Kth Node From End Of Linked List

faq

Q1) How many Approach are there ?
ans : Two
      i) Ue the length of the linked List.
      ii) By maintaining two pointer.
      
Q2) What is the time complexity in this problem?
ans: The time complexity of the above solution is O(n) as we have traversed the linked list once. We have traversed it in two parts. 
     First we traversed k elements and then we traversed the remaining (size minus k) elements.

Q3) What is the space complexity in this problem?
ans: 0(1) as we have not used any extra space for the solution.

Hints

Hint 1: Try to use two pointer or find the size of the array and find the kth node

Hint 2: keep two pointers and update one with k steps ahead.

Hint 3: The kth node is the node where pointer with delay a of k steps is pointing.

Maq

Q1) What is the time complexity of the above code ?
    a) O(1)
    b) O(n^3)
    c) O(n)
    d) O(k)

Ans : c) O(n)

Q2) The three data members present int the Linked List class are :
    a) display, getFirst, getLast
    b) head, size, tail
    c) addAt, addLast, removeLast
    d) size, display, head

Ans : b) head, size, tail

Q3) Fill in the blank:

    Node f=head;
    Node s=head;
    for(int i=0;i<k;i++)
    {
        f=______;
    }
    while(f.next!=null)
    {
        f=______;
        s=s.next;
    }
    return s.data;
    
    a) f.next, f.next.next
    b) f.next.next, f.next
    c) f.next.next, f.next.next
    d) f.next, f.next
  
Ans: d) f.next, f.next