# Insert a node at a specific position in a linked list

### Problem Statement :

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node.

A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is empty.

Example
head refers to the first node in the list 1->2->3
data=4.
position=2

Insert a node at position 2 with data=4. The new list is 1->2->4->3

Function Description Complete the function insertNodeAtPosition in the editor below. It must return a reference to the head node of your finished list.

insertNodeAtPosition has the following parameters:
2. data: an integer value to insert as data in your new node
3. position: an integer position to insert the new node, zero based indexing

Returns:
1. SinglyLinkedListNode pointer: a reference to the head of the revised list

Input Format:

The first line contains an integer n, the number of elements in the linked list.
Each of the next n lines contains an integer SinglyLinkedListNode[i].data.
The next line contains an integer dta, the data of the node that is to be inserted.
The last line contains an integer position.

Constraints:
1.  1<=n<=1000
3.  0<=position<=n

### Solution :

Solution in C :

in C:

//The following function is all you need to complete the challenge in
//hackerrank platform

//iterate to the correct position in the linked list
if((position-1)>0){
insertNodeAtPosition(llist->next, data, position-1);
}
else{
newnode->next = llist->next;
llist->next = newnode;
}
return llist;
}

In C++:

//the following function is all that is needed to complete
//the challenge in hackerrank platform

Node* InsertNth(Node *head, int data, int position)
{
if(position == 0){
Node* a = (Node*)malloc(sizeof(Node));
a->data = data;
return a;
}else{
int i;
for(i = 1; i < position; i++)
a = a->next;
Node* tmp = (Node*)malloc(sizeof(Node));
tmp->data = data;
tmp->next = a->next;
a->next = tmp;
}
}

In Java:

//the following method is all that is needed to complete the
//challenge in hackerrank platform

if(position==0){
node.next = llist.next;
llist=node;
}else{
aux.next = llist;
for(int i=0;i<position;i++)aux=aux.next;
node.next = aux.next;
aux.next=node;
}

return llist;

}

In Python 3:

# the following method is all that is need to complete
# the challenge in hackerrank platform.

i = 0
newNode = Node(data, None)
if position is 0:
while i is not position - 1:
curr = curr.next
i+= 1
prev = curr
next = curr.next
prev.next = newNode
newNode.next = next

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