<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:
    def longestConsecutive(self, nums: List[int]) -> int:
# could sort then just iterate tracking how many in a row. but o of n means no sort
# can also do a heap in o of n then pop and see
        if not nums or len(nums) < 1: return 0
        heapq.heapify(nums) # creates min heap, get smallest first
        prev, consec, maxconsec = float(inf), 1, 1
        while nums:
            n = heapq.heappop(nums)
            if n == prev + 1:
                consec = consec + 1
                maxconsec = max(maxconsec, consec)
            else:
                if n != prev:
                    consec = 1
            print(prev, n, 'consecutive now ', consec)
            prev = n
        return maxconsec
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