<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:
    int minimumHammingDistance(vector<int>& source, vector<int>& target, vector<vector<int>>& allowedSwaps) {
        // use union find to 
        // find out element in connected parts
        // find overlap of corresponding indices in target
        vector<int> parent(source.size(), -1);
        for(auto pair : allowedSwaps){
            union_f(pair[0], pair[1], parent);
        }
        unordered_map<int, vector<int>> mp; // <root, vector of all connected nodes>
        for(int i=0; i<parent.size(); i++){// essentially finding 
        // what is the parent of every idx
            mp[find(i, parent)].push_back(i); 
        }
        // vector<int> s_con_parts = {};
        // vector<int> t_con_parts = {};
        int count_diff = 0;
        for(auto iter=mp.begin(); iter!=mp.end(); iter++){
            //cout<<mp.size()<<endl;
            vector<int> s_con_parts;
            vector<int> t_con_parts;
            for(auto idx : iter->second){
                s_con_parts.push_back(source[idx]);
                t_con_parts.push_back(target[idx]);
            }
            // cout<<'a'<<endl;
            sort(s_con_parts.begin(), s_con_parts.end());
            sort(t_con_parts.begin(), t_con_parts.end());
            count_diff += count_helper(s_con_parts, t_con_parts);
        }
        return count_diff;

    }
    int count_helper(vector<int>& vec1, vector<int>& vec2){
        vector<int> intersection;
        set_intersection(vec1.begin(), vec1.end(), vec2.begin(), vec2.end(), back_inserter(intersection));
        return vec1.size()-intersection.size();
    }
    int find(int i, vector<int>& parent){
        if(parent[i]==-1){
            return i;
        }else{
            return parent[i] = find(parent[i], parent);
        }
    }
    void union_f(int i, int j, vector<int>& parent){
        int i_rep = find(i, parent);
        int j_rep = find(j, parent);
        if(i_rep != j_rep){ // if not connected
             parent[j_rep] = i_rep;
        }
    }
};
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