Given a 2D array** houses[][]** consisting of **N** 2D coordinates **{x, y}** where each coordinate represents the location of each house, the task is to find the minimum cost to connect all the houses of the city.

Cost of connecting two houses is the

Manhattan Distancebetween the two points (x_{i}, y_{i}) and (x_{j}, y_{j}) i.e.,|x, where |p| denotes the absolute value of p._{i}– x_{j}| + |y_{i}– y_{j}|

**Examples:**

Input:houses[][] = [[0, 0], [2, 2], [3, 10], [5, 2], [7, 0]]Output:20Explanation:Connect house 1 (0, 0) with house 2 (2, 2) with cost = 4

Connect house 2 (2, 2) with house 3 (3, 10) with cost =9

Connect house 2 (2, 2) with house 4 (5, 2) with cost =3

At last, connect house 4 (5, 2) with house 5 (7, 0) with cost 4.

All the houses are connected now.

The overall minimum cost is 4 + 9 + 3 + 4 = 20.

Input:houses[][] = [[3, 12], [-2, 5], [-4, 1]]Output:18Explanation:

Connect house 1 (3, 12) with house 2 (-2, 5) with cost = 12

Connect house 2 (-2, 5) with house 3 (-4, 1) with cost = 6

All the houses are connected now.

The overall minimum cost is 12 + 6 = 18.

**Approach:** The idea is to create a weighted graph from the given information with weights between any pair of edges equal to the cost of connecting them, say **C _{i}** i.e., the Manhattan distance between the two coordinates. Once the graph is generated, apply Kruskal’s Algorithm to find the Minimum Spanning Tree of the graph using Disjoint Set. Finally, print the minimum cost.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `vector<` `int` `> parent, size;` `// Utility function to find set of an` `// element v using path compression` `// technique` `int` `find_set(` `int` `v)` `{` ` ` `// If v is the parent` ` ` `if` `(v == parent[v])` ` ` `return` `v;` ` ` `// Otherwsie, recursively` ` ` `// find its parent` ` ` `return` `parent[v]` ` ` `= find_set(parent[v]);` `}` `// Function to perform union` `// of the sets a and b` `int` `union_sets(` `int` `a, ` `int` `b)` `{` ` ` `// Find parent of a and b` ` ` `a = find_set(a);` ` ` `b = find_set(b);` ` ` `// If parent are different` ` ` `if` `(a != b) {` ` ` `// Swap Operation` ` ` `if` `(size[a] < size[b])` ` ` `swap(a, b);` ` ` `// Update parent of b as a` ` ` `parent[b] = a;` ` ` `size[a] += size[b];` ` ` `return` `1;` ` ` `}` ` ` `// Otherwise, return 0` ` ` `return` `0;` `}` `// Function to create a Minimum Cost` `// Spanning tree for given houses` `void` `MST(` `int` `houses[][2], ` `int` `n)` `{` ` ` `// Stores adjacency list of graph` ` ` `vector<pair<` `int` `, pair<` `int` `, ` `int` `> > > v;` ` ` `// Traverse each coordinate` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `for` `(` `int` `j = i + 1; j < n; j++) {` ` ` `// Find the Manhattan distance` ` ` `int` `p = ` `abs` `(houses[i][0]` ` ` `- houses[j][0]);` ` ` `p += ` `abs` `(houses[i][1]` ` ` `- houses[j][1]);` ` ` `// Add the edges` ` ` `v.push_back({ p, { i, j } });` ` ` `}` ` ` `}` ` ` `parent.resize(n);` ` ` `size.resize(n);` ` ` `// Sort all the edges` ` ` `sort(v.begin(), v.end());` ` ` `// Initialize parent[] and size[]` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `parent[i] = i, size[i] = 1;` ` ` `}` ` ` `/// Stores the minimum cost` ` ` `int` `ans = 0;` ` ` `// Finding the minimum cost` ` ` `for` `(` `auto` `x : v) {` ` ` `// Perform the unioun operation` ` ` `if` `(union_sets(x.second.first,` ` ` `x.second.second)) {` ` ` `ans += x.first;` ` ` `}` ` ` `}` ` ` `// Print the minimum cost` ` ` `cout << ans;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given houses` ` ` `int` `houses[][2] = { { 0, 0 }, { 2, 2 },` ` ` `{ 3, 10 }, { 5, 2 },` ` ` `{ 7, 0 }};` ` ` `int` `N = ` `sizeof` `(houses)` ` ` `/ ` `sizeof` `(houses[0]);` ` ` `// Function Call` ` ` `MST(houses, N);` ` ` `return` `0;` `}` |

## Python3

`# Python3 program for the above approach` `parent ` `=` `[` `0` `] ` `*` `100` `size ` `=` `[` `0` `] ` `*` `100` `# Utility function to find set of an` `# element v using path compression` `# technique` `def` `find_set(v):` ` ` ` ` `# If v is the parent` ` ` `if` `(v ` `=` `=` `parent[v]):` ` ` `return` `v` ` ` `# Otherwsie, recursively` ` ` `# find its parent` ` ` `parent[v] ` `=` `find_set(parent[v])` ` ` `return` `parent[v]` `# Function to perform union` `# of the sets a and b` `def` `union_sets(a, b):` ` ` ` ` `# Find parent of a and b` ` ` `a ` `=` `find_set(a)` ` ` `b ` `=` `find_set(b)` ` ` `# If parent are different` ` ` `if` `(a !` `=` `b):` ` ` ` ` `# Swap Operation` ` ` `if` `(size[a] < size[b]):` ` ` `a, b ` `=` `b, a` ` ` `# Update parent of b as a` ` ` `parent[b] ` `=` `a` ` ` `size[a] ` `+` `=` `size[b]` ` ` `return` `1` ` ` `# Otherwise, return 0` ` ` `return` `0` `# Function to create a Minimum Cost` `# Spanning tree for given houses` `def` `MST(houses, n):` ` ` ` ` `# Stores adjacency list of graph` ` ` `v ` `=` `[]` ` ` `# Traverse each coordinate` ` ` `for` `i ` `in` `range` `(n):` ` ` `for` `j ` `in` `range` `(i ` `+` `1` `, n):` ` ` `# Find the Manhattan distance` ` ` `p ` `=` `abs` `(houses[i][` `0` `] ` `-` ` ` `houses[j][` `0` `])` ` ` `p ` `+` `=` `abs` `(houses[i][` `1` `] ` `-` ` ` `houses[j][` `1` `])` ` ` `# Add the edges` ` ` `v.append([p, i, j])` ` ` `# Sort all the edges` ` ` `v ` `=` `sorted` `(v)` ` ` `# Initialize parent[] and size[]` ` ` `for` `i ` `in` `range` `(n):` ` ` `parent[i] ` `=` `i` ` ` `size[i] ` `=` `1` ` ` `# Stores the minimum cost` ` ` `ans ` `=` `0` ` ` `# Finding the minimum cost` ` ` `for` `x ` `in` `v:` ` ` `# Perform the unioun operation` ` ` `if` `(union_sets(x[` `1` `], x[` `2` `])):` ` ` `ans ` `+` `=` `x[` `0` `]` ` ` ` ` `# Print the minimum cost` ` ` `print` `(ans)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `# Given houses` ` ` `houses ` `=` `[ [ ` `0` `, ` `0` `], [ ` `2` `, ` `2` `],` ` ` `[ ` `3` `, ` `10` `], [ ` `5` `, ` `2` `],` ` ` `[ ` `7` `, ` `0` `] ]` ` ` `N ` `=` `len` `(houses)` ` ` `# Function Call` ` ` `MST(houses, N)` `# This code is contributed by mohit kumar 29` |

**Output:**

20

**Time Complexity:** O(N^{2})**Auxiliary Space:** O(N^{2})