<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();
        
        pair<int, int>fm= {INT_MIN, -1};
        pair<int, int>sm= {INT_MIN, -1};
        pair<int, int>tm= {INT_MIN, -1};
        
        for(int i=0; i<n; i++)
        {
            if(nums[i]>fm.first||(nums[i]==INT_MIN&&fm.second==-1))
            {
                if(sm.second!=-1)
                {
                    tm.first= sm.first;
                    tm.second=1;
                }
                if(fm.second!=-1)
                {
                    sm.first= fm.first;
                    sm.second=1;
                }
                fm.first= nums[i];
                 fm.second=1;
            }
            if(fm.first==nums[i])
            {
                continue;
            }
            else if(nums[i]>sm.first||(nums[i]==INT_MIN&&sm.second==-1))
            {
                if(sm.second!=-1)
                {
                    tm= sm;
                    tm.second=1;
                }
                sm.first= nums[i];
                sm.second=1;
            }
            if(sm.first==nums[i])
            {
                continue;
            }
            else if(nums[i]>tm.first||(nums[i]==INT_MIN&&tm.second==-1))
            {
                tm.first= nums[i];
                tm.second= 1;
            }
        }
        if(tm.second==-1) return fm.first;
        
        return tm.first;
    }
};
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