Adjacency List Representation in DSA Python - Time & Space Complexity

Time Complexity: Adjacency List Representation

O(V + E)

Understanding Time Complexity

We want to understand how fast we can build and use an adjacency list for a graph.

How does the time needed grow as the graph gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Graph:
    def __init__(self, vertices):
        self.V = vertices
        self.adj = [[] for _ in range(vertices)]

    def add_edge(self, u, v):
        self.adj[u].append(v)
        self.adj[v].append(u)  # For undirected graph

This code creates an adjacency list for a graph and adds edges between nodes.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding an edge involves appending to two lists.
• How many times: Once per edge added.

How Execution Grows With Input

Each edge addition takes constant time, but total time grows with number of edges.

Input Size (Edges)	Approx. Operations
10	20 appends (2 per edge)
100	200 appends
1000	2000 appends

Pattern observation: Operations grow linearly with the number of edges.

Final Time Complexity

Time Complexity: O(V + E)

This means building the adjacency list takes time proportional to the number of vertices plus the number of edges.

Common Mistake

[X] Wrong: "Adding edges takes O(V) time each because we have to check all vertices."

[OK] Correct: We directly add edges to the lists without scanning all vertices, so each edge addition is constant time.

Interview Connect

Understanding adjacency lists helps you work efficiently with graphs, a common topic in interviews and real projects.

Self-Check

"What if the graph was directed and we only added one direction per edge? How would the time complexity change?"

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