<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() <= 1)
        {
            return nums.size();
        }
        // max heap
        priority_queue<int> pq(nums.begin(), nums.end());

        int prev = pq.top();
        pq.pop();
        int maxLen = 1;
        int currLen = 1;

        while(!pq.empty())
        {
            auto& top = pq.top();
            if(top + 1 == prev)
            {
                currLen++;
            }
            else if(top == prev)
            {
                pq.pop();
                continue;
            }
            else
            {
                maxLen = std::max(maxLen, currLen);
                currLen = 1;
            }
            prev = top;
            pq.pop();
        }
        maxLen = std::max(maxLen, currLen);


        return maxLen;
    }
};
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