LC 708: Insert into a Sorted Circular Linked List

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 {

    # val = 0
    Node* next;

    Node() {}

    Node(int _val) {
        val = _val;
        next = NULL;
    }

    Node(int _val, Node* _next) {
        val = _val;
        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
        toInsert = False
        
        while True:
            # Normal case: insert between curr and curr.next
            if curr.val <= insertVal <= curr.next.val:
                toInsert = True
            # Wrap-around case: curr.next.val < curr.val indicates the wrap point
            elif curr.next.val < curr.val:
                # Insert at wrap-around (largest or smallest value)
                if insertVal >= curr.val or insertVal <= curr.next.val:
                    toInsert = True
            
            if toInsert:
                newNode = Node(insertVal)
                newNode.next = curr.next
                curr.next = newNode
                return head
            
            curr = curr.next
            if curr == head:
                break

        # All values are the same or insert at current position
        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

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

        # Insert at the end if value is too large
        if insertVal >= maxNode.val or insertVal <= maxNode.next.val:
            newNode = Node(insertVal, maxNode.next)
            maxNode.next = newNode
            return head

        # Find the correct insertion point
        curr = maxNode.next
        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:

    Node* insert(Node* head, int insertVal) {
        Node* newNode = new Node(insertVal);
        
        // Empty list
        if(!head) {
            newNode->next = newNode;
            return newNode;
        }

        curr = head
        
        while curr.next != head:
            # Normal insertion point
            if curr.val <= insertVal <= curr.next.val:
                break
            # Wrap-around insertion point
            if curr.val > curr.next.val and (insertVal >= curr.val or insertVal <= curr.next.val):
                break
            curr = curr.next
        
        # Insert at current position
        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(!head) {
    Node * newNode = new Node(insertVal);
    newNode->next = newNode;  // Self-referencing
    return newNode;
}

Normal Insertion Case

// Insert between curr and curr->next
if(curr->val <= insertVal  curr->next->val >= insertVal) {
    Node* newNode = new Node(insertVal, curr->next);
    curr->next = newNode;
    return head;
}

Wrap-Around Insertion Case

// At the wrap point (largest to smallest)
if(curr->next->val < curr->val) {
    // Insert if value is larger than max OR smaller than min
    if(insertVal >= curr->val || insertVal <= curr->next->val) {
        // Insert here
    }
}

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.