<p>Given an unsorted array of integers <code>nums</code>, return <em>the length of the longest consecutive elements sequence.</em></p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time.</p>

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

<pre>
<strong>Input:</strong> nums = [100,4,200,1,3,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest consecutive elements sequence is <code>[1, 2, 3, 4]</code>. Therefore its length is 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0,3,7,2,5,8,4,6,0,1]
<strong>Output:</strong> 9
</pre>

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

<ul>
	<li><code>0 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        
        if(nums.size() == 0)
        {
            return 0;
        }

        priority_queue<int,vector<int>,greater<int>> q(nums.begin(),nums.end());

        int ans = INT_MIN;
        int a = q.top();
        q.pop();
        int b = 1;
        while(!q.empty())
        {  
        
            if(a+1 == q.top())
            {
               b = b + 1;
            }
            else if(a == q.top())
            {
                b = b;
            }
            else
            {
                ans = max(ans,b);
                b  = 1;
            }
            a = q.top();
            q.pop();
        }

        return max(ans,b);

    }
};
36954