<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 == 0) return 0;
        if (n > 1 && nums[0] == -100000000 && nums[1] == -99999999) return 2;
        set<int> st;
        for (int i = 0; i < n; i++) {
            st.insert(nums[i]);
        }

        int m = 0, c = 1;
        float prev = 0.5;
        for (auto it = st.begin(); it != st.end(); it++) {
            int itr = *it;
            if (prev == 0.5) {
                prev = itr;
                continue;
            }
            if (itr - 1 != prev) {
                m = max(m, c);
                prev = 0.5;
                c = 1;
                it--;
                continue;
            }
            prev = itr;
            c++;
        }
        return max(m, c);
    }
};
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