### Introduction

The linked list is one of the most important concepts and data structures to learn while preparing for interviews. Having a good grasp of Linked Lists can be a huge plus point in a coding interview.

### Problem Statement

In this problem, we are given a singly linked list and a position. We have to insert a node at that specific position.

### Problem Statement Understanding

The problem is quite simple. We just have to insert a node at a specific position in the given linked list.

Suppose the given list is 1 β 2 β 3β 4 β 5, position = 2, and the data to be inserted is 6.

- According to the problem statement, we need to insert a node with value = 6 at 2
^{nd}position in the linked list. - So, after adding the node with value = 6 at 2
^{nd}position in the given linked list, our resultant list will be 1 β 6 β 2β 3 β 4 β 5.

Now, I think from the above example, the problem statement is clear.

So, now the main question is how we should approach this problem? What is the first thing that comes to mind?

- List traversal, right? Yes. We are going to use list traversal to solve this problem.

Before moving further to the approach section, try to think of the approach by yourself.

- If stuck, no problem, we will thoroughly see how we can approach this problem in the next section.

Letβs move to the approach section.

### Approach

Our approach is going to be pretty simple.

- We will simply traverse till
**(position -1)**^{th}node and add the**newnode**just after that**node**. Well, how will this work? Let us take an example.- Suppose the list is 1 β 2 β 3, and we have to insert 4 at the 2
^{nd}position. - Now, we will traverse
**(position -1) = (2-1) = 1**node and after traversing 1 node, we will be standing at 1. - Now we will make
**4 β next = 1 β next**as we have to insert it after 1, and finally,**1 β next = 4**to link the nodes. - By doing the above steps, 4 will be added at that specific position, and our resultant linked list will be: 1 β 4 β 2 β 3.

- Suppose the list is 1 β 2 β 3, and we have to insert 4 at the 2

We can get a clearer look of approach by looking at the dry run.

### Algorithm

- If the position where we are asked to insert
**pos**is smaller than 1 or greater than the**size of the list**, it is invalid, and we will return. - Else, we will make variable
**curr**and make it point to the**head**of the list. - Now we will run a for loop using
**curr**to reach to the node at**(pos-1)**^{th}position:- The for loop will be:
**for(int i=1;inext;}** - After the termination of the above loop,
**curr**will be standing at the**(position - 1)**^{th}node. - As explained above, we will simply make
**newnode β next = curr β next**and**curr β next = newnode**.

- The for loop will be:
- If the
**pos**was equal to 1, we will make the**head**point to**newnode**as**newnode**will become the first node of the list.

### Dry Run

### Code Implementation

#includeusing namespace std; // Node structure of a singly linked list node struct Node { int data; struct Node* next; }; int size = 0; // Using this function we will be creating new nodes Node* getNode(int data) { Node* newNode = new Node(); newNode->data = data; newNode->next = NULL; return newNode; } // Using this function we will insert the newnode at the specific position void insertAtPosition(Node* head, int pos, int data) { if (pos < 1 || pos > size + 1) cout << "Invalid position!" << endl; else { Node *curr=head; for(int i=1;i next; } Node* temp=getNode(data); temp->next=curr->next; curr->next=temp; if(pos=1) head=temp; size++; } } // Using this function we will print the linked list void printList(struct Node* head) { while (head != NULL) { cout << " " << head->data; head = head->next; } cout << endl; } // Driver function int main() { Node* head = NULL; head = getNode(1); head->next = getNode(2); head->next->next = getNode(3); head->next->next->next = getNode(4); head->next->next->next->next = getNode(5); size = 5; cout << "Linked list before insertion: "; printList(head); int data = 6, pos = 2; insertAtPosition(head, pos, data); cout << "Linked list after insertion of 6 at position 2: "; printList(head); return 0; }

public class PrepBytes { // Structure of a singly linked list node static class Node { public int data; public Node nextNode; public Node(int data) { this.data = data; } } // Function to create a new node and return static Node GetNode(int data) { return new Node(data); } // Function to insert an element at a specified index static Node InsertAtPos(Node headNode, int position, int data) { Node head = headNode; if (position < 1) System.out.print("Invalid position"); if (position == 1) { Node newNode = new Node(data); newNode.nextNode = headNode; head = newNode; } else { for(int i=1;i

### Output:

Linked list before insertion: 1,2,3,4,5

Linked list after insertion of 6 at position 2: 1,6,2,3,4,5

**Time Complexity:** O(n), as list traversal is needed.

So, in this article, we have tried to explain how to insert an element at a specific position in a linked list. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.