<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>

# mantain the longest consecutive sequence for each number upto that point
# when we find a new number we need to update every number after it

# if we already knew about the successor we need to update it

# store interval for each number

# there is a problem with the order

# invariant: the borders of intervals are ALWAYS up to date
class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        # nums.sort()
        intervals = {}
        max_length = 0
        for num in nums:
            if num in intervals:
                continue
            left = num
            right = num
            if num - 1 in intervals:
                left, _ = intervals[num - 1]
            if num + 1 in intervals:
                _, right = intervals[num + 1]
            intervals[left] = left, right
            intervals[right] = left, right
            intervals[num] = left, right
            max_length = max(max_length, right - left + 1)
        return max_length

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