C# || How To Find The Minimum Fuel Cost To Report To The Capital Using C#

May 24, 2023 admin No comments

The following is a module with functions which demonstrates how to find the minimum fuel cost to report to the capital using C#.

1. Minimum Fuel Cost – Problem Statement

There is a tree (i.e., a connected, undirected graph with no cycles) structure country network consisting of n cities numbered from 0 to n – 1 and exactly n – 1 roads. The capital city is city 0. You are given a 2D integer array roads where roads[i] = [a_i, b_i] denotes that there exists a bidirectional road connecting cities a_i and b_i.

There is a meeting for the representatives of each city. The meeting is in the capital city.

There is a car in each city. You are given an integer seats that indicates the number of seats in each car.

A representative can use the car in their city to travel or change the car and ride with another representative. The cost of traveling between two cities is one liter of fuel.

Return the minimum number of liters of fuel to reach the capital city.

Example 1:

Input: roads = [[0,1],[0,2],[0,3]], seats = 5 Output: 3 Explanation: - Representative₁ goes directly to the capital with 1 liter of fuel. - Representative₂ goes directly to the capital with 1 liter of fuel. - Representative₃ goes directly to the capital with 1 liter of fuel. It costs 3 liters of fuel at minimum. It can be proven that 3 is the minimum number of liters of fuel needed.

Example 2:

Input: roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], seats = 2 Output: 7 Explanation: - Representative₂ goes directly to city 3 with 1 liter of fuel. - Representative₂ and representative₃ go together to city 1 with 1 liter of fuel. - Representative₂ and representative₃ go together to the capital with 1 liter of fuel. - Representative₁ goes directly to the capital with 1 liter of fuel. - Representative₅ goes directly to the capital with 1 liter of fuel. - Representative₆ goes directly to city 4 with 1 liter of fuel. - Representative₄ and representative₆ go together to the capital with 1 liter of fuel. It costs 7 liters of fuel at minimum. It can be proven that 7 is the minimum number of liters of fuel needed.

Example 3:

Input: roads = [], seats = 1 Output: 0 Explanation: No representatives need to travel to the capital city.

2. Minimum Fuel Cost – Solution

The following is a solution which demonstrates how to find the minimum fuel cost to report to the capital.





		
		
			2. Minimum Fuel Cost - Solution
			
C#
			
			// ============================================================================
//    Author: Kenneth Perkins
//    Date:   May 24, 2023
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to find minimum fuel cost to the capital
// ============================================================================
public class Solution {
    public long MinimumFuelCost(int[][] roads, int seats) {
        int n = roads.Length + 1;
        var adj = new Dictionary<int, List<int>>();
        int[] degree = new int[n];

        foreach (int[] road in roads) {
            if (!adj.ContainsKey(road[0])) {
                adj[road[0]] = new List<int>();
            }
            adj[road[0]].Add(road[1]);

            if (!adj.ContainsKey(road[1])) {
                adj[road[1]] = new List<int>();
            }
            adj[road[1]].Add(road[0]);

            ++degree[road[0]];
            ++degree[road[1]];
        }

        return BFS(n, adj, degree, seats);
    }

    public long BFS(int n, Dictionary<int, List<int>> adj, int[] degree, int seats) {
        var q = new Queue<int>();
        for (int i = 1; i < n; i++) {
            if (degree[i] == 1) {
                q.Enqueue(i);
            }
        }

        int[] representatives = new int[n];
        Array.Fill(representatives, 1);
        long fuel = 0;

        while (q.Count > 0) {
            int node = q.Dequeue();
            fuel += (long)Math.Ceiling((double) representatives[node] / seats);

            foreach (int neighbor in adj[node]) {
                --degree[neighbor];
                representatives[neighbor] += representatives[node];
                if (degree[neighbor] == 1 && neighbor != 0) {
                    q.Enqueue(neighbor);
                }
            }
        }
        return fuel;
    }
}// http://programmingnotes.org/
			
				
					
				
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						// ============================================================================
//    Author: Kenneth Perkins
//    Date:   May 24, 2023
//    Taken From: http://programmingnotes.org/
//    File:  Solution.cs
//    Description: Demonstrates how to find minimum fuel cost to the capital
// ============================================================================
public class Solution {
    public long MinimumFuelCost(int[][] roads, int seats) {
        int n = roads.Length + 1;
        var adj = new Dictionary<int, List<int>>();
        int[] degree = new int[n];
 
        foreach (int[] road in roads) {
            if (!adj.ContainsKey(road[0])) {
                adj[road[0]] = new List<int>();
            }
            adj[road[0]].Add(road[1]);
 
            if (!adj.ContainsKey(road[1])) {
                adj[road[1]] = new List<int>();
            }
            adj[road[1]].Add(road[0]);
 
            ++degree[road[0]];
            ++degree[road[1]];
        }
 
        return BFS(n, adj, degree, seats);
    }
 
    public long BFS(int n, Dictionary<int, List<int>> adj, int[] degree, int seats) {
        var q = new Queue<int>();
        for (int i = 1; i < n; i++) {
            if (degree[i] == 1) {
                q.Enqueue(i);
            }
        }
 
        int[] representatives = new int[n];
        Array.Fill(representatives, 1);
        long fuel = 0;
 
        while (q.Count > 0) {
            int node = q.Dequeue();
            fuel += (long)Math.Ceiling((double) representatives[node] / seats);
 
            foreach (int neighbor in adj[node]) {
                --degree[neighbor];
                representatives[neighbor] += representatives[node];
                if (degree[neighbor] == 1 && neighbor != 0) {
                    q.Enqueue(neighbor);
                }
            }
        }
        return fuel;
    }
}// http://programmingnotes.org/

QUICK NOTES:
The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

3 7 0

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Posted in: C#, Console Tags: breadth first search, c-sharp, leetcode

C# || How To Find The Minimum Fuel Cost To Report To The Capital Using C#

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