Adjacency List Representation in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of adjacency list after adding edges

What is the printed adjacency list after running the following code?

DSA Python

graph = {0: [], 1: [], 2: [], 3: []}
graph[0].append(1)
graph[0].append(2)
graph[1].append(2)
graph[2].append(0)
graph[2].append(3)
print(graph)

A {"0": [1, 2], "1": [2], "2": [0, 3], "3": []}

B {0: [1, 2], 1: [2], 2: [3, 0], 3: []}

C {0: [1, 2], 1: [2], 2: [0, 3], 3: []}

D {0: [2, 1], 1: [2], 2: [0, 3], 3: []}

💡 Hint

Remember that append adds elements at the end of the list in the order they are called.

The adjacency list stores neighbors in the order they are added. The keys are integers, so printed dict keys are integers, not strings. The order of neighbors matters.

🧠 Conceptual

intermediate

Number of edges in an undirected graph adjacency list

If an undirected graph has the adjacency list below, how many edges does the graph have? graph = {0: [1, 2], 1: [0, 2], 2: [0, 1, 3], 3: [2]}

A 5

B 4

C 6

D 7

💡 Hint

In an undirected graph, each edge appears twice in the adjacency list.

Count all neighbors: 2 + 2 + 3 + 1 = 8 entries. Each edge counted twice, so edges = 8 / 2 = 4.

🔧 Debug

advanced

Identify the error in adjacency list creation

What error will this code produce when run? nodes = 3 graph = [[]] * nodes graph[0].append(1) print(graph)

A Output: [[1], [1], [1]]

B Output: [[1], [], []]

C TypeError: 'int' object is not callable

D IndexError: list index out of range

💡 Hint

Check how list multiplication affects nested lists.

Using [[]] * nodes creates multiple references to the same list. Appending to one affects all.

❓ Predict Output

advanced

Output after removing an edge from adjacency list

What is the adjacency list after removing edge 2->3 from the graph below? graph = {0: [1, 2], 1: [2], 2: [0, 3], 3: []} graph[2].remove(3) print(graph)

A {0: [1, 2], 1: [2], 2: [0], 3: []}

B {0: [1, 2], 1: [2], 2: [3], 3: []}

C {0: [1, 2], 1: [2], 2: [], 3: []}

D ValueError: list.remove(x): x not in list

💡 Hint

The remove method deletes the first occurrence of the value.

Removing 3 from graph[2] leaves [0]. Other lists remain unchanged.

🚀 Application

expert

Find if a path exists using adjacency list

Given the adjacency list below, which option correctly returns True if there is a path from node 0 to node 3 using depth-first search (DFS)?

DSA Python

graph = {0: [1, 2], 1: [2], 2: [3], 3: []}
visited = set()
def dfs(node):
    if node == 3:
        return True
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            if dfs(neighbor):
                return True
    return False
result = dfs(0)
print(result)

A False

B None

C KeyError

D True

💡 Hint

DFS explores neighbors recursively and returns True when target is found.

The DFS visits nodes 0 -> 1 -> 2 -> 3 and returns True when node 3 is reached.

Practice

1. What does an adjacency list represent in graph data structures?

easy

A. A list of all nodes without connections

B. A matrix showing distances between nodes

C. A list of nodes sorted by their values

D. A list where each node stores its connected neighbors

Solution

1. Step 1: Understand adjacency list concept

   An adjacency list stores each node with a list of nodes it connects to directly.

2. Step 2: Compare options with definition

   Only A list where each node stores its connected neighbors correctly describes this structure as neighbors stored per node.

3. Final Answer:

   A list where each node stores its connected neighbors -> Option D

4. Quick Check:

   Adjacency list = neighbors per node [OK]

Hint: Adjacency list = nodes with their direct neighbors [OK]

Common Mistakes:

• Confusing adjacency list with adjacency matrix
• Thinking it stores distances instead of neighbors
• Assuming it lists nodes without connections

2. Which of the following is the correct Python syntax to initialize an empty adjacency list for a graph with 3 nodes?

easy

A. graph = {0: [], 1: [], 2: []}

B. graph = [[], [], []]

C. graph = {0: None, 1: None, 2: None}

D. graph = [0, 1, 2]

Solution

1. Step 1: Understand adjacency list initialization

   An adjacency list is usually a dictionary with keys as nodes and values as lists of neighbors.

2. Step 2: Check each option

   graph = {0: [], 1: [], 2: []} correctly creates a dict with empty lists for each node. graph = [[], [], []] is a list of lists but lacks node keys. graph = {0: None, 1: None, 2: None} uses None instead of lists. graph = [0, 1, 2] is just a list of nodes, not adjacency lists.

3. Final Answer:

   graph = {0: [], 1: [], 2: []} -> Option A

4. Quick Check:

   Dict with empty lists = adjacency list init [OK]

Hint: Use dict with nodes as keys and empty lists as values [OK]

Common Mistakes:

• Using None instead of empty lists for neighbors
• Using list without node keys
• Confusing adjacency list with simple node list

3. What is the output of the following code?

graph = {0: [1, 2], 1: [2], 2: [0, 3], 3: []}
print(graph[2])

medium

A. [1, 2]

B. [0, 3]

C. [2]

D. []

Solution

1. Step 1: Identify the adjacency list for node 2

   In the dictionary, key 2 maps to the list [0, 3].

2. Step 2: Print the value for key 2

   Printing graph[2] outputs [0, 3].

3. Final Answer:

   [0, 3] -> Option B

4. Quick Check:

   graph[2] = [0, 3] [OK]

Hint: Look up the key in dict to get neighbors list [OK]

Common Mistakes:

• Confusing keys and values
• Printing wrong node's neighbors
• Misreading list contents

4. The following code tries to add an edge from node 1 to node 3 in the adjacency list but causes an error. What is the error?

graph = {0: [1], 1: [2], 2: [0]}
graph[1].append(3)
print(graph)

medium

A. KeyError because node 3 does not exist in graph

B. IndexError because list index out of range

C. No error, code runs correctly

D. TypeError because append is used on a dict

Solution

1. Step 1: Check if node 1 exists in graph

   Node 1 is a key in graph with value [2], so graph[1] is a list.

2. Step 2: Append 3 to graph[1]

   Appending 3 to the list [2] is valid and updates graph[1] to [2, 3].

3. Final Answer:

   No error, code runs correctly -> Option C

4. Quick Check:

   Appending to list in dict value works [OK]

Hint: Appending to existing list key is valid, no error [OK]

Common Mistakes:

• Thinking node 3 must exist as key before adding
• Confusing append on list vs dict
• Expecting error when adding new neighbor

5. Given a graph represented by the adjacency list:

graph = {0: [1, 2], 1: [2], 2: [0], 3: [2]}

Which node(s) have no outgoing edges?

hard

A. None, all have outgoing edges

B. Node 1

C. Node 2

D. Node 3

Solution

1. Step 1: Check neighbors list for each node

   Node 0 connects to [1, 2], node 1 to [2], node 2 to [0], node 3 to [2].

2. Step 2: Identify nodes with empty neighbor lists

   Node 3 has neighbors [2], so it has outgoing edges. No node has an empty list, so none have no outgoing edges.

3. Final Answer:

   None, all have outgoing edges -> Option A

4. Quick Check:

   All adjacency lists non-empty = no node without edges [OK]

Hint: Check for empty neighbor lists to find no outgoing edges [OK]

Common Mistakes:

• Assuming node 3 has no edges because it appears last
• Confusing incoming edges with outgoing edges
• Ignoring non-empty lists