<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() < 2) { return nums.size(); }
        nums = remove_dupes(nums);
        vector<int> counts;
        int count = 1, max_count = 1;
        for (int i = 1; i < nums.size(); i++) {
            int prev = nums[i-1], curr = nums[i];
            if (prev == curr-1) { count ++; }
            else {
                counts.push_back(count);
                count = 1;
            }
        }
        counts.push_back(count);
        sort(counts.begin(), counts.end());
        return counts[counts.size()-1];  
    }

    vector<int> remove_dupes(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<int> new_nums;
        int val = nums[0];
        new_nums.push_back(val);
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] != val) {
                new_nums.push_back(nums[i]);
                val = nums[i];
            }
        }
        return new_nums;
    }
};
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