Question

ngu Cons eY Ja Question 1 a) Write pseudo code to output a singly-linked list in reverse order when you are NOT allowed to allocate memory dynamically. What is the running time of the algorithm? b) Write pseudo code to output a singly-linked list in reverse order when you are ALLOWED to allocate memory dynamically. What is the running time of the algorithm? c) You have an increasingly-sorted circular list (using an array) of n elements that is full. The front and rear pointers have been lost. Explain how you can find the smallest element in better then O(n).

Answer 1

(a) Node reverseWIthoutDynamic(Node node) {
   Node prev = null;
   Node current = node;
   Node next = null;
   while (current != null) {
   next = current.next;
   current.next = prev;
   prev = current;
   current = next;
   }
   node = prev;
   return node;
   }
   // Time complexity = O(N)
(b) Node reverseWIthDynamic(Node node) {
   Node newList = null;
   Node current = node;
  
   while (current != null) {
   Node newnode = new Node(current.data);
   newnode.next = newList;
   newList = newnode;
   }
  
   return newList;
   }
   // Time complexity = O(N)

(c)

We can find minimum element in O(Logn) using Binary Search.

The minimum element is the only element whose previous is greater than it. We check this condition for middle element by comparing it with (mid-1)’th and (mid+1)’th elements.
If minimum element is not at middle (neither mid nor mid + 1), then minimum element lies in either left half or right half.
If middle element is smaller than last element, then the minimum element lies in left half.
Else minimum element lies in right half.