<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) {
        // int n = nums.size();
        // if(n==1)return 1;
        // sort(nums.begin(),nums.end());
        // int count=0;
        // int maxi = 0;
        // for(int i=0;i<n-1;i++)
        // {
        //     if(nums[i+1]-nums[i]==1)count++;
            
        //     else if(nums[i+1]-nums[i]!=1){count =0;}
        //     maxi = max(maxi,count+1);
        // }
        // return maxi;
        vector<int> ans;
        int counter = 1;
        int maxi = INT_MIN;

        if(nums.size() == 0)
            return 0;
        else if(nums.size() == 1)
            return 1;
        
        sort(nums.begin(),nums.end());

        ans.push_back(nums[0]);

        for(int i = 1;i < nums.size();i++){
            if(ans.back() + 1 == nums[i]){
                ans.push_back(nums[i]);
                counter++;
            }else if(ans.back()+1 != nums[i] && ans.back() != nums[i]){
                ans.clear();
                ans.push_back(nums[i]);
                maxi = max(maxi,counter);
                counter = 1;
            }
            maxi = max(maxi, counter);
        }

        return maxi;
    }
};
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