Finding missing array : PHP

Question

I am trying to find missing in array.

with this code

<?php                       
     $no = array(1,2,3,5,6,7);
     $max=max($no);
     for($x=0; $x<=$max; $x++){
       if(!in_array($x,$no)){ 
       $id = $x;
       }else{
       $id = $x+1;
       }                        
     }
     echo '<pre>'; print_r($id);
?>

but the result is

8

someone can help me?

your trying to store the missing number to another variable right? — Jenuel Ganawed
– Jenuel Ganawed, Commented Jan 18, 2020 at 1:52
Do you want to find the first missing value or all the missing values? Do the missing values start at 0 or the minimum value in the array? — Nick
– Nick, Commented Jan 18, 2020 at 2:07

Nick · Accepted Answer · 2020-01-18 06:03:11Z

1

You are executing your loop to the end, regardless of whether you find a missing value or not. Thus $id will always be $max+1. You need to break out of the loop when you find a missing value (or if you want all missing values, push the missing value to an array). However, the code can be more simply implemented using array_diff on a range from 0 to the maximum value in $no:

$no = array(1,2,3,5,6,7);
$max = max($no);
// you may want to use min($no) here
$min = 0;
$missing = array_diff(range($min, $max), $no);
// print all missing values in the range
print_r($missing);
// if you only want the first missing value
echo min($missing);

Output:

Array
(
    [0] => 0
    [4] => 4
)
0

Demo on 3v4l.org

edited Jan 18, 2020 at 6:03

answered Jan 18, 2020 at 1:25

Nick

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