<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:
        n=len(nums)
        count=1
        a=0
        start=0
        end=n
        nums.sort()
        if len(nums)<1:
            return 0
        while start<end-1:
            i=start+1
            
            # while i<end-1:
            if abs(nums[start]-nums[start+1])==1 :
                count+=1
                nums.pop(start)
                # nums.pop(start
                # end=end-1
                # i+=1
                print(count)
            elif abs(nums[start]-nums[start+1])==0:
                nums.pop(start)
            else:
                start+=1
                count=1

            if count>a:
                a=count
            if a>=n:
                return a
            end=len(nums)
        if count>a:
            return count  
        return a        
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