What is the time complexity of Maximal Network Rank?

The optimal solution has a time complexity of O(n^2 + m).

What is the space complexity of Maximal Network Rank?

The space complexity is O(n^2).

What data structures are used to solve Maximal Network Rank?

This problem uses Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximal Network Rank?

Insight 1: The network rank of two cities is the sum of their degrees, minus 1 if they are directly connected. Insight 2: We need to efficiently determine the degree of each city (number of directly connected roads) and whether two cities are directly connected.

Which companies ask LeetCode 1615. Maximal Network Rank in interviews?

This problem is frequently asked at DRW, smartnews.

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Maximal Network Rank - LeetCode 1615 Solution

Maximal Network Rank - Complete Solution Guide

Maximal Network Rank is LeetCode problem 1615, 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 an infrastructure of n cities with some number of roads connecting these cities. Each roads[i] = [a i , b i ] indicates that there is a bidirectional road between cities a i and b i . The network rank of two different cities is defined as the total number of directly connected roads to either city. If a road is directly connected to both cities, it is only counted once . The maximal network rank of the infrastructure is the maximum network rank of all pairs of different cities. Given th

Detailed Explanation

The problem asks us to find the 'maximal network rank' of a network of cities and roads. We're given the number of cities (n) and a list of roads, where each road connects two cities bidirectionally. The network rank of *two distinct cities* is the total number of roads directly connected to either of those cities. If there's a road connecting both cities, it's counted only once. The goal is to find the highest possible network rank among all pairs of distinct cities in the network. Constraints include the number of cities being between 2 and 100, and roads being represented as pairs of city indices from 0 to n-1.

Solution Approach

The solution uses a brute-force approach to check all possible pairs of cities. It first calculates the degree of each city by iterating through the `roads` array. It also builds an adjacency matrix `adj` to quickly determine if two cities are directly connected. Then, it iterates through all pairs of distinct cities, calculates their network rank (sum of degrees, minus 1 if connected), and keeps track of the maximum rank found so far.

Step-by-Step Algorithm

Step 1: Initialize `degree` array of size `n` to store the degree of each city (number of connected roads). Initialize `adj` (adjacency matrix) as an `n x n` boolean matrix to indicate road connections between cities.
Step 2: Iterate through the `roads` array. For each road `[u, v]`, increment `degree[u]` and `degree[v]`. Set `adj[u][v]` and `adj[v][u]` to `true` to indicate a connection between cities `u` and `v`.
Step 3: Initialize `max_rank` to 0.
Step 4: Iterate through all possible pairs of cities `(i, j)` where `0 <= i < n` and `i + 1 <= j < n`. This avoids duplicate pairs and considers only distinct cities.
Step 5: Calculate the network rank for the current pair `(i, j)`: `current_rank = degree[i] + degree[j]`. If `adj[i][j]` is `true` (cities are directly connected), subtract 1 from `current_rank`.
Step 6: Update `max_rank` to the maximum of `max_rank` and `current_rank`.
Step 7: After iterating through all pairs, return `max_rank`.

Key Insights

Insight 1: The network rank of two cities is the sum of their degrees, minus 1 if they are directly connected.
Insight 2: We need to efficiently determine the degree of each city (number of directly connected roads) and whether two cities are directly connected.
Insight 3: Brute-force checking all pairs of distinct cities is feasible given the constraint on the number of cities (n <= 100).

Complexity Analysis

Time Complexity: O(n^2 + m)

Space Complexity: O(n^2)

Topics

This problem involves: Graph.

Companies

Asked at: DRW, smartnews.

Maximal Network Rank | LeetCode 1615

Medium

Graph

Time: O(n^2 + m)Space: O(n^2)

Solution Code

Available in:

class Solution {
    public int maximalNetworkRank(int n, int[][] roads) {
        int[] degree = new int[n];
        boolean[][] adj = new boolean[n][n];

        for (int[] road : roads) {
            int u = road[0];
            int v = road[1];
            degree[u]++;
            degree[v]++;
            adj[u][v] = true;
            adj[v][u] = true;
        }

        int max_rank = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int current_rank = degree[i] + degree[j];
                if (adj[i][j]) {
                    current_rank--;
                }

                max_rank = Math.max(max_rank, current_rank);
            }
        }

        return max_rank;
    }
}

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Problem Info

Description

Problem 1615: Maximal Network Rank - Detailed explanation and solution approaches.

Topics

Graph

Companies

DRWsmartnews

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1615

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

7 steps

1
Step 1: Initialize `degree` array of size `n` to store the degree of each city (number of connected roads). Initialize `adj` (adjacency matrix) as an `n x n` boolean matrix to indicate road connections between cities.
2
Step 2: Iterate through the `roads` array. For each road `[u, v]`, increment `degree[u]` and `degree[v]`. Set `adj[u][v]` and `adj[v][u]` to `true` to indicate a connection between cities `u` and `v`.
3
Step 3: Initialize `max_rank` to 0.
4
Step 4: Iterate through all possible pairs of cities `(i, j)` where `0 <= i < n` and `i + 1 <= j < n`. This avoids duplicate pairs and considers only distinct cities.
5
Step 5: Calculate the network rank for the current pair `(i, j)`: `current_rank = degree[i] + degree[j]`. If `adj[i][j]` is `true` (cities are directly connected), subtract 1 from `current_rank`.
6
Step 6: Update `max_rank` to the maximum of `max_rank` and `current_rank`.
7
Step 7: After iterating through all pairs, return `max_rank`.

💡

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^2 + m)

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is derived from the nested loops used to compare all pairs of cities, which results in O(n^2). The space complexity is determined by the size of the adjacency matrix, which is needed to efficiently check if two cities are directly connected.

📖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

Sorting Degrees and Greedy Approach

Sort the degrees of all cities in descending order. Then, consider the two cities with the highest degrees. If they are connected, decrement the rank by 1. This approach attempts to find the optimal pair by prioritizing high-degree cities.

Time: O(n log n + m), where n is the number of cities and m is the number of roads. Sorting takes O(n log n), and iterating through the roads takes O(m).Space: O(n + m). O(n) to store degrees and O(m) to optionally store edges

Trade-offs: This approach is harder to implement correctly. The brute-force approach is simpler and efficient enough given the constraints. The sorting approach *might* improve performance if there were significantly more cities and fewer roads, but it's more complex and not guaranteed to be faster in all cases.

✨

Key Insights

1Insight 1: The network rank of two cities is the sum of their degrees, minus 1 if they are directly connected.
2Insight 2: We need to efficiently determine the degree of each city (number of directly connected roads) and whether two cities are directly connected.
3Insight 3: Brute-force checking all pairs of distinct cities is feasible given the constraint on the number of cities (n <= 100).

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: No roads are present (roads.length == 0). In this case, the degree of each city will be 0, and the maximal network rank will also be 0. The code handles this correctly as the loop won't execute and initial max_rank will be returned.
Edge case 2: All cities are connected to each other (complete graph). In this case, the algorithm still correctly calculates the degrees and adjacency matrix and finds the maximal network rank which will be 2 * (n-1) - 1 = 2n - 3.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where two cities are directly connected and counting the shared road twice. The solution correctly subtracts 1 from the rank if they are adjacent.

✗Common mistake 2: Considering the same city twice (i == j) when calculating the network rank. The solution avoids this by iterating with `j = i + 1`.

✅

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: Analyzing social networks. Cities could represent people, and roads could represent friendships. Finding the maximal network rank helps identify the two individuals with the most connections in their combined networks.

🏦Real-world scenario 2: Analyzing transportation networks. Cities could represent locations, and roads could represent transportation routes. The maximal network rank identifies the two locations that have the most transportation links to other locations, indicating potentially important hubs.

💡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

Before coding, decide between BFS (shortest path, level-order) and DFS (exhaustive search, connected components). Draw the graph on paper first.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Forgetting to mark nodes as visited, causing infinite loops. Also watch for disconnected components.

👨‍💻 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: November 23, 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

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 "Maximal Network Rank"?

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

🤖 Need help with this problem?