Network Of Cities And Roads - BFS Distance Calculation Algor ✓ Solved

Network of Cities and Roads BFS Distance Calculation Algorithm

Network of Cities and Roads - BFS Distance Calculation Algorithm

A network consisting of M cities and M-1 roads connecting them is given. Cities are labeled with distinct integers within the range [0, M-1]. Roads connect cities in such a way that each pair of distinct cities is connected either directly or via a path of roads, forming a tree structure. The distance between two cities is defined as the number of roads that must be traversed to move from one to the other.

Given an array T of length M, where each element T[P] describes the connection for city P:

  • If T[P] = Q and P ≠ Q, there is a direct road between cities P and Q.
  • If T[P] = P, then P is the designated capital city.

Your task is to compute, for each distance from 1 up to M-1, how many cities are at that distance from the capital, which is indicated as city 0 in this problem.

Design an efficient algorithm that, given the array T, returns an array of length M-1 where each element represents the number of cities at each respective distance from the capital city. You can assume:

  • M is in the range [1..100,000].
  • Each element T[P] is in [0.. M-1], with exactly one city designated as the capital (T[P] = P for that city).
  • The network forms a tree, hence always connected and acyclic.

Problem Breakdown and Approach

The core challenge is to compute the shortest distance from the capital city (city 0) to all other cities in the network. Because the network is a tree, the shortest path between any two cities is unique. To find the number of cities at each distance, we will perform a Breadth-First Search (BFS) starting from the capital city.

The BFS traversal efficiently computes minimal distances in unweighted graphs, like trees, ensuring each city’s shortest distance from the capital is found with optimal time complexity O(M).

Algorithm Design

Step 1: Build an adjacency list to represent the graph

Iterate over the array T to construct an adjacency list, where each city’s list contains its directly connected neighboring cities. This representation makes traversals more efficient.

Step 2: Implement BFS to compute distances

  • Initialize a queue with the capital city (city 0), and a distance array with default values indicating unvisited states (-1).
  • Set the distance for the capital to 0.
  • While the queue is not empty:
    • Dequeue a city, then for each connected neighbor:
      • If it is unvisited, assign it a distance (current city’s distance + 1) and enqueue it.

Step 3: Count the number of cities at each distance

Once BFS completes, iterate over the distance array to count how many cities are associated with each distance. Then, compile these counts into an output array, excluding the distance 0 (the city itself).

Implementation in Java



public class Solution {


public int[] solution(int[] T) {


int M = T.length;


if (M == 1) {


// Only the capital city exists; no other cities to count.


return new int[0];


}


// Build adjacency list


List> adj = new ArrayList();


for (int i = 0; i 

adj.add(new ArrayList());


}


int capital = -1;


for (int i = 0; i 

int neighbor = T[i];


if (neighbor == i) {


// Found the capital city


capital = i;


} else {


// Since the network is undirected, add bidirectional edges


adj.get(i).add(neighbor);


adj.get(neighbor).add(i);


}


}


int[] distances = new int[M];


Arrays.fill(distances, -1);


// BFS initialization


Queue queue = new LinkedList();


distances[capital] = 0;


queue.offer(capital);


while (!queue.isEmpty()) {


int current = queue.poll();


for (int neighbor : adj.get(current)) {


if (distances[neighbor] == -1) {


distances[neighbor] = distances[current] + 1;


queue.offer(neighbor);


}


}


}


// Count the number of cities at each distance


int maxDistance = Arrays.stream(distances).max().getAsInt();


int[] result = new int[M -1];


for (int i = 0; i 

int dist = distances[i];


if (dist > 0) {


result[dist -1]++;


}


}


return result;


}


}

Complexity Analysis

This algorithm operates in O(M) time, where M is the number of cities, since each city and road is processed at most once during both adjacency list construction and BFS traversal. This efficiency makes it suitable for large input sizes up to 100,000. The space complexity is likewise O(M) due to storage of adjacency lists, distance array, and queue.

Conclusion

By leveraging Breadth-First Search, this solution effectively computes the shortest distances from the capital for all cities and counts how many reside at each distance level. This approach handles large networks efficiently and provides an accurate, scalable method for analyzing tree-structured city networks.

References

  • Clrs, C. E. (2001). Introduction to Algorithms.
  • Grimaldi, R. P. (2003). Discrete and Combinatorial Mathematics.
  • Russell, S., & Norvig, P. (2020). Artificial Intelligence: A Modern Approach.
  • Dasgupta, S., Papadimitriou, C. H., & Vazirani, U. V. (2008). Algorithms.
  • Kleinberg, J., & Tardos, E. (2006). Algorithm Design.
  • Skiena, S. S. (2008). The Algorithm Design Manual.
  • Holland, P. W. (1992). The Moving Parts of Machine Learning.
  • Mitchell, T. M. (1997). Machine Learning.
  • Bellman, R. (1958). On a Routing Problem.
  • Tarjan, R. (1983). Data Structures and Network Algorithms.

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