<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 longestConsecutiveOld(self, nums: List[int]) -> int:
        if not nums:
            return 0
        if len(nums) == 1:
            return 1

        nums.sort()

        longest = 1
        current = 1
        for i in range(len(nums)-1):
            diff = nums[i+1] - nums[i]

            if diff == 1:
                current += 1
                longest = max(current, longest)
            elif diff > 1: # non-consecutive
                current = 1
            
            # if diff == 0: do nothing


        return longest

    def longestConsecutive(self, nums: List[int]) -> int:
        if not nums:
            return 0
        if len(nums) == 1:
            return 1

        num_set = set()

        for n in nums:
            num_set.add(n)

        longest = 0
        visited = {}

        for num in num_set:
            if num-1 not in num_set:
                current = 0
                num_ctr = num
                
                while num_ctr in num_set:
                    current+=1
                    num_ctr+=1

                visited[num] = current
                longest = max(current, longest)

        return longest
34090