<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>& arr) {
        int n=arr.size();
        if(n<=1)
        return n;
        sort(arr.begin(),arr.end());
    int i=0;int j=0,maxc=1,c=1;
    vector<int> v=arr;
    arr.clear();
    for(int i=0;i<n-1;i++)
    {
        if(v[i]!=v[i+1])
        {
            arr.push_back(v[i]);
        }
    }
    arr.push_back(v[v.size()-1]);
    while(i<arr.size()-1)
    {
        if((arr[i]+1)==arr[i+1])
        {
            j=i;
            while(j<arr.size()-1 &&((arr[j]+1)==arr[j+1]))
            {

            c+=1;
            j++;
            }
        i=j;
        maxc=max(c,maxc);
        c=1;
        }
        else
        i++;
    }
    return maxc;
    }
};
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