Why Graphs Exist and What Trees Cannot Model in DSA Java - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a graph adjacency list in Java.

DSA Java

Map<Integer, List<Integer>> graph = new [1]<>();

Drag options to blanks, or click blank then click option'

AHashMap

BArrayList

CLinkedList

DTreeMap

Attempts:

3 left

💡 Hint

Common Mistakes

Using ArrayList instead of HashMap for the main graph structure.

We use HashMap to store adjacency lists for each node in a graph.

2fill in blank

medium

Complete the code to add an edge from node u to node v in the graph.

DSA Java

graph.get(u).[1](v);

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Aput

Bset

Cinsert

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using put() which is for maps, not lists.

We use add() to append v to the adjacency list of u.

3fill in blank

hard

Fix the error in the code to check if the graph contains a cycle.

DSA Java

if (visited[node] && [1][node]) { return true; }

Drag options to blanks, or click blank then click option'

AadjList

Bgraph

CrecStack

Dqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using graph or adjList which are not boolean arrays.

recStack tracks nodes in the current recursion stack to detect cycles.

4fill in blank

hard

Fill both blanks to complete the BFS traversal starting from node s.

DSA Java

Queue<Integer> queue = new [1]<>();
queue.[2](s);

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ALinkedList

Badd

Cpush

DArrayDeque

Attempts:

3 left

💡 Hint

Common Mistakes

Using push() which is for stacks, not queues.

We use LinkedList as a queue and add() to enqueue the start node.

5fill in blank

hard

Fill all three blanks to complete the code that models a graph with cycles, which trees cannot represent.

DSA Java

graph.putIfAbsent([1], new ArrayList<>());
graph.get([2]).add([3]);

Drag options to blanks, or click blank then click option'

A1

B2

C3

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing node names or forgetting to initialize adjacency list.

We ensure node 1 exists, then add node 2 as its neighbor, allowing cycles.

Practice

1. Why do we need graphs when we already have trees as data structures?

easy

A. Because graphs use less memory than trees.

B. Because graphs can represent connections between nodes in any pattern, including cycles and multiple parents, which trees cannot.

C. Because trees cannot store any data in nodes.

D. Because trees are slower to traverse than graphs.

Solution

1. Step 1: Understand tree structure limitations

   Trees have a strict hierarchy where each node has exactly one parent except the root, and no cycles are allowed.

2. Step 2: Understand graph flexibility

   Graphs allow nodes to connect in any way, including cycles and multiple parents, which trees cannot model.

3. Final Answer:

   Because graphs can represent connections between nodes in any pattern, including cycles and multiple parents, which trees cannot. -> Option B

4. Quick Check:

   Graphs model complex connections = D [OK]

Hint: Graphs allow cycles and multiple parents; trees do not [OK]

Common Mistakes:

• Thinking trees can have cycles
• Confusing memory use between trees and graphs
• Believing trees cannot store data

2. Which of the following is the correct way to declare a graph adjacency list in Java using a HashMap?

easy

A. HashMap graph = new HashMap<>();

B. HashMap graph = new HashMap<>();

C. HashMap graph = new HashMap<>();

D. HashMap graph = new HashMap<>();

Solution

1. Step 1: Identify correct key and value types for adjacency list

   Keys are node identifiers (Integer), values are lists of connected nodes (List<Integer>).

2. Step 2: Check syntax correctness

   HashMap graph = new HashMap<>(); uses HashMap which correctly represents adjacency list.

3. Final Answer:

   HashMap graph = new HashMap<>(); -> Option D

4. Quick Check:

   Adjacency list uses map of node to list [OK]

Hint: Adjacency list maps node to list of neighbors [OK]

Common Mistakes:

• Using primitive int as generic type
• Mapping node to single Integer instead of list
• Using wrong generic types like String when nodes are integers

3. Consider the following Java code snippet that builds a graph and prints its adjacency list:

import java.util.*;
public class GraphExample {
    public static void main(String[] args) {
        HashMap graph = new HashMap<>();
        graph.put(1, Arrays.asList(2, 3));
        graph.put(2, Arrays.asList(1, 4));
        graph.put(3, Arrays.asList(1));
        graph.put(4, Arrays.asList(2));
        for (int node : graph.keySet()) {
            System.out.print(node + " -> ");
            for (int neighbor : graph.get(node)) {
                System.out.print(neighbor + " ");
            }
            System.out.println();
        }
    }
}

What is the output of this program?

medium

A. 1 -> 2 3
2 -> 1 4
3 -> 1
4 -> 2

B. 1 -> 3 2
2 -> 4 1
3 -> 1
4 -> 2

C. 1 -> 2 3
2 -> 4 1
3 -> 1
4 -> 2

D. 1 -> 2 3
2 -> 1 4
3 -> 4
4 -> 2

Solution

1. Step 1: Analyze graph construction

   Node 1 connects to 2 and 3; 2 connects to 1 and 4; 3 connects to 1; 4 connects to 2.

2. Step 2: Trace print output

   For each node, print node followed by neighbors in order stored in list.

3. Final Answer:

   1 -> 2 3 2 -> 1 4 3 -> 1 4 -> 2 -> Option A

4. Quick Check:

   Printed adjacency matches stored lists = A [OK]

Hint: Print neighbors in stored order for each node [OK]

Common Mistakes:

• Mixing neighbor order
• Printing wrong neighbors for nodes
• Confusing node keys with neighbor values

4. The following Java code attempts to add an edge between two nodes in an undirected graph represented by an adjacency list. Find the error:

void addEdge(HashMap graph, int u, int v) {
    graph.get(u).add(v);
    graph.get(v).add(u);
}

medium

A. The code uses List<Integer> instead of Set<Integer> for neighbors.

B. The code should remove edges instead of adding them.

C. The code does not check if nodes u and v exist in the graph before adding edges.

D. The code incorrectly adds edges only in one direction.

Solution

1. Step 1: Check for node existence before adding edges

   graph.get(u) or graph.get(v) may return null if nodes are not present, causing NullPointerException.

2. Step 2: Identify missing null checks or initialization

   Code should verify nodes exist or initialize adjacency lists before adding neighbors.

3. Final Answer:

   The code does not check if nodes u and v exist in the graph before adding edges. -> Option C

4. Quick Check:

   Missing node existence check causes error = B [OK]

Hint: Always check nodes exist before adding edges [OK]

Common Mistakes:

• Assuming nodes always exist in graph
• Confusing undirected with directed edges
• Ignoring possibility of NullPointerException

5. You want to model a social network where users can follow each other in any pattern, including mutual follows and cycles. Why is a graph a better choice than a tree for this model, and how would you represent it in Java?

hard

A. Because trees cannot represent cycles or mutual connections; use a HashMap to store user IDs and their followers.

B. Because trees are faster to search; use a TreeMap to store user IDs.

C. Because trees allow multiple parents; use a HashSet<Integer> to store users.

D. Because trees prevent cycles; use an array to store user connections.

Solution

1. Step 1: Understand social network connections

   Users can follow each other in any pattern, including cycles and mutual follows, which trees cannot model.

2. Step 2: Choose graph representation in Java

   Use HashMap to map each user ID to a list of users they follow.

3. Final Answer:

   Because trees cannot represent cycles or mutual connections; use a HashMap to store user IDs and their followers. -> Option A

4. Quick Check:

   Graphs model cycles; HashMap with lists stores connections [OK]

Hint: Use graph with adjacency list for mutual and cyclic connections [OK]

Common Mistakes:

• Choosing tree which disallows cycles
• Using TreeMap or array which limits connection types
• Confusing followers with parents in tree