<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;
        if(nums.size()==1)return 1;
        int cnt=1;
        vector<int>v;
         sort(nums.begin(),nums.end());
        v.push_back(nums[0]);
        int maxi= INT_MIN;
       
        for(int i=1;i<nums.size();i++)
        {
            if(v.back()+1==nums[i])
            {
                v.push_back(nums[i]);
                cnt++;
            }
            else if(v.back()+1!=nums[i] && v.back()!=nums[i])
            {
                v.clear();
                v.push_back(nums[i]);
                maxi=max(maxi,cnt);
                cnt=1;
            }
        }
        maxi=max(maxi,cnt);
        
        return maxi;
    }
};

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