<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:
        if len(nums) == 0:
            return 0
        heapify(nums)
        max_len = 1
        curr_len = 1
        prev = heappop(nums)
        while(len(nums) > 0):
            num = heappop(nums)
            print(prev,num,curr_len,max_len)
            if num == prev:
                continue
            elif num == prev + 1:
                curr_len+=1
            else:
                max_len = max(max_len,curr_len)
                curr_len = 1
            prev = num 
        return max(max_len,curr_len)


        '''import heapq
        l = min(1,len(nums))
        hash_map = {}
        
        for i in range(len(nums)):
            hash_map[nums[i]] = 1
        
        heapq.heapify(nums)
        while(len(hash_map)>l):
            num = heapq.heappop(nums)
            next = num
            local_len = 0
            while(next in hash_map):
                local_len+=1
                del hash_map[next]
                next = next + 1
            l = max(l,local_len)
            
        return l'''
            

             
        
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