Minimum Fuel Cost to Reach the Capital City

Problem Description

Given a tree structure representing a country's road network with n cities, where city 0 is the capital. Roads are bidirectional, and each city has a representative who needs to attend a meeting in the capital. Each city has a car with a limited number of seats. The goal is to find the minimum liters of fuel needed for all representatives to reach the capital city. Representatives can share cars.

Naive Approach

A brute-force approach would involve exploring all possible combinations of carpooling. For each city, we can decide whether the representative travels alone or joins another group. This quickly becomes computationally infeasible as the number of cities increases due to the exponential number of possible combinations.

Optimal Solution

A more efficient approach uses Depth-First Search (DFS) to traverse the tree. We calculate the number of representatives that need to travel from each city to the capital. Then, based on the seats capacity, determine how many cars are needed for each segment of the journey.

1. Build the Adjacency List: Represent the road network as an adjacency list, where each city has a list of its neighboring cities.
2. DFS Traversal: Start a DFS from the capital city (city 0). For each city, calculate the number of representatives from its subtree (including itself) that need to travel to the capital.
3. Calculate Fuel: For each edge traversed during DFS (from a child to its parent), determine the number of cars needed. This can be calculated as ceil(representatives / seats). Multiply this by 1 (as each car trip uses 1 liter of fuel) and add it to the total fuel.

Edge Cases

• If roads is empty, it means there's only one city (the capital), so no fuel is needed. Return 0.
• Ensure that seats is a positive integer. If not, throw an error.
• Handle disconnected graphs (though the problem states it's a tree, so this isn't strictly needed).

Code Implementation (Python)

import math
from collections import defaultdict

def minimumFuelCost(roads, seats):
    n = len(roads) + 1
    if n == 1:
        return 0

    adj = defaultdict(list)
    for a, b in roads:
        adj[a].append(b)
        adj[b].append(a)

    total_fuel = 0

    def dfs(node, parent):
        nonlocal total_fuel
        representatives = 1  # Start with the representative from the current city

        for neighbor in adj[node]:
            if neighbor != parent:
                reps = dfs(neighbor, node)
                representatives += reps
                total_fuel += math.ceil(reps / seats)

        return representatives

    dfs(0, -1)
    return total_fuel

Example

roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]]
seats = 2
print(minimumFuelCost(roads, seats))  # Output: 7

Time and Space Complexity

• Time Complexity: O(N), where N is the number of cities. The DFS traversal visits each city and edge once.
• Space Complexity: O(N). The adjacency list takes O(N) space to store the graph. The DFS call stack can also grow to O(N) in the worst-case (a skewed tree).