#2477

Medium

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Minimum Fuel Cost to Report to the Capital

Calculate the minimum fuel needed for all city representatives to reach the capital using DFS on a tree graph efficiently.

tree depth first search breadth first search graph

Problem Statement

You are given a country with n cities connected by n-1 bidirectional roads forming a tree. City 0 is the capital. Each city has one representative who must attend a meeting in the capital. Each city also has one car, and cars have a fixed number of seats.

Given a 2D array roads where roads[i] = [ai, bi] represents a road between cities ai and bi, and an integer seats indicating the seat capacity per car, calculate the minimum fuel required to bring all representatives to the capital. Representatives can share cars, but each car can only carry up to seats representatives per trip.

Examples

Example 1

Input: roads = [[0,1],[0,2],[0,3]], seats = 5

Output: 3

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

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

No representatives need to travel to the capital city.

Constraints

1 <= n <= 105
roads.length == n - 1
roads[i].length == 2
0 <= ai, bi < n
ai != bi
roads represents a valid tree.
1 <= seats <= 105

Solution Approach

Depth-First Search Aggregation

Use DFS starting from the capital to traverse the tree. At each node, count the number of representatives in its subtree. Return this count to the parent while calculating the trips needed to move them towards the capital.

Fuel Calculation per Subtree

For each child node, compute the number of trips required using ceil(subtree_size / seats). Each trip consumes one unit of fuel per road. Accumulate the total fuel at each step while returning counts to the parent node.

Avoid Redundant Computation

Track visited nodes to prevent revisiting and double counting. Merge representatives efficiently using the seat constraint to minimize trips. This ensures optimal fuel usage without unnecessary traversal or miscounting representatives.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) since each node and edge is visited once during DFS. Space complexity is O(n) for adjacency lists and recursive DFS stack.

What Interviewers Usually Probe

Asks how to aggregate representatives efficiently in the tree.
Hints at considering car seat constraints and combining trips.
Probes for edge cases like single city or empty road list.

Common Pitfalls or Variants

Common pitfalls

Forgetting to ceil the division when calculating trips, causing underestimation of fuel.
Ignoring subtree aggregation, leading to redundant trips counted.
Assuming bidirectional edges without properly marking visited nodes, causing infinite recursion.

Follow-up variants

Change seat capacity per car dynamically and recompute fuel.
Add weighted roads where fuel per road varies, requiring modified DFS.
Limit the number of trips per car per day, adding additional constraints on aggregation.

FAQ

What is the main pattern for solving Minimum Fuel Cost to Report to the Capital?

The problem follows a graph traversal with depth-first search pattern, aggregating representatives in subtrees to calculate minimal fuel.

How do I calculate trips for each node?

Use ceil(subtree_size / seats) to determine the number of trips from a node to its parent, then sum fuel across all edges.

Can representatives share cars from different cities?

Yes, as long as the total number of representatives in a car does not exceed the seat limit.

What edge cases should I consider?

Consider empty roads, single city, or when seat capacity exceeds the total representatives to ensure correct fuel calculation.

Is DFS the only way to solve this efficiently?

DFS is the most direct method for subtree aggregation and minimal fuel computation, although BFS can be adapted but is less straightforward.

terminal

Solution

Solution 1: Greedy + DFS

According to the problem description, we can find that all cars will only drive towards the capital (node $0$).

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class Solution:
    def minimumFuelCost(self, roads: List[List[int]], seats: int) -> int:
        def dfs(a: int, fa: int) -> int:
            nonlocal ans
            sz = 1
            for b in g[a]:
                if b != fa:
                    t = dfs(b, a)
                    ans += ceil(t / seats)
                    sz += t
            return sz

        g = defaultdict(list)
        for a, b in roads:
            g[a].append(b)
            g[b].append(a)
        ans = 0
        dfs(0, -1)
        return ans

Continue Practicing

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