Problem Description

This problem involves a special kind of linked list called a Circular Linked List, where the last element points back to the first, creating a closed loop. Your task is to insert a new value into the list so that the list's non-descending (sorted) order is maintained.

The nuances of this task are:

1. The node you are given could be any node in the loop, not necessarily the smallest value.
2. You may insert the new value into any position where it maintains the sort order.
3. If the list is empty (i.e., the given node is null), you must create a new circular list that contains the new value and return it.

You're expected to return a reference to any node in the modified list.

Intuition

The intuition behind the solution starts with understanding the properties of a sorted circular linked list. In such a list, values gradually increase as you traverse the list, before suddenly "resetting" to the smallest value once the loop completes.

Given these properties, the solution involves handling three distinct scenarios:

1. The list is empty: If the list is empty, you create a new node that points to itself and return it as the new list.

2. Normal insertion: In a populated list, you traverse the nodes looking for the right place to insert the new value. The right place would be between two nodes where the previous node's value is less than or equal to the new value, and the next node's value is greater than or equal to the new value.

3. Edge Insertion: If you don't find such a position, it means the new value either goes before the smallest value in the list or after the largest one. This specific case occurs at the "reset" point where the last node in sort order (with the largest value) points to the first node (with the smallest value).

The algorithm involves iterating through the list, comparing the insertVal with the current and next node values to find the correct insertion point. Once found, adjust the necessary pointers to insert the new node.

We also guarantee that we loop no more than once by comparing the current node to the initial head node. If we reach the head again, it means we've checked every node, which handles cases where all nodes in the list have the same value and the insertVal could go anywhere.

So the high-level steps are:

• Handle the edge case where the list is empty.
• Traverse the list looking for the insertion point.
• Insert the node when the correct spot is found.
• If the loop completes without finding the exact position, insert the node just after the largest value or before the smallest value.

With this approach, we insert the new value while maintaining the list's sorted order.

Learn more about Linked List patterns.

Solution Approach

The solution implementation involves a straightforward but careful traversal of the circular linked list to maintain its sorted property after insertion.

First, we need to handle a special case:

• Empty list: If the list is empty (head is None), we create a new node that points to itself and return it as the new list head (circular in nature).

For a non-empty list, we follow these steps:

1. Initialize two pointers, prev and curr - starting from head and head.next respectively. We’ll move these pointers forward until we find the correct insert location.

2. Loop through the circular list and look for the correct position to insert the insertVal. The loop will terminate if:

   • We find a spot where the value directly after prev (curr) is greater than or equal to insertVal, and the value at prev is less than or equal to insertVal. This means insertVal fits right between prev and curr.
   • We find the point where the values "reset," indicating prev holds the maximum value and curr holds the minimum value in the list. If insertVal is greater than or equal to prev.val (greater than the list's max value) or insertVal is less than or equal to curr.val (less than the list's min value), it means insertVal must be inserted here.

3. For insertion:

   • Create a new node with insertVal.
   • Adjust the next pointer of prev to point to this new node.
   • Set the new node's next pointer to curr.

4. Return the head of the list.

Here is the algorithm in Python code snippet from the Reference Solution Approach:

class Node:
    def __init__(self, val=None, next=None):
        self.val = val
        self.next = next

class Solution:
    def insert(self, head: 'Optional[Node]', insertVal: int) -> 'Node':
        node = Node(insertVal)
        if head is None:
            node.next = node
            return node
        prev, curr = head, head.next
        while curr != head:
            if prev.val <= insertVal <= curr.val or (
                prev.val > curr.val and (insertVal >= prev.val or insertVal <= curr.val)
            ):
                break
            prev, curr = curr, curr.next
        prev.next = node
        node.next = curr
        return head

By utilizing the Node class to instantiate the new element to be inserted and updating pointers in a singly linked list fashion, you efficiently maintain both the circular nature of the list and the sorted order with a simple yet effective algorithm that walks through the list once at most.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Suppose we have a circular sorted linked list and we need to insert a value insertVal into the list. Our circular linked list is as follows:

3 -> 4 -> 1
^         |
|_________|

Here, 3 is the smallest element before the reset and 1 is the last element pointing back to 3. Now let's say we want to insert 5. We must find the correct spot for 2 so that the list remains sorted in non-descending order. The step will be as follows:

1. Since the list is not empty, we initialize prev to this node (3) and curr to the next node (4).

2. We start traversing the list. As insertVal (5) is not between 3 (prev.val) and 4 (curr.val), we move forward.

3. Now, prev is 4 and curr is 1. We check:

   • Is 4 (prev.val) less than 5 (insertVal) and 5 (insertVal) less than 1 (curr.val)? No.
   • Does 4 (prev.val) > 1 (curr.val) indicating a reset point in the list ordering? Yes.
   • Is 5 (insertVal) >= 4 (prev.val) or 5 (insertVal) <= 1 (curr.val)? The first condition is true.

4. We have reached the reset point of our list, and since 5 (insertVal) is less than 1 (curr.val), it must be inserted between 4 (prev) and 1 (curr).

5. We create a new node with the value 5, update the prev.next to point to this new node, and set the new node's next pointer to curr.

Now our list looks like this:

3 -> 4 -> 5 -> 1
^               |
|_______________|

6. We return the head of the list, which can be any node in the list; in this case, we still consider the node with value 3 as the head.

Through this example, we can see how the list is effectively iterated once to find the correct spot, and how the pointers are adjusted to maintain the order after insertion. The critical part of the approach is handling the edge cases where the insertVal is greater than the maximal value or smaller than the minimal value in the list.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(n), where n is the number of nodes in the circular linked list. This complexity arises because in the worst-case scenario, the code would iterate through all the elements of the linked list once to find the correct position to insert the new value. The while loop continues until it returns to the head, meaning it could traverse the entire list.

Space Complexity

The space complexity of the given code is O(1). The only extra space used is for creating the new node to be inserted, and no additional space that grows with the size of the input linked list is used.

Learn more about how to find time and space complexity quickly using problem constraints.