<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:
        numSet = set()
        groupMap = {}  # maps start to end index of each group of consecutive elements
        groupMapInv = {}  # inverse of groupMap

        ans = 0
        for n in nums:
            if n in numSet:
                continue
            numSet.add(n)
            x = (n-1 in groupMapInv)
            y = (n+1 in groupMap)
            if x and y:
                tmp = (groupMapInv[n-1], groupMap[n+1])
                groupMap[tmp[0]] = tmp[1]
                groupMapInv[tmp[1]] = tmp[0]
                groupMapInv.pop(n-1)
                groupMap.pop(n+1)
                ans = max(ans, tmp[1] - tmp[0] + 1)
            elif x:
                tmp = groupMapInv[n-1]
                groupMap[tmp] = n
                groupMapInv.pop(n-1)
                groupMapInv[n] = tmp
                ans = max(ans, n - tmp + 1)
            elif y:
                tmp = groupMap[n+1]
                groupMapInv[tmp] = n
                groupMap.pop(n+1)
                groupMap[n] = tmp
                ans = max(ans, tmp - n + 1)
            else:
                groupMap[n] = n
                groupMapInv[n] = n
                ans = max(ans, 1)

        return ans

37870