<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 unione(vector<int>& p,int u,int v,vector<int>& rank) {
        int a=find(p,u);
        int b=find(p,v);
        if(a==b) return;
// Optimising by union by rank method
        if(rank[a]>rank[b]) {
            rank[b]++;
            p[b]=a;
        } else if(rank[a]<rank[b]) {
            rank[a]++;   
            p[a]=b;
        } else {
            rank[a]++;
            p[a]=b;
        }
    }

    int find(vector<int>& p,int u) {
        if(p[u]==u) return u;
        return p[u]=find(p,p[u]);
    }


    int minimumHammingDistance(vector<int>& s, vector<int>& t, vector<vector<int>>& a) {
        int n=a.size();
        vector<int> p(s.size());
        for(int i=0;i<s.size();i++) p[i]=i;
        vector<int> rank(s.size(),1);

       for(auto i:a) unione(p,i[0],i[1],rank);
//--------------------------
       map<int,multiset<int > > m;
       for(int i=0;i<s.size();i++)  {
           m[find(p,i)].insert(i);
       }
//--------------------------  This part creates the disjoint sets of unique indices belonging to that disjoint set
       int ans=0;
       for(auto &[k,v]:m) {
//-----------------------------------------
           unordered_map<int,int> m1,m2;
           for(auto i:v) {
               m1[s[i]]++;
               m2[t[i]]++;
           }
//-----------------------------------------  Keeping track of each element belonging to that group of indices from both source and target arrays
           int d=0;
//-----------------------------------------
           for(auto &[k1,v1]:m1) {
               cout<<k1<<" "<<v1<<"\n";
               d+=max(0,v1-m2[k1]);
           }
//-----------------------------------------  This part has found the maximum difference possible for the count of each element within that group indices between the source and the target array
           ans+=d;
       }

       return ans;

    }
};
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