<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:
        newnums = []

        

        sortedNums = list(set(nums))
        sortedNums.sort()
        if len(sortedNums) == 1:
            return 1
        print(sortedNums)

        sequenceCount = 0
        finalCount = 0
        for i in range(len(sortedNums)):
            if i == len(sortedNums) -1 :
                break
            currVal = sortedNums[i + 1] - sortedNums[i]
            print(currVal)
            if currVal == 1:
                sequenceCount = sequenceCount + 1
            else:
                sequenceCount = 0
            
            if finalCount < sequenceCount:
                finalCount = sequenceCount
        
        if finalCount > 0:
            return finalCount + 1
        else:
            if len(nums) > len(sortedNums):
                return 1
            return finalCount
        
        
33730