Program for n’th node from the end of a Linked List in C program

Given with n nodes the task is to print the nth node from the end of a linked list. The program must not change the order of nodes in a list instead it should only print the nth node from the last of a linked list.
Input -: 10 20 30 40 50 60
   N=3
Output -: 40
In the above example, starting from first node the nodes till count-n nodes are traversed i.e 10,20 30,40, 50,60 and so the third node from the last is 40.
Instead of traversing the entire list this efficient approach can be followed −
Take a temporary pointer, let’s say, temp of type node
Set this temp pointer to first node which is pointed by head pointer
Set counter to the number of nodes in a list
Move temp to temp → next till count-n
Display temp → data
If we use this approach, than count will be 5 and program will iterate the loop till 5-3 i.e. 2, so starting from 10 on 0th location than to 20 on 1st location and 30 on 2nd location which is the result. So by this approach there is no need to traverse the entire list till end which will save space and memory.
Start
Step 1 -> create structure of a node and temp, next and head as pointer to a structure node
   struct node
      int data
      struct node *next, *head, *temp
   End
Step 2 -> declare function to insert a node in a list
   void insert(int val)
      struct node* newnode = (struct node*)malloc(sizeof(struct node))
      newnode->data = val
      IF head= NULL
         set head = newnode
         set head->next = NULL
      End
      Else
         Set temp=head
         Loop While temp->next!=NULL
            Set temp=temp->next
         End
         Set newnode->next=NULL
         Set temp->next=newnode
      End
Step 3 -> Declare a function to display list
   void display()
      IF head=NULL
         Print no node
      End
      Else
         Set temp=head
         Loop While temp!=NULL
            Print temp->data
            Set temp=temp->next
         End
      End
Step 4 -> declare a function to find nth node from last of a linked list
   void last(int n)
      declare int product=1, i
      Set temp=head
      Loop For i=0 and i<count-n and i++
         Set temp=temp->next
      End
      Print temp->data
Step 5 -> in main()
   Create nodes using struct node* head = NULL
   Declare variable n as nth to 3
   Call function insert(10) to insert a node
   Call display() to display the list
   Call last(n) to find nth node from last of a list
Stop
 Live Demo
#include<stdio.h>
#include<stdlib.h>
//structure of a node
struct node{
   int data;
   struct node *next;
}*head,*temp;
int count=0;
//function for inserting nodes into a list
void insert(int val){
   struct node* newnode = (struct node*)malloc(sizeof(struct node));
   newnode->data = val;
   newnode->next = NULL;
   if(head == NULL){
      head = newnode;
      temp = head;
      count++;
   } else {
      temp->next=newnode;
      temp=temp->next;
      count++;
   }
}
//function for displaying a list
void display(){
   if(head==NULL)
      printf("no node ");
   else {
      temp=head;
      while(temp!=NULL) {
         printf("%d ",temp->data);
         temp=temp->next;
      }
   }
}
//function for finding 3rd node from the last of a linked list
void last(int n){
   int i;
   temp=head;
   for(i=0;i<count-n;i++){
      temp=temp->next;
   }
   printf("\n%drd node from the end of linked list is : %d" ,n,temp->data);
}
int main(){
   //creating list
   struct node* head = NULL;
   int n=3;
   //inserting elements into a list
   insert(1);
   insert(2);
   insert(3);
   insert(4);
   insert(5);
   insert(6);
   //displaying the list
   printf("\nlinked list is : ");
   display();
   //calling function for finding nth element in a list from last
   last(n);
   return 0;
}
linked list is : 1 2 3 4 5 6
3rd node from the end of linked list is : 4