<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.empty()) return 0;
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>> > pq;
        for(const auto i: nums) pq.push({i, i});
        auto t = pq.top();
        pq.pop();
        int ml = 0;
        while(!pq.empty()){
            if(pq.top().first <= t.second || (t.second+1==pq.top().first)){
                t.second = pq.top().second;
                pq.pop();
            }
            else{
                if(ml < t.second-t.first) ml=t.second -t.first;
                t=pq.top();
                pq.pop();
            }
        }
        return 1+max(ml, t.second-t.first);

    }
};
56878