CareerCup

Better would be to keep 2 pointers ptr1 ptr2.....start ptr1 and move upto k elements....then start the ptr2....when ptr1 reaches the end....ptr2 will be in kth position from end....since they both have a difference of k....

1. Find the length n of list by traversing it. O(n) time
2. Again travese the list upto n-k+1 th element. You will get the answer. O(n) time

Use recursion count nodes,
& print kth node from the last while returning back from the list.

Very hard to do it not in O(n).

Really good solution.

struct node *findElement(struct node *head, int k) {
struct node *slow, *fast;
slow = fast = head;
while(fast->next != NULL) {
slow = slow->next;
for(i = 0; i < k ; i++) {
fast = fast->next;
}//end of for loop
}//end of while loop
return slow;
}