<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>

import queue

class Solution:
    def bfs(self, num, setNums, visited):
        q = queue.Queue()
        count = 0

        q.put(num)
        visited.add(num)

        while q.qsize() > 0:
            nextNum = q.get()
            count += 1
            for neigh in [nextNum - 1, nextNum + 1]:
                if neigh in setNums and neigh not in visited:
                    visited.add(neigh)
                    q.put(neigh)
        return count
    def longestConsecutive(self, nums: List[int]) -> int:
        if not nums:
            return 0 
        
        setNums = set(nums)

        # O(n) auxiliary memory 
        visited = set() 
        maxResult = 0
        for num in nums: 
            if num not in visited: 
                maxResult = max(maxResult, self.bfs(num, setNums, visited))
        
        return maxResult
        
37690