<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:
    struct Union{
        vector<int>parent;
        vector<int>sz;
        Union(int nodes){
            parent.resize(nodes);
            sz.resize(nodes,1);
            for(int i=0;i<nodes;i++){
                parent[i]=i;
            }
        }
        int findParent(int u){
            if(parent[u]==u)return u;
            else return findParent(parent[u]);
        }
        void Unite(int x,int y){
            int u=findParent(x);
            int v=findParent(y);
            if(u==v)return;
            if(sz[u]>sz[v]){
                sz[u]+=sz[v];
                parent[v]=u;
            }
            else{
                sz[v]+=sz[u];
                parent[u]=v;
            }
        }
    };
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        // so let's make the component first right
        int n=source.size();
        Union dsu(n);
        for(int i=0;i<(int)allowedSwaps.size();i++){
            int x=dsu.findParent(allowedSwaps[i][0]);
            int y=dsu.findParent(allowedSwaps[i][1]);
            if(x==y)continue;
            else{
                dsu.Unite(x,y);
            }
        }
        map<int,multiset<int>>v;
        for(int i=0;i<n;i++){
            int par=dsu.findParent(i);
            v[par].insert(source[i]);
        }
        int cnt=0;
        for(int i=0;i<(int)target.size();i++){
            int uv=dsu.findParent(i);
            if(v[uv].find(target[i])==v[uv].end()){
                cnt++;
            }
            else{
                v[uv].erase(v[uv].find(target[i]));
            }
        }
        return cnt;
    }
};
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