<p>You are given two integer arrays, <code>source</code> and <code>target</code>, both of length <code>n</code>. You are also given an array <code>allowedSwaps</code> where each <code>allowedSwaps[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that you are allowed to swap the elements at index <code>a<sub>i</sub></code> and index <code>b<sub>i</sub></code> <strong>(0-indexed)</strong> of array <code>source</code>. Note that you can swap elements at a specific pair of indices <strong>multiple</strong> times and in <strong>any</strong> order.</p>

<p>The <strong>Hamming distance</strong> of two arrays of the same length, <code>source</code> and <code>target</code>, is the number of positions where the elements are different. Formally, it is the number of indices <code>i</code> for <code>0 &lt;= i &lt;= n-1</code> where <code>source[i] != target[i]</code> <strong>(0-indexed)</strong>.</p>

<p>Return <em>the <strong>minimum Hamming distance</strong> of </em><code>source</code><em> and </em><code>target</code><em> after performing <strong>any</strong> amount of swap operations on array </em><code>source</code><em>.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> source = [1,2,3,4], target = [2,1,4,5], allowedSwaps = [[0,1],[2,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> source can be transformed the following way:
- Swap indices 0 and 1: source = [<u>2</u>,<u>1</u>,3,4]
- Swap indices 2 and 3: source = [2,1,<u>4</u>,<u>3</u>]
The Hamming distance of source and target is 1 as they differ in 1 position: index 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> source = [1,2,3,4], target = [1,3,2,4], allowedSwaps = []
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are no allowed swaps.
The Hamming distance of source and target is 2 as they differ in 2 positions: index 1 and index 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> source = [5,1,2,4,3], target = [1,5,4,2,3], allowedSwaps = [[0,4],[4,2],[1,3],[1,4]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == source.length == target.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= source[i], target[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= allowedSwaps.length &lt;= 10<sup>5</sup></code></li>
	<li><code>allowedSwaps[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
</ul>

class Solution {
public:
    // Weighted Union-Find with path compression 
    struct UnionFind {
    private:
        vector<int> weights;
        vector<int> connections;
    public:
        UnionFind(int size) {
            weights.resize(size);
            connections.resize(size);

            for (int i = 0; i < size; i++) {
                connections[i] = i;
                weights[i] = 1;
            }
        }

        int root(int id) {
            while (id != connections[id]) {
                // Flatten tree by setting root of current node from its parent to its grandparent
                connections[id] = connections[connections[id]];
                id = connections[id];
            }

            return id;
        }

        bool connected(int a, int b) {
            return root(a) == root(b);
        }

        bool connect(int a, int b) {
            a = root(a);
            b = root(b);

            // Check if nodes already connected
            if (a == b)
                return false;

            // Always connect root with smaller weight to root with bigger weight to keep trees more flat
            if (weights[a] > weights[b])
                swap(a, b);

            // Set b as root of a
            connections[a] = b;
            // Increase weight of b by weight of a subtree
            weights[b] += weights[a];
            return true;
        }
    };

    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        UnionFind uf = UnionFind(source.size());
        for (auto& swap : allowedSwaps)
            uf.connect(swap[0], swap[1]);
            
        vector<map<int, int>> cycles(source.size());
        for (int i = 0; i < source.size(); i++) {
            cycles[uf.root(i)][source[i]]++;
            cycles[uf.root(i)][target[i]]--;
        }

        int hamming = 0;
        for (auto& cycle : cycles)
            for (auto& num : cycle)
                hamming += num.second > 0 ? num.second : 0;
        return hamming;
    }
};
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