<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){
        for(int i=0;i<n;i++){
            parent.push_back(i);
            size.push_back(1);
        }
    }
    int findPar(int u){
        if(u == parent[u]) return u;
        return parent[u] = findPar(parent[u]);
    }
    void unionBySize(int u, int v){
        int ulp_u = findPar(u);
        int ulp_v = findPar(v);
        if(ulp_u == ulp_v) return;
        if(size[ulp_u] < size[ulp_v]){
            size[ulp_v] += size[ulp_u];
            parent[ulp_u] = ulp_v ;
        } 
        else{
            size[ulp_u] += size[ulp_v];
            parent[ulp_v] = ulp_u ;
        }
    }
};


class Solution {
public:
    int minimumHammingDistance(vector<int>& s, vector<int>& t, vector<vector<int>>& as) {
        int n = s.size();
        DisjointSet d1(n);
        for(int i=0;i<as.size();i++){
            d1.unionBySize(as[i][0], as[i][1]);
        }
        map<int, vector<int>> map1;
        for(int i=0;i<n;i++){
            map1[d1.findPar(i)].push_back(i);
        }
        int ans = 0;
        for(auto &it : map1){
            map<int, int> map2;
            for(int i=0;i<it.second.size();i++){
                map2[s[it.second[i]]]++;
                map2[t[it.second[i]]]--;
            }
            int sum = 0;
            for(auto it1 : map2){
                if(it1.second > 0){
                    sum += it1.second;
                }
            }
            ans += sum;
        }
        return ans;
    }
};
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