<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:
    
    // Graph DFS to find the connected components
    void dfs(int pos, vector<vector<int>>& adj, vector<int> &vis, vector<int> &group, int top){
        group[pos] = top;
        vis[pos] = 1;
        for(int i = 0 ; i < adj[pos].size() ; i++){
            int cur = adj[pos][i];
            if(vis[cur] == 0){
                dfs(cur, adj, vis, group, top);
            }
        }
    }
    
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& as) {
        int n = source.size();
        vector<vector<int>> adj(n);
        
        // First made the allowed swaps turned into graph edge with source size as nodes
        for(int i = 0 ; i < as.size() ; i++){
            adj[as[i][0]].push_back(as[i][1]);
            adj[as[i][1]].push_back(as[i][0]);
        }
        vector<int> vis(n , 0);
        vector<int> group(n);
        for(int i = 0 ; i < n ; i++){
            group[i] = i;
        }
        
        // Then performed DFS to find the connected components.
        for(int i = 0 ; i < n ; i++){
            if(vis[i] == 0){
                dfs(i, adj, vis, group , group[i]);
            }
        }
        
        
        // Similar groups are stored into the map 
        map<int,vector<int>> mp;
        for(int i=0 ; i < n ; i++){
            mp[group[i]].push_back(i);
        }
        
        // now traverse in each group, and check how many elements are common in each connected components.
        // rest of the non-common components result into the hamming distance
        int ans = 0;
        for(auto i : mp){
            vector<int> pos = i.second;
            map<int, int> check;
            for(int j = 0 ; j < pos.size() ; j++){
                check[source[pos[j]]]++;
            }
            
            for(int j = 0 ; j < pos.size() ; j++){
                if(check[target[pos[j]]] > 0){
                    check[target[pos[j]]]--;
                }else{
                    ans++;
                }
            }
            
        }
        return ans;
    }
};
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