<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:
        # Understand: O(n) time with O(n) space
            # Consecutive => strictly increasing by 1

        # Match: Hash-map
        # Plan/Pseudocode:
            # 1. Run through nums array, add to hash_map
            # 2. MAYBE: First check freq
                # a. Only add unique ones
            # 3. Run through hash_map again,
                # a. Now for every element check if there is a val,
                    # such that it's 
        min_heap = nums
        heapq.heapify(min_heap)
        result = []
        max_count, counter = 1, 1

        while len(min_heap) > 0:
            val = heapq.heappop(min_heap)
            if not result or result[-1] != val: # Avoid duplicates
                result.append(val)

        if len(result) == 0:
            return 0

        print(result)

        for i in range(len(result) - 1):
            if result[i] + 1 == result[i + 1]:
                counter += 1
            else:
                counter = 1
            max_count = max(counter, max_count)

        return max_count
            
27790