<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;
        }
        int maxi=INT_MIN;

        priority_queue<int,vector<int>,greater<int>> pq;
        for(int i=0;i<nums.size();i++){
            pq.push(nums[i]);
        }
        vector<int> temp1;
        while(!pq.empty()){
            temp1.push_back(pq.top());
            pq.pop();
        }
        int i=0;
        vector<int> temp;
        while(i<temp1.size()-1){
            if(temp1[i]!=temp1[i+1]){
                temp.push_back(temp1[i]);
            }
            i++;
        }
        temp.push_back(temp1[temp1.size()-1]);
        
        for(int i=0;i<temp.size();i++){
            int prev=temp[i];
            int j=i+1;
            while(j<temp.size() && temp[j]==prev+1){
                prev=temp[j];
                j++;
            }
            int len=j-i;
            maxi=max(maxi,len);
            i=j-1;
        }
        if(maxi==INT_MIN){
            return 0;
        }
        return maxi;
    }
};
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