Notify When Average of 10 Subsequent Numbers Is Bigger Than Given Value

I found an interesting question at StackOverflow asking for help finding solution to what I have already mentioned in the title, with PHP. I gave it a try before reading answers and came up with the following code:

$avg  = // value we are looking for
$size = count($numbers);

for ($i = 0; $i < $size; $i += 1) {
    if ($i + 9 < 51) {
        $val += $numbers[$i];
        for ($j = $i + 1; $j < 10 + $i; $j += 1) {
            $val += $numbers[$j];
        }
        if ($val / 10 >= $avg) { // hit
            // do something
        }
        $val = 0;
    }
}

That was the first that I could think of. And it worked. The answer given by Dave Chen was much more elegant than my solution (although I think it does something a bit different, but that’s not the point here):

$number = 10; //numbers in a set
$max = 0;
$index = 0;

$size = sizeof($numbers) - $number;
for ($i = 0; $i < $size; $i++) {
    $tmp = array_sum(array_slice($numbers, $i, $number)) / $number;
    if ($tmp > $max) {
        $max = $tmp;
        $index = $i;
    }
}

I made a simple benchmark with microtime() and found out that my solution (ran 100k times) took about ~12.3 seconds while Dave’s took only ~7.4 seconds to finish. That makes his code almost twice faster than mine.

Lesson learned: do not stop learning!