What is the time complexity of Minimum Fuel Cost to Report to the Capital?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Fuel Cost to Report to the Capital?

This problem uses Tree, Depth-First Search, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Fuel Cost to Report to the Capital?

Insight 1: The problem can be modeled as a tree traversal, specifically a Depth-First Search (DFS), to calculate the number of representatives that need to travel from each city to the capital. Insight 2: The fuel needed between a city and its parent is determined by the number of representatives traveling from that city and the `seats` available in each car. We need to round up the number of cars needed using `(representatives + seats - 1) // seats`.

Which companies ask LeetCode 2477. Minimum Fuel Cost to Report to the Capital in interviews?

This problem is frequently asked at American Express, DRW.

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Minimum Fuel Cost to Report to the Capital - LeetCode 2477 Solution

Minimum Fuel Cost to Report to the Capital - Complete Solution Guide

Minimum Fuel Cost to Report to the Capital is LeetCode problem 2477, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 th

Detailed Explanation

The problem describes a country represented as a tree structure where each node is a city. The capital city is city 0. Each city has a representative that needs to attend a meeting in the capital. Each city also has a car with a specified number of `seats`. The objective is to find the minimum amount of fuel needed for all representatives to reach the capital. Representatives can share cars, and the cost of traveling between two adjacent cities is one liter of fuel. The problem gives an array `roads` describing the connections between cities, and an integer `seats` representing the car capacity.

Solution Approach

The provided solutions use a Depth-First Search (DFS) algorithm to traverse the tree structure. The DFS starts from the capital city (city 0) and recursively explores its children. For each city, the algorithm calculates the number of representatives that need to travel to the capital, including the representatives from its children. The fuel cost between a node and its parent is calculated based on the total number of representatives from the child node that need to travel to the capital, divided by the number of seats available in each car (rounding up to account for partially filled cars). The total fuel cost is accumulated during the DFS traversal.

Step-by-Step Algorithm

Step 1: Create an adjacency list representation of the graph from the given `roads` array. This allows efficient access to the neighbors of each city.
Step 2: Define a recursive DFS function that takes the current node and its parent node as input. This is crucial to avoid revisiting nodes and causing infinite loops.
Step 3: In the DFS function, initialize the number of representatives to 1 (representing the representative from the current city).
Step 4: Iterate through the neighbors of the current node. If a neighbor is not the parent node, recursively call the DFS function on that neighbor. Add the number of representatives returned by the recursive call to the current node's representative count.
Step 5: After visiting all children, if the current node is not the capital (node 0), calculate the fuel needed to travel from the current node to its parent node. The fuel is calculated as `(representatives + seats - 1) // seats`. Add this fuel to the total fuel cost.
Step 6: Return the number of representatives from the current node to the parent. This value is then used by the parent node to calculate its fuel cost.

Key Insights

Insight 1: The problem can be modeled as a tree traversal, specifically a Depth-First Search (DFS), to calculate the number of representatives that need to travel from each city to the capital.
Insight 2: The fuel needed between a city and its parent is determined by the number of representatives traveling from that city and the `seats` available in each car. We need to round up the number of cars needed using `(representatives + seats - 1) // seats`.
Insight 3: The base case for the recursion is when we reach a leaf node (a city with only one connection, which is the parent). In the DFS, keep track of the parent to avoid cycles.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Tree, Depth-First Search, Breadth-First Search, Graph.

Companies

Asked at: American Express, DRW.

Minimum Fuel Cost to Report to the Capital | LeetCode 2477

Medium

Tree Depth-First Search Breadth-First Search Graph

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    private long fuel = 0;

    public long minimumFuelCost(int[][] roads, int seats) {
        Map<Integer, List<Integer>> adj = new HashMap<>();
        for (int[] road : roads) {
            int u = road[0];
            int v = road[1];
            adj.computeIfAbsent(u, k -> new ArrayList<>()).add(v);
            adj.computeIfAbsent(v, k -> new ArrayList<>()).add(u);
        }

        this.fuel = 0;

        dfs(0, -1, adj, seats);
        return this.fuel;
    }

    private long dfs(int node, int parent, Map<Integer, List<Integer>> adj, int seats) {
        long representatives = 1;
        if (adj.containsKey(node)) {
            for (int neighbor : adj.get(node)) {
                if (neighbor != parent) {
                    representatives += dfs(neighbor, node, adj, seats);
                }
            }
        }
        

        if (node != 0) {
            this.fuel += (representatives + seats - 1) / seats;
        }

        return representatives;
    }
}

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Problem Info

Description

Problem 2477: Minimum Fuel Cost to Report to the Capital - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Graph

Companies

American ExpressDRW

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2477

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Create an adjacency list representation of the graph from the given `roads` array. This allows efficient access to the neighbors of each city.
2
Step 2: Define a recursive DFS function that takes the current node and its parent node as input. This is crucial to avoid revisiting nodes and causing infinite loops.
3
Step 3: In the DFS function, initialize the number of representatives to 1 (representing the representative from the current city).
4
Step 4: Iterate through the neighbors of the current node. If a neighbor is not the parent node, recursively call the DFS function on that neighbor. Add the number of representatives returned by the recursive call to the current node's representative count.
5
Step 5: After visiting all children, if the current node is not the capital (node 0), calculate the fuel needed to travel from the current node to its parent node. The fuel is calculated as `(representatives + seats - 1) // seats`. Add this fuel to the total fuel cost.
6
Step 6: Return the number of representatives from the current node to the parent. This value is then used by the parent node to calculate its fuel cost.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because each node and edge is visited a constant number of times during the DFS traversal. The space complexity is determined by the adjacency list, which stores the graph structure, and the call stack during recursion. Both of these contribute to a space complexity proportional to the number of cities.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

The problem can also be solved using BFS. However, using a standard BFS, it would be more difficult to track the number of representatives travelling on each edge. Therefore, DFS is more natural for this problem.

Time: O(n)Space: O(n)

Trade-offs: While BFS has the same time and space complexity, the iterative approach of BFS might be slightly more complex to implement in this specific case due to the need to track the number of representatives traveling up the tree.

✨

Key Insights

1Insight 1: The problem can be modeled as a tree traversal, specifically a Depth-First Search (DFS), to calculate the number of representatives that need to travel from each city to the capital.
2Insight 2: The fuel needed between a city and its parent is determined by the number of representatives traveling from that city and the `seats` available in each car. We need to round up the number of cars needed using `(representatives + seats - 1) // seats`.
3Insight 3: The base case for the recursion is when we reach a leaf node (a city with only one connection, which is the parent). In the DFS, keep track of the parent to avoid cycles.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: `roads` is empty (n=1). In this case, the function correctly returns 0, because there are no roads and no need for any travel.
Edge case 2: `seats` is very large, such that all representatives from a subtree can fit in one car. The `(representatives + seats - 1) // seats` calculation handles this correctly, ensuring that only one car is used.
Edge case 3: `seats` is 1. In this extreme case, each representative requires a separate car. The code correctly calculates the fuel cost as the sum of representatives traveling along each road.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where `roads` is empty (n=1). This results in an incorrect calculation or potential errors.

✗Common mistake 2: Forgetting to exclude the parent node when iterating through the neighbors of a node in the DFS. This can lead to infinite recursion or incorrect calculations.

✗Common mistake 3: Incorrectly calculating the number of cars needed. Using simple division instead of `(representatives + seats - 1) // seats` will lead to underestimation of the fuel cost.

✗Common mistake 4: Not using a long long for fuel and representatives count can lead to integer overflow for large values of n.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing transportation costs in a hierarchical organizational structure. Consider a company where employees need to travel to headquarters for a meeting. This problem can be used to minimize the fuel cost by optimizing carpooling based on the organizational hierarchy.

🏦Real-world scenario 2: Network routing. The tree structure and fuel cost minimization can be adapted to optimize the routing of data packets in a network, where fuel represents the cost of transmitting data and seats represent the bandwidth capacity of a link.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 16, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Fuel Cost to Report to the Capital"?

For this medium-level Tree problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?