🚀Day 10. Minimum cost to connect all houses in a city 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a 2D array houses[][] of size n, where each subarray contains the (x, y) coordinates of a house, the goal is to connect all houses such that:

• Each house is connected to at least one other house.

• The total cost of connecting all houses is minimized.

The cost to connect two houses is the Manhattan Distance: $[ \text{Cost} = |x_1 - x_2| + |y_1 - y_2| $]

Note: Sorry for uploading late, my Final Sem exam is going on.

🔍 Example Walkthrough:

Example 1:

Input:

n = 5
houses[][] = [[0, 7], [0, 9], [20, 7], [30, 7], [40, 70]]

Output:

105

Explanation:

• Connect house 1 (0,7) and house 2 (0,9) with cost = 2

• Connect house 1 (0,7) and house 3 (20,7) with cost = 20

• Connect house 3 (20,7) and house 4 (30,7) with cost = 10

• Connect house 4 (30,7) and house 5 (40,70) with cost = 73 Overall minimum cost: 2 + 10 + 20 + 73 = 105.

Example 2:

Input:

n = 4
houses[][] = [[0, 0], [1, 1], [1, 3], [3, 0]]

Output:

7

Explanation:

• Connect house 1 (0,0) and house 2 (1,1) with cost = 2

• Connect house 2 (1,1) and house 3 (1,3) with cost = 2

• Connect house 1 (0,0) and house 4 (3,0) with cost = 3 Overall minimum cost: 2 + 2 + 3 = 7.

Constraints:

• $1 ≤ n ≤ 10^3$

• $0 ≤ houses[i][j] ≤ 10^3$

🎯 My Approach:

Optimized Prim’s Algorithm

1. Overview: We use Prim's algorithm to compute the Minimum Spanning Tree (MST) of a complete graph where each node represents a house and the weight between two houses is their Manhattan distance.

2. Algorithm Steps:

   • Initialize:

     • Create an array d to store the minimum cost (distance) to connect each house; initialize with a large value (infinity), except for the first house set to 0.

     • Maintain an array vis to mark visited houses.

   • Construct the MST:

     • Repeat for all houses:

       • Select the house u not yet visited with the smallest cost d[u].

       • Mark the house u as visited and add d[u] to the total cost.

       • Update the cost of reaching every other house v (not yet visited) as the minimum of its current cost and the Manhattan distance from house u to house v.

   • Return the Total Cost:

     • After processing all houses, the total cost accumulated gives the minimum cost to connect all houses.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n²), as for each of the n houses, we update distances for up to n houses.

• Expected Auxiliary Space Complexity: O(n), as we maintain two arrays (d and vis) of size n.

📝 Solution Code

Code (C++)

class Solution {
  public:
    int minCost(vector<vector<int>>& a) {
        int n = a.size(), ans = 0, u = 0;
        vector<bool> vis(n);
        vector<int> d(n, 1e9);
        d[0] = 0;
        for (int i = 0; i < n; ++i) {
            int m = 1e9, idx = -1;
            for (int j = 0; j < n; ++j)
                if (!vis[j] && d[j] < m) m = d[j], idx = j;
            vis[idx] = 1;
            ans += m;
            for (int j = 0; j < n; ++j)
                if (!vis[j])
                    d[j] = min(d[j], abs(a[idx][0] - a[j][0]) + abs(a[idx][1] - a[j][1]));
        }
        return ans;
    }
};

Code (Java)

class Solution {
    public int minCost(int[][] a) {
        int n = a.length, ans = 0;
        boolean[] vis = new boolean[n];
        int[] d = new int[n];
        Arrays.fill(d, Integer.MAX_VALUE);
        d[0] = 0;
        for (int i = 0; i < n; i++) {
            int m = Integer.MAX_VALUE, u = -1;
            for (int j = 0; j < n; j++)
                if (!vis[j] && d[j] < m) {
                    m = d[j];
                    u = j;
                }
            vis[u] = true;
            ans += m;
            for (int j = 0; j < n; j++)
                if (!vis[j])
                    d[j] = Math.min(d[j], Math.abs(a[u][0] - a[j][0]) + Math.abs(a[u][1] - a[j][1]));
        }
        return ans;
    }
}

Code (Python)

class Solution:
    def minCost(self, a):
        n, ans = len(a), 0
        vis = [0]*n
        d = [float('inf')]*n
        d[0] = 0
        for _ in range(n):
            m, u = float('inf'), -1
            for i in range(n):
                if not vis[i] and d[i] < m:
                    m, u = d[i], i
            vis[u] = 1
            ans += m
            for v in range(n):
                if not vis[v]:
                    d[v] = min(d[v], abs(a[u][0] - a[v][0]) + abs(a[u][1] - a[v][1]))
        return ans

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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