LeetCode 2477 - Minimum Fuel Cost to Report to the Capital

The problem gives us a country road network that forms a tree. A tree is a connected graph with no cycles, which means there is exactly one path between any two cities. The cities are numbered from 0 to n - 1, and city 0 is always the capital.

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LeetCode Problem 2477

Difficulty: 🟡 Medium
Topics: Tree, Depth-First Search, Breadth-First Search, Graph Theory

Solution

Problem Understanding

The problem gives us a country road network that forms a tree. A tree is a connected graph with no cycles, which means there is exactly one path between any two cities. The cities are numbered from 0 to n - 1, and city 0 is always the capital.

Each city has exactly one representative who must travel to the capital city for a meeting. Every city also has a car, and every car has the same number of seats, given by seats.

Traveling along one road costs exactly one liter of fuel, regardless of how many passengers are in the car. Representatives are allowed to switch cars and share rides. The objective is to minimize the total amount of fuel consumed so that every representative reaches the capital.

The input consists of:

  • roads, a list of undirected edges representing the tree
  • seats, the seating capacity of every car

The output is the minimum total liters of fuel required.

The constraints are important:

  • n can be as large as 10^5
  • The graph is guaranteed to be a valid tree
  • A solution slower than linear or near linear time will likely time out

Because the graph is a tree, there is exactly one route from every city to the capital. This removes pathfinding ambiguity and strongly suggests a traversal based solution such as DFS or BFS.

There are several important edge cases:

  • A country with only the capital city, roads = []. No travel is needed, so the answer is 0.
  • seats = 1, meaning nobody can share rides. Every representative effectively travels independently.
  • Very large seat counts, where many representatives can merge into one car before moving upward.
  • Deep tree structures, which can expose recursion depth or traversal issues if implemented carelessly.

The key observation is that every representative must eventually move upward toward the capital along the unique tree path, so the optimization comes entirely from intelligently grouping passengers together.

Approaches

Brute Force Approach

A brute force strategy would attempt to simulate all possible ways representatives could combine cars and share rides while traveling toward the capital.

For every city, we could consider:

  • Whether the representative drives alone
  • Which intermediate city they merge at
  • Which car they continue riding in afterward

This quickly becomes combinatorially explosive because every subtree may combine representatives in many different ways.

Even if we tried dynamic programming over subsets or explored all grouping combinations, the number of possibilities grows exponentially with the number of cities. Such an approach is completely infeasible for n = 10^5.

The brute force method is correct because it explicitly examines all valid transportation arrangements, but it is far too slow.

Optimal Insight

The crucial insight is that in a tree, every representative has exactly one path to the capital.

This means we do not need to decide which route people take. The only decision is how many cars are needed when representatives move upward from a subtree to its parent.

Suppose a subtree contains k representatives. To move all of them one edge upward toward the capital, we need:

$\left\lceil \frac{k}{\text{seats}} \right\rceil$

cars crossing that edge.

Each such crossing costs one liter of fuel.

Therefore, for every edge in the tree, the optimal cost depends only on how many representatives must cross that edge.

This naturally leads to a postorder DFS:

  1. Compute the number of representatives in each subtree.
  2. For every non-capital node:
  • Determine how many cars are needed to move its subtree representatives upward.
  • Add that number to the answer.

This works because representatives inside a subtree can always optimally consolidate before leaving the subtree.

Approach Time Complexity Space Complexity Notes
Brute Force Exponential Exponential Tries all ride-sharing combinations
Optimal DFS O(n) O(n) Aggregates subtree representatives bottom up

Algorithm Walkthrough

Step 1: Build the adjacency list

Convert the roads edge list into an adjacency list representation.

Because the graph is undirected, every road [a, b] is added in both directions.

This allows efficient traversal of neighboring cities.

Step 2: Start DFS from the capital

We perform a depth first traversal beginning at city 0.

The DFS function will return the total number of representatives in the current subtree.

Every city contributes one representative, namely itself.

Step 3: Traverse child subtrees

For each neighboring city:

  • Skip the parent city to avoid revisiting nodes
  • Recursively process the child subtree
  • Receive the number of representatives coming from that subtree

This gives us the total people that eventually need to move upward through the edge connecting the child to the current node.

Step 4: Compute fuel cost for the child subtree

If a child subtree contains people representatives, then the number of cars needed to move them one edge upward is:

$\left\lceil \frac{\text{people}}{\text{seats}} \right\rceil$

Each car crossing one edge consumes one liter of fuel.

Therefore we add this value to the global answer.

Step 5: Accumulate representatives

After processing all children, return:

  • 1 for the current city's representative
  • plus all representatives from child subtrees

This value propagates upward toward the capital.

Step 6: Finish traversal

The capital itself does not need to travel upward, so we never charge fuel for node 0.

The accumulated global answer is the minimum fuel cost.

Why it works

The algorithm relies on the fact that every representative must cross every edge on the unique path to the capital exactly once.

For any subtree, the only thing that matters is how many representatives leave that subtree. Since cars can be reorganized freely inside the subtree before crossing upward, the optimal strategy is always to pack cars as fully as possible.

The DFS computes subtree sizes exactly once, and for every edge calculates the minimum number of cars required to transport those representatives upward. Since every edge contribution is optimal independently, the total fuel cost is globally optimal.

Python Solution

from typing import List
from collections import defaultdict
import math

class Solution:
    def minimumFuelCost(self, roads: List[List[int]], seats: int) -> int:
        n = len(roads) + 1
        
        graph = defaultdict(list)
        
        for a, b in roads:
            graph[a].append(b)
            graph[b].append(a)
        
        fuel_cost = 0
        
        def dfs(node: int, parent: int) -> int:
            nonlocal fuel_cost
            
            representatives = 1
            
            for neighbor in graph[node]:
                if neighbor == parent:
                    continue
                
                people = dfs(neighbor, node)
                
                fuel_cost += math.ceil(people / seats)
                
                representatives += people
            
            return representatives
        
        dfs(0, -1)
        
        return fuel_cost

The implementation begins by constructing an adjacency list using defaultdict(list). Since the graph is undirected, each edge is inserted in both directions.

The variable fuel_cost stores the running total answer.

The DFS function returns the number of representatives contained in the subtree rooted at the current node.

Every node initially contributes one representative, itself. Then we recursively process each child subtree.

For every child:

  • We recursively compute the number of representatives in that subtree.
  • We determine how many cars are needed to move those representatives upward by using ceiling division.
  • We add that fuel cost to the global answer.
  • We accumulate the representatives into the current subtree total.

Finally, the DFS starts at the capital city 0. Since the capital does not move upward, no fuel is charged for it.

Go Solution

package main

func minimumFuelCost(roads [][]int, seats int) int64 {
	n := len(roads) + 1

	graph := make([][]int, n)

	for _, road := range roads {
		a, b := road[0], road[1]

		graph[a] = append(graph[a], b)
		graph[b] = append(graph[b], a)
	}

	var fuelCost int64

	var dfs func(int, int) int

	dfs = func(node int, parent int) int {
		representatives := 1

		for _, neighbor := range graph[node] {
			if neighbor == parent {
				continue
			}

			people := dfs(neighbor, node)

			fuelCost += int64((people + seats - 1) / seats)

			representatives += people
		}

		return representatives
	}

	dfs(0, -1)

	return fuelCost
}

The Go implementation follows the same logic as the Python version.

One important difference is integer handling. The final answer may exceed the range of a 32 bit integer, so the result is stored as int64.

Instead of using math.Ceil, Go performs ceiling division with:

(people + seats - 1) / seats

The adjacency list is represented with a slice of slices rather than a hash map for better performance.

Worked Examples

Example 1

Input:

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

Tree:

    0
   /|\
  1 2 3

Each leaf city has one representative.

DFS processing:

Node Representatives in Subtree Cars Needed Upward Fuel Added
1 1 ceil(1 / 5) = 1 1
2 1 ceil(1 / 5) = 1 1
3 1 ceil(1 / 5) = 1 1

Total fuel:

1 + 1 + 1 = 3

Final answer:

3

Example 2

Input:

roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]]
seats = 2

Tree:

        0
      / | \
     1  4  5
     |   \
     3    6
     |
     2

Process bottom up.

Node 2

Value Result
Representatives 1
Cars upward ceil(1 / 2) = 1
Fuel added 1

Node 3

Receives one representative from node 2.

Value Result
Representatives 1 + 1 = 2
Cars upward ceil(2 / 2) = 1
Fuel added 1

Node 1

Receives two representatives from node 3.

Value Result
Representatives 1 + 2 = 3
Cars upward ceil(3 / 2) = 2
Fuel added 2

Node 6

Value Result
Representatives 1
Cars upward 1
Fuel added 1

Node 4

Receives one representative from node 6.

Value Result
Representatives 2
Cars upward 1
Fuel added 1

Node 5

Value Result
Representatives 1
Cars upward 1
Fuel added 1

Total fuel:

1 + 1 + 2 + 1 + 1 + 1 = 7

Final answer:

7

Example 3

Input:

roads = []
seats = 1

There is only the capital city.

No representatives need to travel.

Total fuel:

0

Complexity Analysis

Measure Complexity Explanation
Time O(n) Every node and edge is visited once
Space O(n) Adjacency list and recursion stack

The DFS traverses each edge exactly twice, once in each direction, which gives linear time complexity.

The adjacency list stores all edges, requiring O(n) space. The recursion stack may also reach O(n) in a worst case skewed tree.

Test Cases

from typing import List

class Solution:
    def minimumFuelCost(self, roads: List[List[int]], seats: int) -> int:
        from collections import defaultdict
        import math

        n = len(roads) + 1
        graph = defaultdict(list)

        for a, b in roads:
            graph[a].append(b)
            graph[b].append(a)

        fuel = 0

        def dfs(node, parent):
            nonlocal fuel

            reps = 1

            for neighbor in graph[node]:
                if neighbor == parent:
                    continue

                people = dfs(neighbor, node)

                fuel += math.ceil(people / seats)

                reps += people

            return reps

        dfs(0, -1)

        return fuel

solution = Solution()

assert solution.minimumFuelCost([[0,1],[0,2],[0,3]], 5) == 3
# basic star graph

assert solution.minimumFuelCost(
    [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], 2
) == 7
# provided complex example

assert solution.minimumFuelCost([], 1) == 0
# only capital city

assert solution.minimumFuelCost([[0,1]], 1) == 1
# smallest nontrivial tree with single seat

assert solution.minimumFuelCost([[0,1]], 10) == 1
# large seat count does not reduce below one trip

assert solution.minimumFuelCost([[0,1],[1,2],[2,3]], 2) == 4
# chain structure

assert solution.minimumFuelCost([[0,1],[0,2],[0,3],[0,4]], 2) == 4
# star graph with multiple leaves

assert solution.minimumFuelCost([[0,1],[1,2],[1,3],[3,4]], 3) == 4
# mixed branching structure

assert solution.minimumFuelCost([[0,1],[1,2],[2,3],[3,4]], 5) == 4
# enough seats to combine everyone in one car eventually

assert solution.minimumFuelCost([[0,1],[1,2],[1,3],[3,4],[3,5]], 1) == 11
# no ride sharing possible
Test Why
[[0,1],[0,2],[0,3]], seats=5 Validates basic star graph
complex example with seats=2 Validates subtree aggregation
roads=[] Tests empty tree edge case
single edge with seats=1 Smallest travel case
single edge with large seats Ensures minimum one crossing
linear chain Tests deep traversal
star graph Tests many independent leaves
mixed branching tree Tests subtree merging
chain with large seats Tests aggressive consolidation
seats=1 on larger tree Tests no sharing behavior

Edge Cases

Single Capital City

When roads = [], the country contains only city 0.

This is easy to mishandle because many graph algorithms assume at least one edge exists. In this case, no representatives travel anywhere, so the answer must be 0.

The implementation handles this naturally because the DFS starts at node 0, finds no neighbors, and never adds fuel cost.

Seat Count Equals One

When seats = 1, ride sharing becomes impossible.

Every representative effectively requires their own car for every edge traversal. This creates the maximum possible fuel consumption and can expose incorrect grouping logic.

The implementation still works because:

$\left\lceil \frac{k}{1} \right\rceil = k$

So every representative crossing an edge contributes exactly one fuel unit.

Deep Linear Tree

A chain shaped tree such as:

0 - 1 - 2 - 3 - 4

can expose recursion depth or accumulation bugs.

In this structure, representatives progressively merge while moving upward. Incorrect subtree counting would produce wrong results.

The DFS implementation correctly aggregates representatives bottom up, ensuring every subtree contributes the correct number of cars to its parent edge.