## The challenge

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

```Input: [3, 2, 1]
Output: 1
Explanation: The third maximum is 1.
```

Example 2:

```Input: [1, 2]
Output: 2
Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
```

Example 3:

```Input: [2, 2, 3, 1]
Output: 1
Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.```

## The solution

I have seen many possible resolutions to this problem, but my favourite has to be the following:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 `````` ``````class Solution { // input a primitive array of ints and return an int public int thirdMax(int[] nums) { // set variables for our 3 max'es Integer max1 = null; Integer max2 = null; Integer max3 = null; // loop through the input array for (Integer n : nums) { // if already set, then skip this loop iteration if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue; // if 1 if (max1 == null || n > max1) { max3 = max2; max2 = max1; max1 = n; // if 2 } else if (max2 == null || n > max2) { max3 = max2; max2 = n; // if 3 } else if (max3 == null || n > max3) { max3 = n; } } // return max3, unless null return max3 == null ? max1 : max3; } } ``````