Problem Description

You have a country network structured as a tree with n cities numbered from 0 to n - 1, connected by exactly n - 1 bidirectional roads. The capital city is city 0.

Each city has one car with a fixed number of seats (given by the parameter seats). There's a meeting in the capital, and one representative from each city needs to attend.

Representatives can travel in the following ways:

• Use the car from their own city
• Switch cars and ride with another representative along the way

The cost of traveling between any two directly connected cities is 1 liter of fuel, regardless of how many people are in the car.

Your task is to find the minimum total fuel consumption needed for all representatives to reach the capital city.

For example, if roads = [[0,1],[0,2],[0,3]] and seats = 2:

• City 1's representative drives to city 0 (1 liter)
• City 2's representative drives to city 0 (1 liter)
• City 3's representative drives to city 0 (1 liter)
• Total: 3 liters

But if roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]] and seats = 2:

• Representatives from cities 2 and 3 can share a car to city 1
• Then representatives from cities 1, 2, 3 can optimally arrange themselves in cars to reach city 0
• The algorithm calculates the minimum fuel needed by considering how many cars are required at each edge based on the number of representatives passing through

The key insight is that representatives should carpool when possible to minimize the number of cars (and thus fuel) used on each road segment.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem explicitly describes a tree structure (a connected, undirected graph with no cycles) consisting of n cities and n-1 roads. The cities are nodes and the roads are edges.

Is it a tree?

• Yes: The problem states "There is a tree (i.e., a connected, undirected graph with no cycles) structure country network." With n cities and exactly n-1 roads forming a connected graph, this is indeed a tree.

DFS

• Conclusion reached: Since we have a tree structure, the flowchart leads us directly to using DFS (Depth-First Search).

Why DFS is the Right Choice

The DFS pattern is particularly suitable for this problem because:

1. Tree Traversal: We need to traverse the entire tree starting from the capital (node 0) to calculate fuel costs from all cities.

2. Bottom-up Calculation: We need to calculate the number of representatives coming from each subtree. DFS allows us to process leaf nodes first, then aggregate information as we backtrack to parent nodes.

3. Subtree Information: For each node, we need to know how many representatives are coming from its subtree to determine the minimum number of cars needed on each edge. DFS naturally provides this through its recursive nature.

4. Parent-Child Relationships: The solution requires tracking which node we came from (parent) to avoid revisiting it. DFS with a parent parameter handles this elegantly.

The DFS approach allows us to calculate sz (subtree size) for each node and determine the minimum fuel cost as ⌈sz/seats⌉ for each edge connecting a subtree to its parent.

Intuition

Let's think about how representatives travel to the capital. Since we have a tree structure, there's only one path from any city to the capital. This means representatives don't have choices about which route to take - they must follow the unique path to city 0.

The key insight is that representatives naturally converge as they move toward the capital. Imagine representatives from leaf cities starting their journey. When multiple representatives meet at a common node on their way to the capital, they can carpool to save fuel.

Consider a simple example: if cities 2 and 3 both connect to city 1, and city 1 connects to the capital, then representatives from cities 2 and 3 will both pass through city 1. If each car has 2 seats, these two representatives can share one car from city 1 to the capital instead of driving separately.

This leads us to think about the problem from each edge's perspective. For any edge in the tree, we need to determine: how many representatives need to cross this edge to reach the capital? If sz representatives need to cross an edge and each car has seats capacity, we need at least ⌈sz/seats⌉ cars. Since each car consumes 1 liter of fuel to traverse an edge, this is also the fuel cost for that edge.

The beauty of using DFS is that it naturally computes subtree sizes. When we're at a node during DFS traversal, we can calculate how many representatives are in its subtree (including itself). This number tells us exactly how many representatives need to cross the edge from this node to its parent.

By traversing the tree from the capital (root) and recursively calculating subtree sizes, we can determine the minimum fuel needed for each edge. The sum of fuel costs across all edges gives us the total minimum fuel consumption.

The greedy aspect is implicit: at each node, we're packing representatives into the minimum number of cars possible before sending them to the parent node. This local optimization at each node leads to the global optimum.

Learn more about Tree, Depth-First Search, Breadth-First Search and Graph patterns.

Solution Approach

The solution implements a DFS traversal starting from the capital (node 0) to calculate the minimum fuel cost. Let's break down the implementation:

Graph Construction: First, we build an adjacency list representation of the tree using a defaultdict(list). Since the roads are bidirectional, we add edges in both directions:

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

DFS Function: The core logic is in the dfs(a, fa) function where:

• a is the current node being visited
• fa is the parent node (to avoid revisiting)
• The function returns sz, the total number of nodes in the subtree rooted at a

Calculating Subtree Sizes and Fuel Cost: For each node a, we:

1. Initialize sz = 1 (counting the current node itself)
2. Traverse all neighboring nodes b (children in the tree context)
3. Skip the parent node fa to avoid going backward
4. Recursively calculate the subtree size t for each child b
5. Calculate the fuel needed from child b to current node a: ⌈t/seats⌉
6. Add this fuel cost to our answer
7. Accumulate the subtree size: sz += t

The mathematical formula ⌈t/seats⌉ is implemented using Python's ceil function, which gives us the minimum number of cars needed to transport t representatives with seats capacity per car.

Why Start from Node 0: Starting DFS from the capital (node 0) with parent -1 ensures we traverse the entire tree. As we recurse down to leaves and backtrack, we calculate fuel costs for each edge exactly once.

Time and Space Complexity:

• Time: O(n) where n is the number of cities, as we visit each node once
• Space: O(n) for the adjacency list and recursion stack

The elegance of this solution lies in how it leverages the tree property: each edge is traversed exactly once, and the subtree size directly tells us how many representatives need to cross that edge toward the capital.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example:

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

This forms a tree where:

    0 (capital)
    |
    1
   / \
  2   3

Step-by-step DFS traversal:

1. Start DFS from node 0 (capital) with parent -1
   • Visit node 0's children: node 1
2. DFS enters node 1 with parent 0
   • Initialize sz = 1 (node 1 itself)
   • Visit node 1's children: nodes 2 and 3 (skip 0 as it's the parent)
3. DFS enters node 2 with parent 1
   • Initialize sz = 1 (node 2 itself)
   • No unvisited children (node 1 is parent)
   • Return sz = 1 to node 1
4. Back at node 1, process edge 2→1
   • Received t = 1 from node 2
   • Calculate cars needed: ⌈1/2⌉ = 1 car
   • Add 1 liter to total fuel
   • Update node 1's subtree size: sz = 1 + 1 = 2
5. DFS enters node 3 with parent 1
   • Initialize sz = 1 (node 3 itself)
   • No unvisited children
   • Return sz = 1 to node 1
6. Back at node 1, process edge 3→1
   • Received t = 1 from node 3
   • Calculate cars needed: ⌈1/2⌉ = 1 car
   • Add 1 liter to total fuel
   • Update node 1's subtree size: sz = 2 + 1 = 3
   • Return sz = 3 to node 0
7. Back at node 0, process edge 1→0
   • Received t = 3 from node 1 (3 representatives total from cities 1, 2, and 3)
   • Calculate cars needed: ⌈3/2⌉ = 2 cars
   • Add 2 liters to total fuel

Total fuel consumption: 1 + 1 + 2 = 4 liters

Understanding the result:

• Representative from city 2 drives to city 1: 1 liter
• Representative from city 3 drives to city 1: 1 liter
• All 3 representatives (from cities 1, 2, 3) need to travel from city 1 to city 0
• With 2 seats per car, they need 2 cars for this journey: 2 liters

The algorithm correctly computes that we need 4 liters total by calculating the minimum cars required for each edge based on how many representatives pass through it.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs a depth-first search (DFS) traversal of the tree. Each node is visited exactly once during the traversal. For each node, we iterate through its adjacent nodes (edges). Since this is a tree structure with n nodes, there are exactly n-1 edges, and each edge is traversed twice (once from each direction). Therefore, the total number of operations is proportional to the number of nodes plus the number of edges, which gives us O(n + (n-1)) = O(n).

Space Complexity: O(n)

The space complexity consists of several components:

• The adjacency list g stores all edges. For a tree with n nodes, there are n-1 edges, and each edge is stored twice (bidirectional), requiring O(2(n-1)) = O(n) space.
• The recursive call stack for DFS can go as deep as the height of the tree. In the worst case (a skewed tree), this could be O(n).
• Other variables like ans, sz, and loop variables use O(1) space.

Therefore, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Handling the Root Node (Capital)

Pitfall: A common mistake is trying to calculate fuel cost for representatives traveling TO the capital from the capital itself, or counting the capital's representative in the fuel calculation.

Why it happens: The problem states each city has a representative that needs to reach the capital. It's easy to misinterpret this as including city 0's representative needing to "travel" to city 0.

Example of incorrect approach:

def dfs(current_node, parent_node):
    subtree_size = 1
    for neighbor in graph[current_node]:
        if neighbor != parent_node:
            people_from_subtree = dfs(neighbor, current_node)
            # WRONG: Calculating fuel even when current_node is 0
            cars_needed = ceil(people_from_subtree / seats)
            total_fuel += cars_needed
            subtree_size += people_from_subtree
  
    # WRONG: Adding fuel cost for root node itself
    if current_node != 0:
        return subtree_size
    else:
        total_fuel += ceil(subtree_size / seats)
        return subtree_size

Solution: The correct implementation already handles this properly - we only calculate fuel for edges between nodes, not for the destination node itself. The capital's representative doesn't need to travel anywhere.

2. Integer Division Instead of Ceiling Division

Pitfall: Using integer division // or regular division / without ceiling when calculating the number of cars needed.

Wrong implementation:

# WRONG: This would undercount cars needed
cars_needed = people_from_subtree // seats  # Loses remainder
# or
cars_needed = int(people_from_subtree / seats)  # Truncates down

Why it's wrong: If 5 people need to travel and each car has 2 seats, we need ⌈5/2⌉ = 3 cars, not 5//2 = 2 cars.

Solution: Always use ceil() function:

from math import ceil
cars_needed = ceil(people_from_subtree / seats)

3. Double-Counting Edges in Undirected Graph

Pitfall: Forgetting to track the parent node during DFS traversal, causing the algorithm to revisit nodes and double-count fuel costs.

Wrong implementation:

def dfs(current_node):  # Missing parent parameter
    subtree_size = 1
    for neighbor in graph[current_node]:
        # WRONG: No check to avoid going back to parent
        people_from_subtree = dfs(neighbor)
        # This would cause infinite recursion or double counting

Solution: Always pass and check the parent node:

def dfs(current_node, parent_node):
    subtree_size = 1
    for neighbor in graph[current_node]:
        if neighbor != parent_node:  # Critical check
            people_from_subtree = dfs(neighbor, current_node)

4. Misunderstanding When to Add Fuel Cost

Pitfall: Adding fuel cost at the wrong point in the recursion, such as when entering a node rather than when returning from children.

Why it matters: The fuel cost for an edge should be calculated based on how many people need to cross that edge toward the capital. This number is only known after exploring the entire subtree.

Correct approach: Add fuel cost after the recursive call returns, using the subtree size information:

people_from_subtree = dfs(neighbor, current_node)
# Now we know how many people need to cross the edge from neighbor to current_node
cars_needed = ceil(people_from_subtree / seats)
total_fuel += cars_needed