<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;
    sort(nums.begin(),nums.end());
    vector<int> v;
    int i=0;
    while( i < nums.size()){
        if( i != 0 && nums[i] == v[v.size()-1]){
        i++;
        continue;}
        v.push_back(nums[i]);
        i++;
    }
    int l=0,r=1,count=1,ans=INT_MIN;
      while(r < v.size() ){
        if( v[r] - v[l] == 1){
             r++;
             l++;
             count++;}
        else{
             ans=max(ans,count);
             count=1;
              l=r;
              r++;
        }
       cout<<count<<" ";
      }
      ans=max(count,ans);
    return ans;}
};
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