What is the time complexity of Maximum Total Importance of Roads?

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

What data structures are used to solve Maximum Total Importance of Roads?

This problem uses Greedy, Graph, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Total Importance of Roads?

Insight 1: The key idea is that cities with higher degrees (i.e., cities connected to more roads) should be assigned higher values (close to `n`) to maximize the overall importance. Insight 2: Sorting the degrees of cities allows us to greedily assign values from 1 to `n` to cities based on their degree, starting with the lowest degree.

Which companies ask LeetCode 2285. Maximum Total Importance of Roads in interviews?

This problem is frequently asked at Hudson River Trading.

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Maximum Total Importance of Roads - LeetCode 2285 Solution

Maximum Total Importance of Roads - Complete Solution Guide

Maximum Total Importance of Roads is LeetCode problem 2285, 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

You are given an integer n denoting the number of cities in a country. The cities are numbered from 0 to n - 1 . You are also 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 . You need to assign each city with an integer value from 1 to n , where each value can only be used once . The importance of a road is then defined as the sum of the values of the two cities it connects. Return the maximum total import

Detailed Explanation

The problem asks us to assign integer values from 1 to `n` (inclusive) to `n` cities in a way that maximizes the total importance of all roads. The importance of a road is the sum of the values assigned to the two cities it connects. We're given the number of cities `n` and a list of roads `roads`, where each road is a pair of city indices. The goal is to find the maximum possible sum of road importances after the optimal value assignment.

Solution Approach

The solution uses a greedy approach. First, it calculates the degree of each city by iterating through the `roads` array and incrementing the degree count for each city connected by a road. Then, it sorts the cities based on their degrees in ascending order. Finally, it iterates through the sorted degrees, assigning values from 1 to `n` to the cities, and calculates the total importance by summing the product of each city's degree and its assigned value. This approach works because assigning larger values to cities with more connections yields a larger overall importance.

Step-by-Step Algorithm

Step 1: Initialize an array `degree` of size `n` with all elements set to 0. This array will store the degree (number of connections) of each city.
Step 2: Iterate through the `roads` array. For each road `(u, v)`, increment `degree[u]` and `degree[v]` by 1. This calculates the degree of each city.
Step 3: Sort the `degree` array in ascending order. This step is crucial for the greedy assignment strategy.
Step 4: Initialize `total_importance` to 0. This variable will store the sum of the importances of all roads.
Step 5: Iterate through the sorted `degree` array from index 0 to `n-1`. For each city `i`, multiply its degree `degree[i]` by `i+1` (its assigned value), and add the result to `total_importance`.
Step 6: Return `total_importance`.

Key Insights

Insight 1: The key idea is that cities with higher degrees (i.e., cities connected to more roads) should be assigned higher values (close to `n`) to maximize the overall importance.
Insight 2: Sorting the degrees of cities allows us to greedily assign values from 1 to `n` to cities based on their degree, starting with the lowest degree.
Insight 3: The problem is inherently greedy because assigning the highest value to the node with the highest degree will always yield a better or equal result than assigning it a lower value.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Greedy, Graph, Sorting, Heap (Priority Queue).

Companies

Asked at: Hudson River Trading.

Maximum Total Importance of Roads | LeetCode 2285

Medium

Greedy Graph Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long maximumImportance(int n, int[][] roads) {
        long[] degree = new long[n];
        for (int[] road : roads) {
            degree[road[0]] += 1;
            degree[road[1]] += 1;
        }
        
        Arrays.sort(degree);
        
        long total_importance = 0;
        for (int i = 0; i < n; i++) {
            total_importance += degree[i] * (i + 1);
        }
        
        return total_importance;
    }
}

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Problem Info

Description

Problem 2285: Maximum Total Importance of Roads - Detailed explanation and solution approaches.

Topics

Greedy Graph Sorting Heap (Priority Queue)

Companies

Hudson River Trading

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2285

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: Initialize an array `degree` of size `n` with all elements set to 0. This array will store the degree (number of connections) of each city.
2
Step 2: Iterate through the `roads` array. For each road `(u, v)`, increment `degree[u]` and `degree[v]` by 1. This calculates the degree of each city.
3
Step 3: Sort the `degree` array in ascending order. This step is crucial for the greedy assignment strategy.
4
Step 4: Initialize `total_importance` to 0. This variable will store the sum of the importances of all roads.
5
Step 5: Iterate through the sorted `degree` array from index 0 to `n-1`. For each city `i`, multiply its degree `degree[i]` by `i+1` (its assigned value), and add the result to `total_importance`.
6
Step 6: Return `total_importance`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the sorting algorithm used to sort the degrees of the cities. The space complexity comes from the auxiliary space used to store the degrees of each city.

📖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

Heap (Priority Queue)

Instead of sorting the degrees directly, we can use a min-heap (priority queue) to store the degrees. We extract elements from the heap one by one and assign values from 1 to n accordingly. This approach has the same time complexity but might be slightly less efficient in practice due to the overhead of heap operations.

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

Trade-offs: The heap approach is conceptually different but achieves the same time complexity. Sorting is generally faster in practice than using a priority queue for this specific problem size due to the overhead of heap operations.

✨

Key Insights

1Insight 1: The key idea is that cities with higher degrees (i.e., cities connected to more roads) should be assigned higher values (close to `n`) to maximize the overall importance.
2Insight 2: Sorting the degrees of cities allows us to greedily assign values from 1 to `n` to cities based on their degree, starting with the lowest degree.
3Insight 3: The problem is inherently greedy because assigning the highest value to the node with the highest degree will always yield a better or equal result than assigning it a lower value.

📚

Key Terms Explained

For Students

Graph

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: A single road exists connecting two cities. The solution still works correctly because it will assign 1 to one city and 2 to the other, calculating the importance accordingly.
Edge case 2: No roads exist. The solution will initialize the degrees to 0, sort them (which has no effect), and then calculate the total importance as 0, which is the correct result.
Edge case 3: A fully connected graph where every city is connected to every other city. The algorithm will correctly calculate the degree of each city (n-1), sort them, and assign the highest values to cities with the highest degree.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not realizing the importance of sorting. Assigning values arbitrarily without considering the degrees of the cities will not lead to the maximum possible total importance.

✗Common mistake 2: Using integers instead of long long for `total_importance` in languages like C++ and Java. The total importance can become very large and overflow if integers are used.

✗Common mistake 3: Forgetting that the importance of a road is the *sum* of the city values, not the product or some other operation.

✅

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 network traffic. Cities can be represented as network nodes, and roads as network links. Assigning higher bandwidth to nodes with more connections can maximize overall network throughput, analogous to maximizing total importance.

🏦Real-world scenario 2: Resource allocation in a transportation network. Cities represent locations, and roads represent transportation routes. Assigning more resources (e.g., trucks, deliveries) to locations with more connecting routes can improve overall efficiency and minimize transportation costs, similar to maximizing importance.

💡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

Greedy works when local optimal leads to global optimal. Always try to prove (or disprove) the greedy choice property before coding.

🎯 Interview Tip

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

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 4, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

Frequently Asked Questions

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

When should I use BFS vs DFS for graph problems?

Use BFS for shortest path in unweighted graphs, level-order processing, or when exploring neighbors systematically. Use DFS for exhaustive search, detecting cycles, topological sort, or when you need to explore complete paths. DFS uses less memory for very wide graphs.

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 "Maximum Total Importance of Roads"?

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

🤖 Need help with this problem?