Insert at Middle Specific Position in DSA C - Build from Scratch

Insert at Middle Specific Position in a Linked List

📖 Scenario: You are managing a list of tasks in a to-do app. Sometimes, you need to add a new task right in the middle of the list at a specific position.

🎯 Goal: Build a program that creates a linked list of tasks, then inserts a new task at a given middle position, and finally prints the updated list.

📋 What You'll Learn

Create a singly linked list with 3 tasks: 'Task1', 'Task2', 'Task3'

Add an integer variable position set to 2 to specify the insertion point

Write a function to insert a new task 'NewTask' at the position in the linked list

Print the linked list after insertion showing all tasks in order

💡 Why This Matters

🌍 Real World

Linked lists are used in many apps to manage ordered data like tasks, playlists, or messages where items can be added or removed dynamically.

💼 Career

Understanding linked list insertion is fundamental for software developers working with low-level data structures, embedded systems, or performance-critical applications.

1

Create the initial linked list with 3 tasks

Create a singly linked list with nodes containing char task[20] and struct Node *next. Initialize the list with these exact tasks in order: "Task1", "Task2", "Task3". Use a pointer head to the first node.

DSA C

# Define the Node struct and create the linked list with 3 tasks
// Your code here

Hint

Define a struct with a char task[20] and a Node *next. Allocate memory for each node and set the tasks exactly as given.

2

Add a variable for the insertion position

Add an integer variable called position and set it to 2 to specify where the new task will be inserted.

DSA C

#include 
#include 
#include 

typedef struct Node {
    char task[20];
    struct Node *next;
} Node;

int main() {
    Node *head = malloc(sizeof(Node));
    strcpy(head->task, "Task1");
    head->next = malloc(sizeof(Node));
    strcpy(head->next->task, "Task2");
    head->next->next = malloc(sizeof(Node));
    strcpy(head->next->next->task, "Task3");
    head->next->next->next = NULL;

    // Add position variable here
    // Your code here

    return 0;
}

Hint

Just add int position = 2; inside main().

3

Insert a new task at the given position

Write code to insert a new node with task "NewTask" at the position in the linked list. Use a pointer current to traverse the list until the node before the insertion point, then link the new node properly.

DSA C

    Node *newNode = malloc(sizeof(Node));
    strcpy(newNode->task, "NewTask");
    Node *current = head;
    for (int i = 1; i next;
    }
    newNode->next = current->next;
    current->next = newNode;
    // Print the linked list

Hint

Traverse the list to the node before position, then insert the new node by adjusting pointers.

4

Print the updated linked list

Use a pointer temp to traverse the linked list from head and print each task on its own line until the end of the list.

DSA C

Hint

Use a while loop to print each task until temp is NULL.

Practice

1. What does inserting a node at a specific middle position in a linked list mean?

easy

A. Adding a new node only at the start of the list

B. Adding a new node exactly at the given position counting from the head

C. Adding a new node only at the end of the list

D. Replacing the node at the given position with a new node

Solution

1. Step 1: Understand the position concept

   Inserting at a specific position means counting nodes from the head and placing the new node exactly there.

2. Step 2: Differentiate from other insertions

   Insertion at start or end is special cases; here we insert in the middle without removing nodes.

3. Final Answer:

   Adding a new node exactly at the given position counting from the head -> Option B

4. Quick Check:

   Insert at position = Insert node at that place [OK]

Hint: Insert at position means count nodes and place new node there [OK]

Common Mistakes:

• Confusing insertion with replacement
• Thinking insertion only happens at start or end
• Ignoring the position count from head

2. Which of the following C code snippets correctly inserts a new node at position 3 in a singly linked list?

easy

A. Traverse to node 2, set newNode->next = current->next; current->next = newNode;

B. Set newNode->next = head; head = newNode;

C. Traverse to node 3, set current->next = newNode; newNode->next = NULL;

D. Set newNode->next = NULL; current->next = newNode;

Solution

1. Step 1: Identify correct traversal

   To insert at position 3, we must reach node 2 (one before position 3).

2. Step 2: Link new node correctly

   Set newNode->next to current->next, then current->next to newNode to insert properly.

3. Final Answer:

   Traverse to node 2, set newNode->next = current->next; current->next = newNode; -> Option A

4. Quick Check:

   Insert at pos 3 = link after node 2 [OK]

Hint: Insert after (position-1)th node by adjusting pointers [OK]

Common Mistakes:

• Inserting at wrong node position
• Not updating newNode->next before linking
• Setting newNode->next to NULL incorrectly

3. What will be the linked list after inserting a node with value 99 at position 3 in the list 10 -> 20 -> 30 -> 40 -> NULL?

medium

A. 99 -> 10 -> 20 -> 30 -> 40 -> NULL

B. 10 -> 20 -> 30 -> 99 -> 40 -> NULL

C. 10 -> 20 -> 30 -> 40 -> 99 -> NULL

D. 10 -> 20 -> 99 -> 30 -> 40 -> NULL

Solution

1. Step 1: Identify position 3 in the list

   Positions: 1=10, 2=20, 3=30, so new node goes before 30.

2. Step 2: Insert new node at position 3

   New node with 99 points to 30, node 20 points to new node.

3. Final Answer:

   10 -> 20 -> 99 -> 30 -> 40 -> NULL -> Option D

4. Quick Check:

   Insert at pos 3 places 99 before 30 [OK]

Hint: Insert at pos 3 means new node is third in list [OK]

Common Mistakes:

• Inserting after instead of at position
• Appending at end instead of middle
• Confusing node values and positions

4. Given the following code snippet to insert at position 2, what is the error?

void insertAtPos(Node** head, int data, int pos) {
  Node* newNode = malloc(sizeof(Node));
  newNode->data = data;
  newNode->next = NULL;
  Node* temp = *head;
  for(int i = 1; i < pos - 1; i++) {
    temp = temp->next;
  }
  newNode->next = temp->next;
  temp->next = newNode;
}

medium

A. Memory for newNode is not allocated

B. newNode->next is not set before linking

C. Loop runs one time too many causing temp to be NULL

D. Head pointer is not updated after insertion

Solution

1. Step 1: Analyze loop for position 2

   Loop should run from i=1 to i < pos-1 to stop at node before insertion point.

2. Step 2: Check temp usage after loop

   Original loop runs one time too many, moving temp to node 2 instead of node 1, causing incorrect insertion or NULL access.

3. Final Answer:

   Loop runs one time too many causing temp to be NULL -> Option C

4. Quick Check:

   Loop overshoots node before position [OK]

Hint: Loop should stop at position-1 to avoid NULL [OK]

Common Mistakes:

• Loop runs to pos instead of pos-1
• Not checking for NULL before accessing temp->next
• Forgetting to allocate memory for new node

5. You have a singly linked list with 5 nodes: 1 -> 2 -> 3 -> 4 -> 5 -> NULL. You want to insert a node with value 99 at position 6. What should happen and what will be the final list?

hard

A. Node 99 is appended at the end; list becomes 1 -> 2 -> 3 -> 4 -> 5 -> 99 -> NULL

B. Node 99 is inserted at position 5; list becomes 1 -> 2 -> 3 -> 4 -> 99 -> 5 -> NULL

C. Node 99 replaces node 5; list becomes 1 -> 2 -> 3 -> 4 -> 99 -> NULL

D. Insertion fails because position 6 is invalid; list remains unchanged

Solution

1. Step 1: Check list length and position

   List has 5 nodes; position 6 means insert after last node (append).

2. Step 2: Insert node at end

   Since position is one more than length, node 99 is added after node 5.

3. Final Answer:

   Node 99 is appended at the end; list becomes 1 -> 2 -> 3 -> 4 -> 5 -> 99 -> NULL -> Option A

4. Quick Check:

   Insert at pos > length means append [OK]

Hint: Position beyond length means append at end [OK]

Common Mistakes:

• Rejecting valid append position
• Replacing last node instead of appending
• Inserting at wrong position due to off-by-one