Return the smallest positive integer - php

Question

function solution($A);
that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.
Given A = [1, 2, 3], the function should return 4.
Given A = [−1, −3], the function should return 1.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [1..100,000];
each element of array A is an integer within the range [−1,000,000..1,000,000].
Below is my attempt:
function solution($A) {
// write your code in PHP7.0
$n=1;
while($n > 0 && $n <= 1000000)
$n ++;
echo $A=$n+1;
}
echo solution;
?>```

Answer 1

Try this, no loop required:
<?php
function solution($set) {
$diff = array_diff(range(1, max($set)), $set);
sort($diff);
return !isset($diff[0]) ? max($set) + 1 : ($diff[0] < 1 ? 1 : $diff[0]);
}
echo solution([39, 68, 47, 2, 19, 64]); // 1
echo solution([1, 3, 6, 4, 1, 2]); // 5
echo solution([1, 2, 3]); // 4
echo solution([-1, -3]); // 1
https://3v4l.org/h28LZ

Answer 2

Here's a one-liner to either impress your professor or get kicked out of the class for not following instructions:
php > function solution(array $A) { return max(array(1,min(array_diff(range(1,100000),$A)))); }
php > echo solution([39, 68, 47, 2, 19, 64]);
1
php > echo solution([1,3,6,4,1,2]);
5
php > echo solution([-1,-3]);
1
php > echo solution([1,2,3]);
4
It generates array N (1-1,000,000) and runs array_diff against your input A, and returns either the lowest result from that comparison or 1 if it's less than or equal to 0.

Answer 3

function solution($A) {
$smallest = 1;
while(in_array($smallest, $A)){
$smallest++;
}
return $smallest;
}
Above is the smallest code but has O(N^2) Complexity with 66% success
function solution($A) {
$flipped = array_flip($A);
$smallest = 1;
while (isset($flipped[$smallest])){
$smallest++;
}
return $smallest;
}
Detected time complexity:
O(N) or O(N * log(N)) with 100% success rate