<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) {
        /* This is a O(n) solution with O(n) space 
        set<int> s(nums.begin(), nums.end());

        int maxlen = 0;
        int length;
        for (int n : s) {
            length = 0;
            if (s.find(n-1) == s.end()) {
                cout << "Number check passed: " << n;
                while (s.find(n) != s.end()) {
                    n += 1;
                    length++;
                }
            }
            maxlen = max(maxlen, length);

        }
        return maxlen;*/
    //}

        // Below is a nlogn solution
        int start = 0;
        int ans = 0;
        set<int> s(nums.begin(), nums.end());
        vector<int> array(s.begin(), s.end());
        sort(array.begin(), array.end());
        while(start < array.size()) {
            int maxlen = 1;
            while (start + 1 < array.size() && array[start+1] == array[start]+1) {
                //cout << "Im here\n";
                start++;
                maxlen++;
            }
            ans = max(ans, maxlen);
            start++;
        }
        return(ans);
    }
    
};
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