Sorted merge of two sorted circular linked lists
The Sorted Merge of Two Sorted Circular Linked Listsproblem involves merging two circular linked lists that are already sorted in ascending order, while maintaining the sorted order of the merged list. A circular linked list is a data structure where the last node points back to the first node, creating a loop. The task is to merge the two circular linked lists into a single circular linked list while keeping the elements sorted.
Problem Statement
Given two circular linked lists that are already sorted, the problem is to merge these two lists into a single circular linked list while maintaining the sorted order.
Example
Consider two sorted circular linked lists:
 First Linked List: 2 > 5 > 7 > 10 > 11 > 17
 Second Linked List: 1 > 3 > 4 > 8 > 20
The merged circular linked list should be: 1 > 2 > 3 > 4 > 5 > 7 > 8 > 10 > 11 > 17 > 20.
Idea to Solve the Problem
To solve the problem of merging two sorted circular linked lists, follow these steps:
 Create a new circular linked list to store the merged result.
 Initialize pointers
a
andb
to the first nodes of the two input circular linked lists.  Compare the values of
a
andb
nodes and append the smaller value node to the merged list.  Move the pointer of the chosen node to its next.
 Repeat steps 34 until both lists are exhausted.
 If one of the lists still has remaining nodes, append the remaining nodes to the merged list.
Pseudocode
function sortedMerge(l1, l2):
a = l1.head
b = l2.head
result = new CircularLinkedList()
auxiliary = null
while a is not null and b is not null:
if a.data < b.data:
auxiliary = a
a = a.next
else:
auxiliary = b
b = b.next
if result is not null:
result.tail.next = auxiliary
result.tail = auxiliary
auxiliary.next = l1.head
auxiliary = null
if b is not null:
result.tail.next = b
while b.next != l2.head:
b = b.next
b.next = l1.head
l1.tail = b
else if a is not null:
result.tail.next = a
return result
Algorithm Explanation
 Initialize two pointers
a
andb
to the first nodes of the input lists.  Create a new circular linked list
result
to store the merged output.  Compare the values of
a
andb
. Append the smaller node to theresult
list.  Move the pointer of the chosen node (
a
orb
) to its next.  Repeat the above steps until either
a
orb
becomes null.  If either
a
orb
is still not null, append the remaining nodes to theresult
list.  Update the
next
pointers to maintain the circular structure.
Code Solution

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Resultant Output Explanation
After merging the given sorted circular linked lists using the provided code, the merged circular linked list is printed. The output demonstrates that the elements from both input lists have been merged and the resulting list is in sorted order.
Time Complexity
The time complexity of merging two sorted circular linked lists is O(n + m), where n and m are the lengths of the input circular linked lists. This is because, in the worst case, we may need to iterate through both input lists once to merge them.
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