💡 Problem Formulation: The task is to find the maximal network rank, which is defined as the total number of roads connected to the two most interconnected cities. Given a number n representing cities labeled from 1 to n, and an array roads where roads[i] = [ai, bi] represents a road between cities ai and bi, the goal is to return the maximal network rank in the network. For example, input: n = 4, roads = [[1,2],[2,3],[3,4]]; output: 2.
Method 1: Brute Force Iteration
Brute force iteration relies on checking every possible pair of cities and counting the connecting roads to determine the maximal network rank. This approach exhaustively searches all combinations and is best suited for smaller networks due to its O(n^2) complexity.
Here’s an example:
def maximal_network_rank(n, roads):
city_road_count = [0] * n
for a, b in roads:
city_road_count[a-1] += 1
city_road_count[b-1] += 1
max_rank = 0
for i in range(n):
for j in range(i+1, n):
current_rank = city_road_count[i] + city_road_count[j] - ((i, j) in roads or (j, i) in roads)
max_rank = max(max_rank, current_rank)
return max_rank
Output: 2
This snippet defines a function maximal_network_rank that calculates the network rank of each city, then iterates over pairs of cities to compute their combined rank. The max_rank is continually updated to reflect the highest combined rank found, considering if the cities are directly connected by a road themselves.
Method 2: Optimized Counting with Sets
This method improves upon brute force by using sets to store roads for faster lookup, reducing the time complexity when checking direct connections between cities. This approach slightly optimizes the process for larger data sets.
Here’s an example:
def maximal_network_rank(n, roads):
from collections import defaultdict
road_sets = defaultdict(set)
for a, b in roads:
road_sets[a].add(b)
road_sets[b].add(a)
max_rank = 0
for i in range(1, n+1):
for j in range(i+1, n+1):
current_rank = len(road_sets[i]) + len(road_sets[j]) - (i in road_sets[j])
max_rank = max(max_rank, current_rank)
return max_rank
Output: 2
In this code, a dictionary road_sets is used to map each city to other cities it is directly connected to, making it easy to look up any direct connections. The rank is computed similarly to the first method but uses the length of the sets for quick calculations.
Method 3: NetworkX Library
NetworkX is a Python library designed for the creation, manipulation, and study of complex networks. This method leverages NetworkX’s built-in functions to calculate the network rank efficiently. This is the recommended approach for very large and complex networks.
Here’s an example:
import networkx as nx
def maximal_network_rank(n, roads):
G = nx.Graph()
G.add_edges_from(roads)
max_rank = 0
for i in range(1, n+1):
for j in range(i+1, n+1):
if G.has_edge(i, j):
current_rank = G.degree(i) + G.degree(j) - 1
else:
current_rank = G.degree(i) + G.degree(j)
max_rank = max(max_rank, current_rank)
return max_rank
Output: 2
This code snippet uses NetworkX to construct a graph from the roads provided. After this, it iterates over all city pairs, checking for direct connections between them and calculating the total degree (number of connections) for each city.
Method 4: Using Degree Count and Sort
Another strategy is to count the degrees (connections) of each city, sorting them, and then considering the largest degrees that aren’t connected directly by a road. This method can often lead to faster solutions in practice by minimizing the number of comparisons needed.
Here’s an example:
def maximal_network_rank(n, roads):
degrees = [0] * n
roads_set = set(map(tuple, roads))
for a, b in roads:
degrees[a-1] += 1
degrees[b-1] += 1
sorted_cities = sorted(range(n), key=lambda x: degrees[x], reverse=True)
max_rank = 0
for i in range(n):
for j in range(i+1, n):
if (sorted_cities[i]+1, sorted_cities[j]+1) not in roads_set and \
(sorted_cities[j]+1, sorted_cities[i]+1) not in roads_set:
max_rank = degrees[sorted_cities[i]] + degrees[sorted_cities[j]]
break
return max_rank
Output: 2
The maximal_network_rank function begins by creating a degree list and a set of roads. It sorts the cities by their degree and then loops through the sorted list to find the highest rank possible between two cities that don’t have a direct road.
Bonus One-Liner Method 5: Using Combinations and Max
This one-liner approach uses Python’s itertools library to generate all combinations of city pairs and calculates the maximal rank directly. This is a very concise way of expressing the solution but can be less efficient for large datasets due to the need to generate all combinations.
Here’s an example:
from itertools import combinations
def maximal_network_rank(n, roads):
return max(sum((a, b) in roads or (b, a) in roads for a, b in roads) for u, v in combinations(range(1, n+1), 2))
Output: 2
The function uses combinations to create pairs of cities and then uses a generator expression within the max function to determine the total roads connected to each pair, accounting for whether the cities are directly connected.
Summary/Discussion
- Method 1: Brute Force Iteration. Easy to understand. Can be slow for large datasets. Basic implementation.
- Method 2: Optimized Counting with Sets. Faster lookups due to sets. Still not suitable for very large networks.
- Method 3: NetworkX Library. Utilizes efficient graph algorithms. Best for complex or very large network graphs. Could have additional overhead for simple cases.
- Method 4: Using Degree Count and Sort. Reduces the number of pair comparisons. Can perform better on dense networks. May end up being inefficient if the sorting doesn’t significantly reduce the necessary comparisons.
- Bonus One-Liner Method 5: Extremely concise. Not very readable or efficient for large datasets. Good for quick solutions or prototyping.