[Medium] LC 708: Insert into a Sorted Circular Linked List

[Medium] LC 708: Insert into a Sorted Circular Linked List

Difficulty: Medium
Category: Linked List, Circular List
Companies: Amazon, Facebook, Google, Microsoft

Problem Statement

Given a circular linked list, represented by a Node class, insert a new value into the list while maintaining the circular and sorted order of the list.

The list is circular, so the last node points back to the first node. The list is sorted in ascending order.

Examples

Example 1:

Input: head = [3,4,1], insertVal = 2
Output: [3,4,1,2]
Explanation: Insert 2 between 1 and 3, maintaining the circular sorted order.

Example 2:

Input: head = [], insertVal = 1
Output: [1]
Explanation: Create a circular list with a single node.

Example 3:

Input: head = [1], insertVal = 0
Output: [1,0]
Explanation: Insert 0 between 1 (tail) and 1 (head), wrapping around.

Constraints

• The number of nodes in the list is in the range [0, 5 * 10^4]
• -10^6 <= Node.val, insertVal <= 10^6
• List is sorted in ascending order
• List is circular

Solution Approaches

Approach 1: One-Pass Insertion (Recommended)

Key Insight: Traverse the circular list once and look for a valid insertion point. Handle edge cases at the wrap-around point.

Algorithm:

1. Handle empty list by creating a self-referencing node
2. Traverse the circular list
3. Find insertion point where curr->val <= insertVal && curr->next->val >= insertVal
4. Handle wrap-around case where curr->next->val < curr->val (wrap point)
   • Insert if insertVal >= curr->val OR insertVal <= curr->next->val
5. If no valid point found after full traversal, insert after current position

Time Complexity: O(n) where n is the number of nodes
Space Complexity: O(1)

# Definition for a Node.
class Node:
    def __init__(self, _val: int = 0, _next: 'Node | None' = None):
        self.val = _val
        self.next = _next


class Solution:
    def insert(self, head: 'Node', insertVal: int) -> 'Node':
        # Empty list case
        if not head:
            newNode = Node(insertVal)
            newNode.next = newNode
            return newNode

        curr = head

        while True:
            # Case 1: Normal position
            if curr.val <= insertVal <= curr.next.val:
                break

            # Case 2: Wrap-around point (max -> min)
            if curr.val > curr.next.val:
                if insertVal >= curr.val or insertVal <= curr.next.val:
                    break

            curr = curr.next

            # Case 3: Completed full loop (all values same or no spot found)
            if curr == head:
                break

        # Insert new node
        newNode = Node(insertVal, curr.next)
        curr.next = newNode

        return head

Approach 2: Two-Pass with Preprocessing

Algorithm:

1. First pass: Find the node with maximum value
2. Second pass: Search from max node’s next to find insertion point
3. Handle all same values case

Time Complexity: O(n)
Space Complexity: O(1)

class Solution:
    def insert(self, head: 'Node', insertVal: int) -> 'Node':
        if not head:
            newNode = Node(insertVal)
            newNode.next = newNode
            return newNode

        # Step 1: Find the maximum node
        maxNode = head
        curr = head.next
        while curr != head:
            if curr.val >= maxNode.val:
                maxNode = curr
            curr = curr.next

        # Step 2: Insert at wrap-around (max → min)
        if insertVal >= maxNode.val or insertVal <= maxNode.next.val:
            newNode = Node(insertVal, maxNode.next)
            maxNode.next = newNode
            return head

        # Step 3: Insert in normal sorted position
        curr = maxNode.next  # smallest node
        while curr.next.val < insertVal:
            curr = curr.next

        newNode = Node(insertVal, curr.next)
        curr.next = newNode

        return head

Approach 3: Simplified Logic

Algorithm: Streamlined version that handles all cases more elegantly.

class Solution:
    def insert(self, head: 'Node', insertVal: int) -> 'Node':
        newNode = Node(insertVal)

        # Case 1: Empty list
        if not head:
            newNode.next = newNode
            return newNode

        curr = head

        while True:
            # Case 2: Normal insertion
            if curr.val <= insertVal <= curr.next.val:
                break

            # Case 3: Wrap-around point
            if curr.val > curr.next.val and (
                insertVal >= curr.val or insertVal <= curr.next.val
            ):
                break

            curr = curr.next

            # Case 4: Full loop (all values same)
            if curr == head:
                break

        # Insert node
        newNode.next = curr.next
        curr.next = newNode

        return head

Algorithm Analysis

Key Insights

1. Circular List Boundary: The “wrap” happens when curr->val > curr->next->val
2. Insertion Point Detection: Need to check both normal range and wrap-around cases
3. Edge Cases: Empty list, single node, all same values
4. Traversal Guard: Use do-while to ensure at least one iteration

Understanding the Wrap-Around Logic

Sorted circular list: [3, 4, 1] → 1 points to 3

When curr = 4, curr->next = 1:
  - We're at the wrap point (largest → smallest)
  - insertVal = 2: insertVal >= 4? No, insertVal <= 1? No → continue
  - insertVal = 5: insertVal >= 4? Yes → insert here
  - insertVal = 0: insertVal <= 1? Yes → insert here

Implementation Details

Empty List Handling

if not head:
    new_node = Node(insertVal)
    new_node.next = new_node  # Self-referencing
    return new_node

Normal Insertion Case

# Insert between curr and curr.next
if curr.val <= insertVal <= curr.next.val:
    new_node = Node(insertVal, curr.next)
    curr.next = new_node
    return head

Wrap-Around Insertion Case

# At the wrap point (largest → smallest)
if curr.next.val < curr.val:
    # Insert if value is larger than max OR smaller than min
    if insertVal >= curr.val or insertVal <= curr.next.val:
        new_node = Node(insertVal, curr.next)
        curr.next = new_node
        return head

Edge Cases

1. Empty List: Create a circular list with single node
2. Single Node: Insert anywhere (trivially maintains order)
3. All Same Values: Insert at any position
4. Insert at Head: Special care needed for wrap logic
5. Insert Largest Value: Should go after maximum node
6. Insert Smallest Value: Should go before minimum node

Follow-up Questions

• What if the list is not guaranteed to be sorted?
• How would you handle duplicate insertion values?
• What if you need to insert multiple values at once?
• How would you delete a value from a circular list?

Related Problems

• LC 23: Swap Nodes in Pairs - Linked list manipulation
• LC 25: Reverse Nodes in k-Group - Linked list reversal
• LC 141: Linked List Cycle - Detect cycle in linked list
• LC 708: Insert into Sorted Circular List (This problem)

Optimization Techniques

1. Single Pass: Most efficient with O(n) time
2. Early Exit: Return immediately after insertion
3. Edge Case Handling: Handle empty list, single node separately
4. Wrap Detection: Identify wrap point by comparing adjacent values

Code Quality Notes

1. Readability: Clear variable names and logic separation
2. Correctness: Handles all edge cases properly
3. Performance: Optimal O(n) time complexity
4. Memory: O(1) space complexity

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This problem demonstrates sophisticated circular list manipulation and requires careful handling of wrap-around cases and edge conditions.