<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>

/// An UnionFind class with union-by-rank and path compression.
class UnionFind {
private:
    vector<int> _p;
    vector<int> _c;
    int _n;

public:
    explicit UnionFind(int n_) : _n(n_) {
        _p.resize(_n, -1);
        _c.resize(_n, 1);
    }
    void reset() {
        for (int i = 0; i < _n; i++) {
            _p[i] = -1;
            _c[i] = 1;
        }
    }
    int find(int x) {
        if (_p[x] == -1) {
            return x;
        }
        // path compression, i.e. path shortening.
        return _p[x] = find(_p[x]);
    }
    /// unite a and b, aka union them.
    /// Union by rank.
    /// \returns true if a and b were originally not in the same group;
    /// false, otherwise.
    bool unite(int a, int b) {
        a = find(a);
        b = find(b);
        if (a == b)
            return false;
        if (_c[b] > _c[a])
            swap(a, b);
        // here it's guaranteed that a is heavier than b.
        // let b the lighter group use a the heavier group as b's parent.
        _p[b] = a;
        _c[a] += _c[b];
        return true;
    }
};

// Idea: Union-Find and Greedy
// Union-Find for the allowed swaps,
// then we can greedily match the nums in a connected group from source, to the
// target.
class Solution {
public:
    int minimumHammingDistance(vector<int>& source, vector<int>& target,
                               vector<vector<int>>& allowedSwaps) {
        const int n = (int)source.size();
        UnionFind uf(n);
        for (auto& pair : allowedSwaps) {
            uf.unite(pair[0], pair[1]);
        }
        // map a group-id to source values in that group
        unordered_map<int, unordered_multiset<int>> groupToVals;
        for (int i = 0; i < n; ++i) {
            groupToVals[uf.find(i)].insert(source[i]);
        }
        // greedily match the nums in a connected group from source, to the
        // target.
        int res = 0;
        for (int i = 0; i < n; ++i) {
            auto& avail = groupToVals[uf.find(i)];
            if (auto iter = avail.find(target[i]); iter != avail.end()) {
                avail.erase(iter);
            } else { // cannot match them
                ++res;
            }
        }
        return res;
    }
};
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