<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:
        startNumLenMap = {}
        numsSet = set()
        visitedSet = set()
        for i in nums:
            numsSet.add(i)
        maxLength = 0
        for i in numsSet:
            startNum = i
            length = 1
            if startNum not in visitedSet:
                while startNum + length in numsSet:
                    if startNum + length in startNumLenMap:
                        length += startNumLenMap[startNum + length]
                        break
                    else:
                        visitedSet.add(startNum + length)
                        length += 1
                startNumLenMap[startNum] = length
            if length > maxLength:
                maxLength = length
        return maxLength
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