<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 UnionFind{
public:
    int *parent;
    int *size;
    UnionFind(int n)
    {
        parent = new int[n];
        size = new int[n];
        for(int i=0; i<n; i++)
        {
            parent[i] = i;
            size[i] = 1;
        }
    }
    int find(int p)
    {
        int src = p;
        while(parent[src]!=src)
        {
            parent[src] = parent[parent[src]]; //compress tree while doing find
            src=parent[src];
        }
        return src;
    }
    bool Union(int p, int q)
    {
        int rootP = find(p);
        int rootQ = find(q);
        if(rootP == rootQ) return false;
        else
        {
            if(size[rootP] > size[rootQ])
            {
                parent[rootQ] = rootP;
                size[rootP]+= size[rootQ];

            }
            else
            {
                parent[rootP] = rootQ;
                size[rootQ]+= size[rootP];
            }
        }
        return true;
    }
    void Print(int n)
    {
        cout<<"parent: ";
        for(int i=0; i<n; i++)
        {
            cout<<parent[i]<<" ";
        }
        cout<<"\n";

        cout<<"size: ";
        for(int i=0; i<n; i++)
        {
            cout<<size[i]<<" ";
        }
        cout<<"\n";        
    }
};
class Solution {
public:
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        int n = source.size();

        UnionFind *uf = new UnionFind(n);        
        for(int i=0; i<allowedSwaps.size(); i++)
        {
            uf->Union(allowedSwaps[i][0], allowedSwaps[i][1]);
        }
        map<int, map<int,int>> m;

        for(int i=0; i<n; i++)
        {
            int iRoot = uf->find(i);
            m[iRoot][source[i]]++;
        }

        int iRet = 0;
        for(int i=0; i<n ;i++)
        {
            int iRoot = uf->find(i);
            if(m[iRoot][target[i]] ==0) iRet++;
            else
            {
                m[iRoot][target[i]]--;
            }

        }


        return iRet;

        
    }
};
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