Adjacency List Representation in DSA C - Execution Trace

Concept Flow - Adjacency List Representation

Start

↓

Create array of lists

↓

For each edge

↓

Add neighbor to source list

↓

Add neighbor to destination list (if undirected)

↓

Repeat for all edges

↓

Graph represented as adjacency lists

↓

End

Build an array where each index holds a list of neighbors for that vertex, adding edges step-by-step.

Execution Sample

DSA C

Node* graph[4];
// graph as array of lists
addEdge(0, 1);
addEdge(0, 2);
addEdge(1, 2);
addEdge(2, 3);

Create adjacency lists for a graph with 4 vertices and add edges between them.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Initialize graph with 4 empty lists	0: [] 1: [] 2: [] 3: []	graph array created	0: null 1: null 2: null 3: null
2	Add edge 0 -> 1	0: [1] 1: [] 2: [] 3: []	Add node 1 to list at index 0	0: 1 -> null 1: null 2: null 3: null
3	Add edge 0 -> 2	0: [2,1] 1: [] 2: [] 3: []	Add node 2 at head of list 0	0: 2 -> 1 -> null 1: null 2: null 3: null
4	Add edge 1 -> 2	0: [2,1] 1: [2] 2: [] 3: []	Add node 2 to list at index 1	0: 2 -> 1 -> null 1: 2 -> null 2: null 3: null
5	Add edge 2 -> 3	0: [2,1] 1: [2] 2: [3] 3: []	Add node 3 to list at index 2	0: 2 -> 1 -> null 1: 2 -> null 2: 3 -> null 3: null
6	End	Graph fully built	No changes	0: 2 -> 1 -> null 1: 2 -> null 2: 3 -> null 3: null

💡 All edges processed, adjacency lists complete

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	Final
graph[0]	null	1 -> null	2 -> 1 -> null	2 -> 1 -> null	2 -> 1 -> null	2 -> 1 -> null
graph[1]	null	null	null	2 -> null	2 -> null	2 -> null
graph[2]	null	null	null	null	3 -> null	3 -> null
graph[3]	null	null	null	null	null	null

Key Moments - 3 Insights

Why does the list at graph[0] show 2 before 1 after adding edge 0->2?

Because new nodes are added at the head of the list, so 2 is inserted before 1, as shown in step 3 of execution_table.

Why is graph[3] still empty after all edges?

No edges start from vertex 3, so its adjacency list remains empty, as seen in the final visual state.

Why do we add neighbors only to the source vertex's list?

Because this example shows a directed graph; edges go one way, so only source lists get updated, as shown in each addEdge operation.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the order of nodes in graph[0]'s list?

A 2 -> 1 -> null

B 1 -> 2 -> null

C null

D 1 -> null

💡 Hint

Check the Visual State column at step 3 in execution_table

At which step does graph[1] first get a neighbor added?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Look at the Nodes in List column for graph[1] in execution_table

If we add edge 3 -> 0, how would graph[3]'s list change?

A It would become 3 -> null

B It would remain null

C It would become 0 -> null

D It would become 1 -> null

💡 Hint

Adding edge 3->0 adds 0 to graph[3]'s adjacency list

Concept Snapshot

Adjacency List Representation:
- Use an array where each index is a vertex.
- Each array element holds a linked list of neighbors.
- Add edges by inserting neighbors in source vertex's list.
- Efficient for sparse graphs.
- Directed edges update only source list.
- Visual: graph[0]: 2 -> 1 -> null means vertex 0 connects to 2 and 1.

Full Transcript

Adjacency List Representation stores graph edges as lists for each vertex. We start with an array of empty lists. For each edge, we add the destination vertex to the source vertex's list. This example shows a directed graph with 4 vertices. Edges are added step-by-step, updating the lists. The lists grow by adding new nodes at the head. The final structure shows which vertices connect to which neighbors. This method is memory efficient for graphs with fewer edges.

Practice

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

easy

A. A list where each vertex stores its adjacent vertices

B. A matrix showing connections between vertices

C. A list of all vertices without edges

D. A list of edges only without vertices

Solution

1. Step 1: Understand adjacency list concept

   An adjacency list stores each vertex and the vertices it connects to directly.

2. Step 2: Compare with other representations

   Unlike adjacency matrix, adjacency list focuses on neighbors per vertex, not a full matrix.

3. Final Answer:

   A list where each vertex stores its adjacent vertices -> Option A

4. Quick Check:

   Adjacency list = vertex with neighbors [OK]

Hint: Adjacency list stores neighbors per vertex, not full matrix [OK]

Common Mistakes:

• Confusing adjacency list with adjacency matrix
• Thinking it stores only edges without vertices
• Assuming it lists vertices without connections

2. Which of the following is the correct C syntax to declare an adjacency list for a graph with 5 vertices using an array of pointers to linked lists?

easy

A. int* adjList[5];

B. int adjList[5][5];

C. struct Node adjList[5];

D. struct Node* adjList[5];

Solution

1. Step 1: Understand adjacency list declaration

   Adjacency list uses array of pointers to linked list nodes, so 'struct Node* adjList[5];' is correct.

2. Step 2: Check other options

   int adjList[5][5]; is adjacency matrix, C is array of structs (not pointers), A is array of int pointers (not nodes).

3. Final Answer:

   struct Node* adjList[5]; -> Option D

4. Quick Check:

   Array of pointers to nodes = correct adjacency list declaration [OK]

Hint: Adjacency list uses array of pointers to node structs [OK]

Common Mistakes:

• Using 2D array instead of array of pointers
• Declaring array of structs instead of pointers
• Using int pointers instead of node pointers

3. Given the following adjacency list representation code snippet for a graph with 3 vertices (0,1,2):

struct Node {
  int vertex;
  struct Node* next;
};

struct Node* adjList[3] = {NULL, NULL, NULL};

// Adding edge 0->1
struct Node* newNode = malloc(sizeof(struct Node));
newNode->vertex = 1;
newNode->next = adjList[0];
adjList[0] = newNode;

// Adding edge 0->2
newNode = malloc(sizeof(struct Node));
newNode->vertex = 2;
newNode->next = adjList[0];
adjList[0] = newNode;

// Adding edge 1->2
newNode = malloc(sizeof(struct Node));
newNode->vertex = 2;
newNode->next = adjList[1];
adjList[1] = newNode;

// Print adjacency list for vertex 0
struct Node* temp = adjList[0];
while(temp != NULL) {
  printf("%d -> ", temp->vertex);
  temp = temp->next;
}
printf("NULL\n");

What is the printed output?

medium

A. 1 -> 2 -> NULL

B. 0 -> 1 -> 2 -> NULL

C. 2 -> 1 -> NULL

D. NULL

Solution

1. Step 1: Trace insertion order for vertex 0

   Edges 0->1 then 0->2 are added at head, so list is 2 -> 1 -> NULL.

2. Step 2: Understand print loop

   Prints vertices in order from head: 2 then 1, then NULL.

3. Final Answer:

   2 -> 1 -> NULL -> Option C

4. Quick Check:

   Last inserted edge appears first in adjacency list [OK]

Hint: New nodes added at head, so last added appears first [OK]

Common Mistakes:

• Assuming order is insertion order instead of reversed
• Printing vertex 0 itself in adjacency list
• Confusing NULL print format

4. Consider this buggy code snippet to add an edge from vertex u to v in an adjacency list:

void addEdge(struct Node* adjList[], int u, int v) {
  struct Node* newNode = malloc(sizeof(struct Node));
  newNode->vertex = v;
  newNode->next = NULL;
  adjList[u] = newNode;
}

What is the main problem with this code?

medium

A. It overwrites existing adjacency list for vertex u, losing previous edges

B. It does not allocate memory for newNode

C. It sets newNode->next incorrectly to itself

D. It does not assign vertex v to newNode

Solution

1. Step 1: Analyze pointer assignment

   Setting adjList[u] = newNode overwrites old list head, losing previous edges.

2. Step 2: Check linked list linking

   newNode->next is NULL, so no link to old list; old edges lost.

3. Final Answer:

   It overwrites existing adjacency list for vertex u, losing previous edges -> Option A

4. Quick Check:

   Overwriting head pointer loses old adjacency list [OK]

Hint: Always link new node to old head to keep edges [OK]

Common Mistakes:

• Not linking new node to existing list
• Forgetting to allocate memory
• Incorrectly setting next pointer

5. You have a directed graph with 4 vertices (0 to 3). You want to represent it using adjacency lists and add edges:
- 0 -> 1
- 0 -> 2
- 1 -> 2
- 2 -> 0
- 2 -> 3
- 3 -> 3

Which data structure and insertion method correctly represents this graph in C?

hard

A. Use 2D int array; set 1 for edges and 0 otherwise

B. Use array of pointers to Node structs; insert new nodes at head linking to old list

C. Use array of Node structs; overwrite each vertex's list on insertion

D. Use array of int pointers; store vertex numbers without linking

Solution

1. Step 1: Choose correct data structure

   Adjacency list uses array of pointers to linked nodes for each vertex.

2. Step 2: Insert edges properly

   Insert new nodes at head, linking to previous list to keep all edges.

3. Step 3: Check other options

   Use 2D int array; set 1 for edges and 0 otherwise is adjacency matrix, C overwrites lists losing edges, D lacks linking.

4. Final Answer:

   Use array of pointers to Node structs; insert new nodes at head linking to old list -> Option B

5. Quick Check:

   Array of pointers + head insertion = correct adjacency list [OK]

Hint: Use pointers array and link new nodes at head to keep edges [OK]

Common Mistakes:

• Using adjacency matrix instead of list
• Overwriting adjacency list on insertion
• Not linking nodes properly