<p>You are given an array <code>points</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>The cost of connecting two points <code>[x<sub>i</sub>, y<sub>i</sub>]</code> and <code>[x<sub>j</sub>, y<sub>j</sub>]</code> is the <strong>manhattan distance</strong> between them: <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>, where <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>Return <em>the minimum cost to make all points connected.</em> All points are connected if there is <strong>exactly one</strong> simple path between any two points.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/d.png" style="width: 214px; height: 268px;" />
<pre>
<strong>Input:</strong> points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
<strong>Output:</strong> 20
<strong>Explanation:</strong> 
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/c.png" style="width: 214px; height: 268px;" />
We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,12],[-2,5],[-4,1]]
<strong>Output:</strong> 18
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= points.length &lt;= 1000</code></li>
	<li><code>-10<sup>6</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.</li>
</ul>

#include<iostream>
#include<cmath>
#include<climits>
#include<algorithm>
#include<vector>
#include<cstring>
#include<deque>
#include<unordered_map>
#include<stack>
#include<queue>
#include<list>
#define ll long long
using namespace std;

class hashClass{
public:
    size_t operator()(const pair<int, int>& a) const {
        return hash<int>()(a.first) ^ (hash<int>()(a.second) << 1);
    }
};

class Solution {
public:
    // int hashFunction(pair<int,int> a){
    //     return hash<int>()(a.first ^ (a.second <<1));
    // }
    static bool arrComp(vector<int>& a,vector<int>& b){
        return a[4] < b[4];
    }

    pair<int,int> find(unordered_map<pair<int,int>,pair<int,int>,hashClass>& arr,pair<int,int> a){
        if(arr[a] == a){
            return a;
        }
        else{
            arr[a] = find(arr,arr[a]);        // this called path comperation optimiztion
            return arr[a];
        }
    }
    bool Union(unordered_map<pair<int,int>,pair<int,int>,hashClass>& parent,      pair<int,int> x ,     pair<int,int> y,      unordered_map<pair<int,int>,int,hashClass>& rank){
        pair<int,int> e = find(parent,x);
        pair<int,int> f = find(parent,y);
        if(e == f){
            return false;
        }   
        
        else{
            // return     
            if(rank[e] < rank[f]){
                parent[e] = f;
            }
            else if(rank[e] > rank[f]){
                parent[f] = e;
            }
            else{
                parent[e] = f;
                rank[f]++;
            }
            return true;
        }
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        vector<vector<int>> edges;
        // int ind = 0;
        // for(int i = 0; i < points.size();i++){
        //     for(int j = i + 1;j < points.size();j++){
        //         edges[ind][0] = points[i][0];
        //         edges[ind][1] = points[i][1];
        //         edges[ind][2] = points[j][0];
        //         edges[ind][3] = points[j][1];
        //         edges[ind][4] = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
        //         ind++;
        //     }
        // }

        int n = points.size();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int cost = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                edges.push_back({points[i][0], points[i][1], points[j][0], points[j][1], cost});
            }
        }

        sort(edges.begin(),edges.end(),arrComp);
        unordered_map<pair<int,int>,pair<int,int>,hashClass> parent;
        unordered_map<pair<int,int>,int,hashClass> rank;
        
        for(int i = 0; i < points.size();i++){
            parent[{points[i][0],points[i][1]}] = {points[i][0],points[i][1]};
            rank[{points[i][0],points[i][1]}] = 1;
        }

        int ans = 0;
        for(int i = 0; i < edges.size();i++){
            if(Union(parent,{edges[i][0],edges[i][1]},{edges[i][2],edges[i][3]},rank)){
                ans = ans + edges[i][4];
            }
        }
        return ans;
    }
};

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