<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:
    vector<int> parent,rank;
    int findupar(int node){
        if(node==parent[node])return node;
        return parent[node]=findupar(parent[node]);
    }
    void unibyrank(int u,int v){
        int pu=findupar(u);
        int pv=findupar(v);
        if(pu==pv)return;
        if(rank[pu]>rank[pv]){
            parent[pv]=pu;
        }
        else if(rank[pu]<rank[pv]){
            parent[pu]=pv;
        }
        else{
            parent[pv]=pu;
            rank[pu]++;
        }
    }
    void dfs(int node,vector<int> &vis,vector<int> adj[],vector<int>&st1,vector<int>&st2,vector<int>&s, vector<int>&t){
        vis[node]=1;
        st1.push_back(s[node]);
        st2.push_back(t[node]);
        for(auto it:adj[node]){
            if(vis[it]==0)dfs(it,vis,adj,st1,st2,s,t);
        }
    }
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        int ans=0;
        int n=source.size();
        parent.resize(n,0);
        rank.resize(n,0);
        for(int i=0;i<n;i++)parent[i]=i;
        int n1=allowedSwaps.size();
        for(int i=0;i<n1;i++){
             int u=allowedSwaps[i][0],v=allowedSwaps[i][1];
             if(findupar(u)==findupar(v))continue;
             else unibyrank(u,v);
        }
        vector<int> adj[n];
        for(int i=0;i<n1;i++){
           int u=allowedSwaps[i][0],v=allowedSwaps[i][1];
           adj[u].push_back(v);
           adj[v].push_back(u);
        }
        vector<int> vis(n);
        for(int i=0;i<n;i++){
            vector<int> v1;
            vector<int> v2;
            if(vis[i]==0){
                dfs(i,vis,adj,v1,v2,source,target);
            }
            unordered_map<int,int> mp;
            for(auto it:v1){
                mp[it]++;
            }
            for(auto it:v2){
                if(mp[it]>0)mp[it]--;
                else ans++;
            }
        }
        return ans;
    }
};
168431