<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() == 1  ){
            return 1;
        }
        if(nums.size() == 0 ){
            return 0;
        }
        sort(nums.begin(),nums.end());
        for(int i=0;i<(nums.size() -1);i++){
            if(nums[i] == nums[i+1]){
                nums.erase(nums.begin() + i + 1);
                i--;
            }
        }
        int c = -1;
        int x = 1;
        vector<int> vec;
        for(int i=0;i<(nums.size()-1);i++){
            if(nums[i]+1 == nums[i+1]){
                x++;
            }
            else{
                vec.push_back(x);
                x=1;
            }
        }
        vec.push_back(x);
        for(int i=0;i<vec.size();i++){
            if(vec[i] > c){
                c= vec[i];
            }
        }
        return c;
    }
};
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