<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) {
        std::unordered_map<int, int> s;
        vector<int> visited;
        visited.resize(nums.size());

        int maxi = 0;
        
        for(int i=0; i<nums.size(); i++){
            auto num = nums[i];
            if(s.find(num) != s.end())
                continue;

            auto left_it = s.find(num-1);
            auto right_it = s.find(num+1);
            
            auto left_move = left_it != s.end() ? left_it->second : 0;
            auto right_move = right_it != s.end() ? right_it->second : 0;

            s[num - left_move] = left_move + right_move + 1;
            s[num + right_move] = left_move + right_move + 1;
            s[num] = left_move + right_move + 1;
            maxi = max(maxi, left_move + right_move + 1);
            //std::cout<<num<<" "<<left_move<<" "<<right_move<<" "<<maxi<<std::endl;
        }

        return maxi;
    }
};
76801