<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) {
        
        int n = nums.size();
        if (n == 1) return nums[0];
        if (n == 2) return max(nums[0], nums[1]);

        priority_queue<int, vector<int>, std::greater<int>> min_heap;
        unordered_set<int> unique;  // get rid of duplicated as we want the third distict maximum

        for (int i = 0; i < n; i++)
            unique.insert(nums[i]);

        // insert all the distnict numbers to min_heap and keep its size = 3
        // when we finish the min heap will include the three maximum numbers
        for (auto v : unique)
        {
            min_heap.push(v);   

            if (min_heap.size() > 3)
                min_heap.pop();
        }

        if (min_heap.size() == 2)
        {
            min_heap.pop();
        }
        // now min heap includes the 3 maximums
        return min_heap.top();
    }
};

/*
int thirdMax(vector<int>& nums) {
        
        int n = nums.size();

        if (n == 1) return nums[0];

        sort(nums.begin(), nums.end());
        
        int ans = nums[n-1];
        int j = 1;
        for (int i = n-2; i >= 0; i--)
        {
            if (nums[i] != nums[i+1])
                j++;
            if (j == 3) return nums[i];
        }

        return nums[n-1];
    }
*/
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