Remove Nth Node from End of List in DSA C - Build from Scratch

Remove Nth Node from End of List

📖 Scenario: Imagine you have a chain of train cars connected one after another. Sometimes, you want to remove a car counting from the end of the chain, not from the front. This project will help you learn how to remove the Nth car from the end of a linked list, which is like that chain of train cars.

🎯 Goal: You will build a program in C that creates a singly linked list, then removes the Nth node from the end of this list, and finally prints the updated list.

📋 What You'll Learn

Create a singly linked list with exactly these nodes and values: 10, 20, 30, 40, 50

Create an integer variable n and set it to 2

Write a function removeNthFromEnd that removes the Nth node from the end of the list

Print the linked list after removal in the format: 10 -> 20 -> 30 -> 50 -> NULL

💡 Why This Matters

🌍 Real World

Removing the Nth node from the end is like removing a specific car from the end of a train. This is useful in many software systems that use linked lists to manage data dynamically.

💼 Career

Understanding linked list manipulation is important for software developers, especially in roles involving system programming, embedded systems, or any place where memory management and efficient data structures are needed.

1

Create the linked list

Create a singly linked list with nodes containing these exact values in order: 10, 20, 30, 40, 50. Use the struct ListNode and create a pointer head pointing to the first node.

DSA C

#include 
#include 

struct ListNode {
    int val;
    struct ListNode *next;
};

int main() {

    // Create nodes and link them: 10 -> 20 -> 30 -> 40 -> 50 -> NULL

    // Your code here

    return 0;
}

Hint

Use malloc to create each node and link them by setting the next pointer.

2

Set the position to remove

Create an integer variable called n and set it to 2. This means you want to remove the 2nd node from the end of the list.

DSA C

#include 
#include 

struct ListNode {
    int val;
    struct ListNode *next;
};

int main() {
    struct ListNode *head = malloc(sizeof(struct ListNode));
    head->val = 10;
    head->next = malloc(sizeof(struct ListNode));
    head->next->val = 20;
    head->next->next = malloc(sizeof(struct ListNode));
    head->next->next->val = 30;
    head->next->next->next = malloc(sizeof(struct ListNode));
    head->next->next->next->val = 40;
    head->next->next->next->next = malloc(sizeof(struct ListNode));
    head->next->next->next->next->val = 50;
    head->next->next->next->next->next = NULL;

    // Create an integer variable n and set it to 2

    // Your code here

    return 0;
}

Hint

Just write int n = 2; inside main.

3

Write the function to remove the Nth node from end

Write a function called removeNthFromEnd that takes two arguments: a pointer to the head of the list (struct ListNode *head) and an integer n. This function should remove the Nth node from the end of the list and return the new head pointer. Use two pointers named fast and slow to find the node to remove.

DSA C

#include 
#include 

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* removeNthFromEnd(struct ListNode *head, int n) {
    // Your code here
}

int main() {
    struct ListNode *head = malloc(sizeof(struct ListNode));
    head->val = 10;
    head->next = malloc(sizeof(struct ListNode));
    head->next->val = 20;
    head->next->next = malloc(sizeof(struct ListNode));
    head->next->next->val = 30;
    head->next->next->next = malloc(sizeof(struct ListNode));
    head->next->next->next->val = 40;
    head->next->next->next->next = malloc(sizeof(struct ListNode));
    head->next->next->next->next->val = 50;
    head->next->next->next->next->next = NULL;

    int n = 2;

    // Call removeNthFromEnd and update head

    return 0;
}

Hint

Use a dummy node before head to handle edge cases. Move fast pointer n+1 steps ahead, then move both fast and slow until fast reaches the end. Remove the node after slow.

4

Print the updated linked list

Write a loop to print the linked list starting from head after removal. Print each node's value followed by ->. End with NULL. The output should be exactly: 10 -> 20 -> 30 -> 50 -> NULL

DSA C

#include 
#include 

struct ListNode {
    int val;
    struct ListNode *next;
};

struct ListNode* removeNthFromEnd(struct ListNode *head, int n) {
    struct ListNode dummy;
    dummy.next = head;
    struct ListNode *fast = &dummy;
    struct ListNode *slow = &dummy;

    for (int i = 0; i next;
    }

    while (fast != NULL) {
        fast = fast->next;
        slow = slow->next;
    }

    struct ListNode *toDelete = slow->next;
    slow->next = slow->next->next;
    free(toDelete);

    return dummy.next;
}

int main() {
    struct ListNode *head = malloc(sizeof(struct ListNode));
    head->val = 10;
    head->next = malloc(sizeof(struct ListNode));
    head->next->val = 20;
    head->next->next = malloc(sizeof(struct ListNode));
    head->next->next->val = 30;
    head->next->next->next = malloc(sizeof(struct ListNode));
    head->next->next->next->val = 40;
    head->next->next->next->next = malloc(sizeof(struct ListNode));
    head->next->next->next->next->val = 50;
    head->next->next->next->next->next = NULL;

    int n = 2;

    head = removeNthFromEnd(head, n);

    // Print the linked list

    // Your code here

    return 0;
}

Hint

Use a while loop to go through each node starting from head. Print the value and -> until you reach the end, then print NULL.