Bird
Raised Fist0
DSA Cprogramming~30 mins

Adjacency List Representation in DSA C - Build from Scratch

Choose your learning style10 modes available
Adjacency List Representation
📖 Scenario: You are working on a simple social network app. Each person can have friends. You want to store who is friends with whom using a graph.
🎯 Goal: Build an adjacency list representation of a graph using arrays of linked lists in C. You will create the graph, add edges (friend connections), and print the adjacency list.
📋 What You'll Learn
Create a graph with 4 vertices numbered 0 to 3
Add edges to represent friendships: 0-1, 0-2, 1-2, 2-3
Use adjacency lists with linked lists for each vertex
Print the adjacency list showing each vertex and its connected vertices
💡 Why This Matters
🌍 Real World
Graphs are used to model networks like social connections, maps, and communication links.
💼 Career
Understanding adjacency lists is key for software engineers working with network data, pathfinding, and graph algorithms.
Progress0 / 4 steps
1
Create the graph data structure
Create a struct called Node with an int vertex and a Node* called next. Then create a struct called Graph with an int numVertices and a Node** adjLists.
DSA C
Hint

Use typedef struct to define the structs. The adjacency list is an array of pointers to Node.

2
Initialize the graph with 4 vertices
Write a function Graph* createGraph(int vertices) that allocates memory for a Graph, sets numVertices to vertices, and allocates memory for adjLists array with vertices elements initialized to NULL.
DSA C
Hint

Use malloc to allocate memory. Initialize each adjacency list head to NULL.

3
Add edges to the graph
Write a function void addEdge(Graph* graph, int src, int dest) that creates a new Node for dest and adds it at the start of src's adjacency list. Also add a new Node for src at the start of dest's adjacency list to make the graph undirected.
DSA C
Hint

Create new nodes for both directions and insert them at the start of the linked lists.

4
Print the adjacency list
Write a function void printGraph(Graph* graph) that prints each vertex and all vertices connected to it by traversing the adjacency lists. Then in main, create a graph with 4 vertices, add edges 0-1, 0-2, 1-2, 2-3, and call printGraph.
DSA C
Hint

Traverse each adjacency list and print the connected vertices. Then add the edges and call printGraph in main.

Practice

(1/5)
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
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard