<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:
    void trav(int ind, vector<vector<int>> &adj, vector<int> &vis, vector<int> &src, vector<int> &target, unordered_map<int,vector<int>> &mpg, int grp){
        vis[ind] = 1;
        mpg[grp].push_back(ind);
        for(int i=0;i<adj[ind].size();i++){
            if(vis[adj[ind][i]]==0){
                trav(adj[ind][i], adj, vis, src, target, mpg, grp);
            }
        }
    }
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        int lns = source.size();
        vector<vector<int>> adj(lns);
        int ln = allowedSwaps.size();
        for(int i=0;i<ln;i++){
            adj[allowedSwaps[i][0]].push_back(allowedSwaps[i][1]);
            adj[allowedSwaps[i][1]].push_back(allowedSwaps[i][0]);
        }
        // for(int i=0;i<lns;i++){
        //     mpt[target[i]][i]=1;
        // }
        unordered_map<int, vector<int>> mpg;
        vector<int> vis(lns, 0);
        int ret=0;
        int grp=0;
        for(int i=0;i<lns;i++){
            if(vis[i]==0){
                trav(i, adj, vis, source, target, mpg, grp);
                grp++;
            }
        }
        vector<int> ps;
        for(auto m: mpg){
            ps = m.second;
            unordered_map<int, int> mpc;
            for(int i=0;i<ps.size();i++){
                mpc[source[ps[i]]]++;
            }
            for(int i=0;i<ps.size();i++){
                if(mpc[target[ps[i]]]>0){
                    mpc[target[ps[i]]]--;
                }
                else
                    ret++;
            }
        }
        return ret;
    }
};
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