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 Distance between the two points (xi, yi) and (xj, yj) i.e., |xi – xj| + |yi – yj|, where |p| denotes the absolute value of p.
Examples:
Input: houses[][] = [[0, 0], [2, 2], [3, 10], [5, 2], [7, 0]]
Output: 20
Explanation: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: 18
Explanation:
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 Ci 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; // Otherwise, 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 union 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; } |
Java
// Java program for the above approach import java.util.*; // Class for DSU implementation class DSU{ int parent[]; int rank[]; // Constructor to initialize DSU DSU( int n) { parent = new int [n]; rank = new int [n]; for ( int i = 0 ; i < n; i++) { parent[i] = - 1 ; rank[i] = 1 ; } } // Utility function to find set of an // element v using path compression // technique int find_set( int v) { // If v is the parent if (parent[v] == - 1 ) return v; // Otherwise, recursively // find its parent return parent[v] = find_set(parent[v]); } // Function to perform union // of the sets a and b void union_sets( int a, int b) { // Find parent of a and b int p1 = find_set(a); int p2 = find_set(b); // If parent are different if (p1 != p2) { // Swap Operation if (rank[p1] > rank[p2]) { parent[p2] = p1; rank[p1] += rank[p2]; } else { parent[p1] = p2; rank[p2] += rank[p1]; } } } } class GFG{ // Function to create a Minimum Cost // Spanning tree for given houses static void MST( int houses[][], int n) { int ans = 0 ; ArrayList< int []> edges = new ArrayList<>(); // Traverse each coordinate for ( int i = 0 ; i < n; i++) { for ( int j = i + 1 ; j < n; j++) { // Find the Manhattan distance int p = Math.abs(houses[i][ 0 ] - houses[j][ 0 ]); p += Math.abs(houses[i][ 1 ] - houses[j][ 1 ]); // Add the edges edges.add( new int []{ p, i, j }); } } // Sorting arraylist using custome comparator // on the basis of weight i.e first element in // array object stored in Arraylist Collections.sort(edges, new Comparator< int []>() { @Override public int compare( int [] o1, int [] o2) { return Integer.compare(o1[ 0 ], o2[ 0 ]); } }); // Calling DSU class DSU d = new DSU(n); for ( int i = 0 ; i < edges.size(); i++) { int from = edges.get(i)[ 1 ]; int to = edges.get(i)[ 2 ]; // Checking if they lie in different component // or not i.e they have same parent or not in // DSU if (d.find_set(from) != d.find_set(to)) { // Calling union_sets d.union_sets(from, to); ans += edges.get(i)[ 0 ]; } } // Printing the minimum cost System.out.println(ans); } // Driver code public static void main(String args[]) { // Graph in form of 2D array int houses[][] = { { 0 , 0 }, { 2 , 2 }, { 3 , 10 }, { 5 , 2 }, { 7 , 0 } }; int n = houses.length; // Function Call MST(houses, n); } } // This code is contributed by Rahul Verma |
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 parent if (v = = parent[v]): return v # Otherwise, 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 union 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 |
Javascript
<script> // JavaScript program for the above approach let parent = new Array(100).fill(0) let size = new Array(100).fill(0) // Utility function to find set of an // element v using path compression // technique function find_set(v){ // If v is the parent if (v == parent[v]) return v // Otherwise, recursively // find its parent parent[v] = find_set(parent[v]) return parent[v] } // Function to perform union // of the sets a and b function 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 function MST(houses, n){ // Stores adjacency list of graph let v = [] // Traverse each coordinate for (let i=0;i<n;i++){ for (let j=i+1;j<n;j++){ // Find the Manhattan distance let p = Math.abs(houses[i][0] - houses[j][0]) p += Math.abs(houses[i][1] - houses[j][1]) // Add the edges v.push([p, i, j]) } } // Sort all the edges v.sort((a,b)=>a[0]-b[0]) // Initialize parent[] and size[] for (let i=0;i<n;i++){ parent[i] = i size[i] = 1 } // Stores the minimum cost let ans = 0 // Finding the minimum cost for (let x of v){ // Perform the union operation if (union_sets(x[1], x[2])) ans += x[0] } // Print the minimum cost document.write(ans, "</br>" ) } // Driver Code // Given houses let houses = [ [ 0, 0 ], [ 2, 2 ], [ 3, 10 ], [ 5, 2 ],[ 7, 0 ] ] let N = houses.length // Function Call MST(houses, N) // This code is contributed by shinjanpatra </script> |
C#
using System; class UnionFind { private int [] parent; private int [] size; public UnionFind( int n) { parent = new int [n]; size = new int [n]; for ( int i = 0; i < n; i++) { parent[i] = i; size[i] = 1; } } // Utility function to find set of an //element v using path compression // technique public int FindSet( int v) { //if v is parent if (v == parent[v]) { return v; } //Otherwise, recursively //find its parent parent[v] = FindSet(parent[v]); return parent[v]; } // Function to perform union // of the sets a and b public int UnionSets( int a, int b) { // Find parent of a and b a = FindSet(a); b = FindSet(b); // If parent are different if (a != b) { if (size[a] < size[b]) { // Swap Operation int temp = a; a = b; b = temp; } //Update parent of b as a parent[b] = a; size[a] += size[b]; return 1; } return 0; } } class Program { static void Main( string [] args) { int [][] houses = new int [][] { new int [] { 0, 0 }, new int [] { 2, 2 }, new int [] { 3, 10 }, new int [] { 5, 2 }, new int [] { 7, 0 } }; int n = houses.Length; Tuple< int , int , int >[] edges = new Tuple< int , int , int >[(n * (n - 1)) / 2]; int count = 0; // Traverse each coordinate for ( int i = 0; i < n; i++) { for ( int j = i + 1; j < n; j++) { int p = Math.Abs(houses[i][0] - houses[j][0]) + Math.Abs(houses[i][1] - houses[j][1]); edges[count++] = Tuple.Create(p, i, j); } } // Sort all the edges Array.Sort(edges, (a, b) => a.Item1.CompareTo(b.Item1)); UnionFind uf = new UnionFind(n); int ans = 0; foreach ( var edge in edges) { if (uf.UnionSets(edge.Item2, edge.Item3) == 1) { ans += edge.Item1; } } Console.WriteLine(ans); } } |
20
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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