<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 {
    vector<int> rank, parent;
    int count;

    int findParent(int x){
        if(x == parent[x])
            return x;
            
        // Path Compression happening here, as I am storing each elements 
        // value in parent[x]
        return parent[x] = findParent(parent[x]); 
    }

    void unionByRank(int x, int y){
        int par_x = findParent(x);
        int par_y = findParent(y);
                    
        if(par_x == par_y){
                return;
        // Jiski jyada rank vo uska baap bnega
        } else if(rank[par_x] < rank[par_y]){ 
            parent[par_x] = par_y;
        } else { // By default, below nature will be followed
            parent[par_y] = par_x;
            rank[par_x]++;
        }
    }
    
public:
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        
        int n = source.size();
        rank.resize(n, 0);
        parent.resize(n, 0);        

        for(int i=0; i<n; i++){
            parent[i] = i;
        }

        for(auto& swap:allowedSwaps){
            unionByRank(swap[0], swap[1]); // index ko hi union krdo
        }

        unordered_map<int, vector<int>> mp;
        for(int i=0; i<n; i++){
            mp[findParent(i)].push_back(i);
        }

        int dis = 0;
        for(auto& idx:mp){
            unordered_map<int, int> mp1, mp2;

            // map me daal do jo index se humko group mila tha
            for(auto& i:idx.second){
                mp1[source[i]]++;
                mp2[target[i]]++;
            }

            int temp = 0;
            for(auto& [value, count]:mp1){
                temp += max(0, count-mp2[value]);
            }
            dis += temp;
        }

        return dis;
    }   
};
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