<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 DSU {
public:
      int capacity;
      vector<int>par;     //for the leader of the componenet
      vector<int>rank;    //comprasion factor (kaun aage chal raha h)
      vector<int>size;    //for the acutal size of the component
      DSU(int cap)
      {
          capacity = cap;
          rank.resize(capacity,0);
          size.resize(capacity,1);
          for(int i = 0 ; i < capacity ; i++){
              par.push_back(i);
          }
      }
      int find(int u)     //mark
      {
           if(par[u] == u)
           {
               return u;
           }
           return par[u] = find(par[u]);     //branch path compression
      }
      void union_rank(int u , int v)
      {
           int paru = find(u);
           int parv = find(v);

           if(paru == parv) return;

           if(rank[paru] > rank[parv]){
                par[parv] = paru;   //mark
           }
           else if(rank[paru] < rank[parv]){
                par[paru] = parv;   
           }   
           else if(rank[paru] == rank[parv]){
                par[paru] = parv;
                rank[parv]++;
           } 
      }
      void union_size(int u , int v){
          
          int paru = find(u);
          int parv = find(v);

          if(paru == parv) return;

          if(size[paru] > size[parv]){
              par[parv] = paru;
              size[paru]+=size[parv];
          }
          else if(size[paru] <= size[parv]){
              par[paru] = parv;
              size[parv]+=size[paru];
          }
      }

};
class Solution {
public:
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& arr) {
        DSU *dsu = new DSU(source.size());
        for(int i = 0 ; i < arr.size() ; i++){
            int u = arr[i][0];
            int v = arr[i][1];
            
            dsu->union_rank(u,v);
        }
        for(int i = 0 ; i < source.size() ; i++) int par = dsu->find(i);

        //leader vs component mermber list
        unordered_map<int,vector<int>>mp;   //every component
        for(int i = 0 ; i < source.size() ; i++)
        {
              mp[dsu->par[i]].push_back(i);
        }
        cout<<"components = "<<mp.size()<<endl;
         
        int ans = 0;
        for(auto it = mp.begin() ; it!=mp.end() ; it++)
        {
            vector<int>v1,v2;
            for(auto &i : it->second){
                v1.push_back(source[i]);
                v2.push_back(target[i]);
            }
            sort(v1.begin(),v1.end());
            sort(v2.begin(),v2.end());
            ans = ans + fun(v1,v2);
        }
        return ans;
    }
    int fun(vector<int>&v1,vector<int>&v2){
        int total = v1.size();
        int count = 0;
        int idx1 = 0;
        int idx2 = 0;
        while(idx1 < v1.size() and idx2 < v2.size()){
            if(v1[idx1] == v2[idx2])
            {
                count++;
                idx1++;
                idx2++;
            }
            else if(v1[idx1] < v2[idx2])
            {
                idx1++;
            }
            else
            {
                idx2++;
            }
        }
        return total - count;
    }
};
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