Find first duplicate in an array

Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.

My code:

function firstDuplicate($a) {
    $unique = array_unique($a);

    foreach ($a as $key => $val) {
        if ($unique[$key] !== $val){
            return $key;
        }else{
            return -1;
        }
    }
}

The code above will be OK when the input is [2, 4, 3, 5, 1] but if the input is [2, 1, 3, 5, 3, 2] the output is wrong. The second duplicate occurrence has a smaller index. The expected output should be 3.

How can I fix my code to output the correct result?

Answer

$arr = array(2,1,3,5,3,2);
function firstDuplicate($a) {
    $res = -1;
    for ($i = count($a); $i >= 1; --$i) {
        for ($j = 0; $j < $i; ++$j) {
            if ($a[$j] === $a[$i]) {
                $res = $a[$j];
            }
        }
    }
    return $res;
}
var_dump(firstDuplicate($arr));

By traversing the array backwards, you will overwrite any previous duplicates with the new, lower-indexed one.

Note: this returns the value (not index), unless no duplicate is found. In that case, it returns -1.