<p>Given an integer array <code>nums</code>, return <em>the <strong>third distinct maximum</strong> number in this array. If the third maximum does not exist, return the <strong>maximum</strong> number</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
The first distinct maximum is 2.
The second distinct maximum is 1.
The third distinct maximum does not exist, so the maximum (2) is returned instead.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,3,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2&#39;s are counted together since they have the same value).
The third distinct maximum is 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you find an <code>O(n)</code> solution?
class Solution {
public:
    int thirdMax(vector<int>& nums) {
        
        unordered_set<int> s(nums.begin(), nums.end());

        // Initialize three variables to track the top three distinct maximums
        long first = LONG_MIN;
        long second = LONG_MIN;
        long third = LONG_MIN;

        // Traverse the set and update the three maximum values
        for (const int& num : s) {
            if (num > first) {
                third = second;
                second = first;
                first = num;
            } else if (num > second && num != first) {
                third = second;
                second = num;
            } else if (num > third && num != second && num != first) {
                third = num;
            }
        }

        // If we have found a valid third maximum, return it. Otherwise, return the maximum.
        return (third == LONG_MIN) ? first : third;
    
    }
};
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