$count to the operation.
$store = array(0, 1, 0, 1, 0, 0);
$j=0;
$count=0;
$check = array();
for($j=0;$j+$count<count($store); $j++) {
$check[] = $store[$j + $count];
if($store[$j+$count]) {
$count = $count + 1;
}
}
Output:
Array
(
[0] => 0
[1] => 1
[2] => 1
[3] => 0
)

Answer 3

As I understood the question you just need the preceeding element to decide whether a 0 is included in the result or not:
<?php
$store=[0, 1, 0, 1, 0, 0];
$prev=0; $result = array();
foreach($store as $e) {
if ( $e || !$prev ) {
$result[] = $e;
}
$prev = $e;
}
var_export($result);
...there's most likely a more elegant solution but this one prints
array (
0 => 0,
1 => 1,
2 => 1,
3 => 0,
)