Group movies via graph traversal
Company: Google
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Onsite
You are given n movies labeled 0..n-1 and a list of undirected pairs (a, b) meaning movies a and b are similar. Group all movies into categories where each category is a maximal set of movies connected via similarity links (i.e., connected components). Return the categories as lists of movie IDs, each list sorted, and the collection sorted by the smallest ID in each category. Implement solutions using both BFS and DFS, explain their time/space complexity, and compare iterative vs. recursive DFS. Then discuss an alternative Union-Find approach and when it would be preferable. Dry-run on n=8 with pairs: (0,
1),(1,
2),(3,
4),(5,
6),(6,
7).
Overview: This question evaluates graph traversal and connectivity skills, specifically competence with BFS, DFS (iterative and recursive), complexity analysis, and Union-Find for identifying connected components in an undirected graph.
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Community answers
Answer by mojahidislam221
import java.util.*;
public class Solution {
public List> solution(
int n,
List> pairs) {
List> graph = new ArrayList<>();
for (int i = 0; i < n; i++) {
graph.add(new ArrayList<>());
}
// Build undirected graph
for (List pair : pairs) {
int a = pair.get(0);
int b = pair.get(1);
graph.get(a).add(b);
graph.get(b).add(a);
}
boolean[] visited = new boolean[n];
List> result = new ArrayList<>();
for (int i = 0; i < n; i++) {
if (visited[i]) {
continue;
}
List component = new ArrayList<>();
Deque stack = new ArrayDeque<>();
stack.push(i);
visited[i] = true;
while (!stack.isEmpty()) {
int u = stack.pop();
component.add(u);
for (int v : graph.get(u)) {
if (!visited[v]) {
visited[v] = true;
stack.push(v);
}
}
}
// Sort IDs inside component
Collections.sort(component);
result.add(component);
}
// Sort components by their smallest ID
result.sort(Comparator.comparingInt(a -> a.get(0)));
return result;
}
}