PHP find missing numbers in array of hyphenated numbers - php

Question

I have an array of numbers, these numbers are sometimes hyphenated, à la software version numbers. What I'm trying to do is echo "Missing!" or run a specific function when a number is missing.
For example:
$numbers = array('1', '2', '3', '5', '6', '8');
Prints:
1
2
3
Missing!
5
6
Missing!
8
I'm running into problems with the hyphens.
For example:
$numbers = array('1', '1-1', '1-3', '3-1-1', '3-1-3');
Prints:
1
1-1
Missing!
1-3
Missing!
3-1-1
Missing!
3-1-3
Plus my code seems awfully long/doing too many things for what -- seems to me -- should be a simple task. Is there a method or algorithm for this sort of thing?
Here's my code:
<?php
$numbers = array(
'1',
'1-1',
'1-3',
'3-1-1',
'3-1-3'
);
foreach ($numbers as $number) {
if (isset($prev_number)) {
$curr_number = explode('-', $number);
$prev_levels = explode('-', $prev_number);
if (preg_match('/-/', $number) and !preg_match('/-/', $prev_number)) {
if (current() - $prev_levels[0] >= 1) {
echo 'Missing!<br>' . PHP_EOL;
}
}
for ($missing = 1; ((count($curr_number) - count($prev_levels)) - $missing) >= 1; $missing++) {
echo 'Missing!<br>' . PHP_EOL;
}
foreach ($curr_number as $hyphen => $part) {
for ($missing = 1; ($part - $missing) - $prev_levels[$hyphen] >= 1; $missing++) {
echo 'Missing!<br>' . PHP_EOL;
}
}
} else {
if ($number != '1') {
echo 'Missing!<br>' . PHP_EOL;
foreach ($curr_number as $part) {
for ($missing = 1; $part > $missing; $missing++) {
echo 'Missing!<br>' . PHP_EOL;
}
}
}
}
echo $number . '<br>' . PHP_EOL;
$prev_number = $number;
}
?>

Answer 1

Only a partial answer.
Plus my code seems awfully long/doing too many things for what -- seems to me -- should be a simple task.
Right observation. If it's doing too many things, try to split up the task into pieces. Modularize.
For example: I see 6 explode calls in your code. You constantly struggling with transforming the input into a usable data format. Do a pre-process stage, convert the strings into arrays with explode once, and then work on that data format.

Answer 2

You can iterate through the list and for each pair you attempt to reach the second by applying a transformation on the first:
Increase the last value, e.g. "1" becomes "2" and "1-1" becomes "1-2".
Add a new sub value, e.g. "1" becomes "1-1" and "1-1" becomes "1-1-1".
#1 can be expanded by increasing from right to left.
If none of the transformations match the number is considered missing. In code:
$numbers = array('1', '1-1', '1-2', '1-3', '3-1-1', '3-1-3');
$first = array_shift($numbers);
echo "$first\n";
while (($second = array_shift($numbers)) !== null) {
if (find_next($first, $second) === false) {
echo "Missing\n";
}
echo "$second\n";
$first = $second;
}
// attempt to transform between current and next
function find_next($current, $next)
{
if (increment_last($current) == $next) {
return $next; // first transformation worked
} elseif (add_suffix($current) == $next) {
return $next; // second transformation worked
}
return false; // nothing worked
}
// transformation 1
function increment_last($value)
{
if (($pos = strpos($value, '-')) !== false) {
$last = substr($value, $pos + 1) + 1;
return substr_replace($value, $last, $pos + 1, strlen($last));
} else {
return $value + 1;
}
}
// transformation 2
function add_suffix($value)
{
return "$value-1";
}
This is not an algorithm, but a heuristic; it does a best effort at doing what you want but there's no formal proof that it works.