<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>

#include <iostream>
#include <algorithm>
#include <set>
class Solution {
public:
    int longestConsecutive(vector<int>& nums) {
        set<int> actualSet;
        set<int> startingSeqSet;
        int longest = 0;
        // can't sort, cause we need o(n)
        // use a set and remove all duplicate
        for (auto num : nums){
            actualSet.insert(num);
        }
        for (auto num : actualSet){
            int val = num -1;
            if (actualSet.count(val) == 0){
                startingSeqSet.insert(num);
            }
        }

        for (auto num : startingSeqSet){
            int count = 1;
            int nextNum = num+1;
            while (actualSet.count(nextNum) != 0){
                nextNum++;
                count++;
            }
            if (count > longest){
                longest = count;
            }
        }
        return longest;
    }
};
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