<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:
        class UnionFind:
            def __init__(self, n):
                self.ranks = [1] * n
                self.root = [i for i in range(n)]
                self.size = [1] * n  # For tracking component sizes
            
            def find(self, x):
                if self.root[x] != x:
                    self.root[x] = self.find(self.root[x])  # Path Compression
                return self.root[x]
            
            def union(self, x, y):
                rootX = self.find(x)
                rootY = self.find(y)
                if rootX == rootY:
                    return False  # Already connected
                else:
                    # Union by Rank
                    if self.ranks[rootX] < self.ranks[rootY]:
                        self.root[rootX] = rootY
                        self.size[rootY] += self.size[rootX]
                    elif self.ranks[rootX] > self.ranks[rootY]:
                        self.root[rootY] = rootX
                        self.size[rootX] += self.size[rootY]
                    else:
                        self.root[rootY] = rootX
                        self.ranks[rootX] += 1
                        self.size[rootX] += self.size[rootY]
                return True
            
        if not nums:
            return 0 

        # Assign unique indices to each unique number
        elem_to_idx = {}
        idx = 0
        for num in nums:
            if num not in elem_to_idx:
                elem_to_idx[num] = idx
                idx += 1

        # Initialize UnionFind with the number of unique elements
        uf = UnionFind(len(elem_to_idx))

        # Perform unions using unique indices
        for num in elem_to_idx:
            current_idx = elem_to_idx[num]
            smaller_num = num - 1
            if smaller_num in elem_to_idx:
                uf.union(current_idx, elem_to_idx[smaller_num])
            larger_num = num + 1
            if larger_num in elem_to_idx:
                uf.union(current_idx, elem_to_idx[larger_num])

        # Calculate the maximum size of connected components
        max_size = 0
        for idx in elem_to_idx.values():
            root = uf.find(idx)
            max_size = max(max_size, uf.size[root])

        return max_size

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