Delete Node at Specific Position in DSA C - Build from Scratch

Delete Node at Specific Position

📖 Scenario: Imagine you have a list of tasks to do, and you want to remove a task at a certain position from your list.

🎯 Goal: You will create a simple linked list of tasks, then write code to delete a node at a specific position, and finally print the updated list.

📋 What You'll Learn

Create a linked list with exactly 5 nodes containing the values 10, 20, 30, 40, 50

Create an integer variable called position and set it to 3

Write a function called deleteNodeAtPosition that deletes the node at the given position

Print the linked list after deletion in the format: 10 -> 20 -> 40 -> 50 -> NULL

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like task scheduling, undo functionality, and dynamic memory management where items need to be added or removed efficiently.

💼 Career

Understanding linked list operations like deletion is essential for software developers working with low-level data structures, embedded systems, or performance-critical applications.

1

Create the linked list with 5 nodes

Create a linked list with 5 nodes containing the values 10, 20, 30, 40, and 50. Use a struct called Node with an integer data and a pointer next. Create a pointer called head that points to the first node.

DSA C

# Define the Node struct and create the linked list with 5 nodes
// Your code here

Hint

Start by defining the Node struct with data and next. Then create 5 nodes using malloc and link them in order.

2

Set the position to delete

Create an integer variable called position and set it to 3 to represent the node position to delete.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

int main() {
    Node* head = (Node*)malloc(sizeof(Node));
    Node* second = (Node*)malloc(sizeof(Node));
    Node* third = (Node*)malloc(sizeof(Node));
    Node* fourth = (Node*)malloc(sizeof(Node));
    Node* fifth = (Node*)malloc(sizeof(Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = fourth;

    fourth->data = 40;
    fourth->next = fifth;

    fifth->data = 50;
    fifth->next = NULL;

    // Create an integer variable called position and set it to 3
    // Your code here

    return 0;
}

Hint

Just create an integer variable named position and assign it the value 3.

3

Write the function to delete node at position

Write a function called deleteNodeAtPosition that takes Node** head and int position as parameters. It deletes the node at the given position from the linked list. Call this function from main using &head and position.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Write deleteNodeAtPosition function here

int main() {
    Node* head = (Node*)malloc(sizeof(Node));
    Node* second = (Node*)malloc(sizeof(Node));
    Node* third = (Node*)malloc(sizeof(Node));
    Node* fourth = (Node*)malloc(sizeof(Node));
    Node* fifth = (Node*)malloc(sizeof(Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = fourth;

    fourth->data = 40;
    fourth->next = fifth;

    fifth->data = 50;
    fifth->next = NULL;

    int position = 3;

    // Call deleteNodeAtPosition(&head, position);

    return 0;
}

Hint

Write a function that handles deleting the first node separately. For other positions, find the node before the one to delete, then adjust pointers and free memory.

4

Print the linked list after deletion

Write a loop to print the linked list starting from head. Print each node's data followed by ->. After the last node, print NULL. This will show the list after deleting the node at position 3.

DSA C

#include 
#include 

typedef struct Node {
    int data;
    struct Node* next;
} Node;

void deleteNodeAtPosition(Node** head, int position) {
    if (*head == NULL) {
        return;
    }

    Node* temp = *head;

    if (position == 1) {
        *head = temp->next;
        free(temp);
        return;
    }

    for (int i = 1; temp != NULL && i next;
    }

    if (temp == NULL || temp->next == NULL) {
        return;
    }

    Node* next = temp->next->next;
    free(temp->next);
    temp->next = next;
}

int main() {
    Node* head = (Node*)malloc(sizeof(Node));
    Node* second = (Node*)malloc(sizeof(Node));
    Node* third = (Node*)malloc(sizeof(Node));
    Node* fourth = (Node*)malloc(sizeof(Node));
    Node* fifth = (Node*)malloc(sizeof(Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = fourth;

    fourth->data = 40;
    fourth->next = fifth;

    fifth->data = 50;
    fifth->next = NULL;

    int position = 3;

    deleteNodeAtPosition(&head, position);

    // Print the linked list
    // Your code here

    return 0;
}

Hint

Use a while loop to go through each node starting from head. Print the data and -> if there is a next node. End with NULL.