# Third Maximum Number with Java ## 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:

```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;

}
}
```
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