How to find missing values in a sequence with PHP?

Question

Suppose you have an array \"value => timestamp\". The values are increasing with the time but they can be reset at any moment.

For example :

$array = arr         


        

Answer 1

You can do the following:

• Keep comparing adjacent array keys. If they are consecutive you do nothing.
• If they are not consecutive then you check their values, if the value has dropped from a higher value to a lower value, it means there was a reset so you do nothing.
• If the value has not dropped then it is a case of missing key(s). All the numbers between the two keys(extremes not included) are part of the result.

Translated in code:

$array = array( 1 => 6000, 2 => 7000, 3 => 8000, 7 => 9000, 8 => 10000, 
                9 => 11000,55 => 1000, 56 => 2000, 57 => 3000, 59 => 4000, 
                60 => 5000,);

$keys = array_keys($array);
for($i=0;$i<count($array)-1;$i++) {
  if($array[$keys[$i]] < $array[$keys[$i+1]] && ($keys[$i+1]-$keys[$i] != 1) ) {
           print(implode(' ',range($keys[$i]+1,$keys[$i+1]-1)));
           print "\n";
   }   
}

Working link

Answer 2

Here’s another solution:

$prev = null;
$missing = array();
foreach ($array as $curr => $value) {
    if (!is_null($prev)) {
        if ($curr > $prev+1 && $value > $array[$prev]) {
            $missing = array_merge($missing, range($prev+1, $curr-1));
        }
    }
    $prev = $curr;
}

Answer 3

This gives the desired result array(4,5,6,58):

$previous_value = NULL;
$temp_store = array();
$missing = array();
$keys = array_keys($array);

for($i = min($keys); $i <= max($keys); $i++)
{
    if(!array_key_exists($i, $array))
    {
        $temp_store[] = $i;
    }
    else
    {
        if($previous_value < $array[$i])
        {
            $missing = array_merge($missing, $temp_store);
        }
        $temp_store = array();
        $previous_value = $array[$i];
    }
}

var_dump($missing);

Or just use Gumbo's very smart solution ;-)