<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 DisjointSet{
    public:
    vector<int>parent, size;

    DisjointSet(int n){
        parent.resize(n);
        for(int i = 0; i<n;i++) parent[i] = i;
        size.resize(n,1);
    }

    int findParent(int node){
        if(parent[node]==node) return node;
        return parent[node] = findParent(parent[node]);
    }
    void unionBySize(int u, int v){
        int upar = findParent(u), vpar = findParent(v);
        if(upar==vpar) return;
        if(size[upar]<=size[vpar]){
            parent[upar] = vpar;
            size[vpar]+=size[upar];
        }
        else{
            parent[vpar] = upar;
            size[upar]+=size[vpar];
        }
    }
};

class Solution {
public:
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
       int n = source.size();
       DisjointSet ds(n);
       for(auto i: allowedSwaps){
        int u = i[0], v = i[1];
        ds.unionBySize(u,v);
       }
       //same component sttore it in map on the basis of their parent

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

       int ans = 0;

       for(auto i: mp){
        unordered_map<int,int>mp1,mp2;
        for(auto j: i.second){
            mp1[source[j]]++;
            mp2[target[j]]++;
        }
        for(auto j: mp1){
            ans+=max(0, j.second-mp2[j.first]);
        }

       }
       return ans;

        
    }
};
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