Flatten Multilevel List II in O(n) Time Using JavaScript

Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Output: [1,2,3,7,8,11,12,9,10,4,5,6]
Explanation:

The multilevel linked list in the input is as follows:



After flattening the multilevel linked list it becomes:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation:

The input multilevel linked list is as follows:

  1---2---NULL
  |
  3---NULL

Input: head = []
Output: []

 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]

[1,2,3,4,5,6,null]
[null,null,7,8,9,10,null]
[null,11,12,null]

[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]

Your algorithm should be iterative (non-recursive) and run in O(n) time.

Understanding the Problem

The core challenge of this problem is to flatten a multilevel doubly linked list into a single-level doubly linked list. This involves traversing through each node and ensuring that any child nodes are integrated into the main list in the correct order.

Common applications of this problem include scenarios where hierarchical data structures need to be linearized for easier processing or storage.

Potential pitfalls include missing nodes during traversal or incorrectly handling the child pointers, leading to an incorrect flattened list.

Approach

To solve this problem, we need to traverse the list iteratively and handle each node's child pointers appropriately. A naive solution might involve recursive traversal, but this is not optimal due to potential stack overflow issues with deep recursion. Instead, we will use an iterative approach with a stack to manage the nodes.

Naive Solution

A naive solution might involve recursively traversing the list and flattening each child list as it is encountered. However, this approach is not optimal due to the potential for deep recursion and stack overflow.

Optimized Solution

An optimized solution involves using an iterative approach with a stack. This allows us to manage the nodes without the risk of stack overflow and ensures that we can flatten the list in O(n) time.

Algorithm

Here is a step-by-step breakdown of the algorithm:

1. Initialize a stack and push the head of the list onto the stack.
2. While the stack is not empty, pop a node from the stack.
3. If the node has a next pointer, push the next node onto the stack.
4. If the node has a child pointer, push the child node onto the stack and set the child pointer to null.
5. Link the current node to the previous node in the flattened list.
6. Continue this process until the stack is empty.

Code Implementation

/**
 * Definition for a Node.
 * function Node(val, prev, next, child) {
 *    this.val = val;
 *    this.prev = prev;
 *    this.next = next;
 *    this.child = child;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
var flatten = function(head) {
    if (!head) return head;

    // Initialize the stack and push the head of the list
    let stack = [];
    stack.push(head);

    // Pointer to keep track of the previous node
    let prev = null;

    while (stack.length > 0) {
        // Pop a node from the stack
        let curr = stack.pop();

        // If there is a previous node, link it to the current node
        if (prev) {
            prev.next = curr;
            curr.prev = prev;
        }

        // If the current node has a next node, push it onto the stack
        if (curr.next) {
            stack.push(curr.next);
        }

        // If the current node has a child node, push it onto the stack
        if (curr.child) {
            stack.push(curr.child);
            // Don't forget to remove the child pointer
            curr.child = null;
        }

        // Move the previous pointer to the current node
        prev = curr;
    }

    return head;
};

Complexity Analysis

The time complexity of this approach is O(n), where n is the total number of nodes in the multilevel linked list. This is because we visit each node exactly once.

The space complexity is O(n) in the worst case, where all nodes are pushed onto the stack.

Edge Cases

Potential edge cases include:

• An empty list (head is null).
• A list with no child pointers.
• A list where all nodes have child pointers.

Each of these cases is handled by the algorithm, ensuring that the list is correctly flattened regardless of its initial structure.

Testing

To test the solution comprehensively, we should include a variety of test cases, from simple to complex. This includes lists with varying structures and depths.

We can use testing frameworks like Jest or Mocha to automate the testing process and ensure that our solution works correctly for all edge cases.

Thinking and Problem-Solving Tips

When approaching such problems, it's important to break down the problem into smaller, manageable parts. Visualizing the problem with diagrams can also help in understanding the structure and flow of the algorithm.

Practicing similar problems and studying different algorithms can help in developing problem-solving skills and improving efficiency.

Conclusion

In this blog post, we discussed how to flatten a multilevel doubly linked list into a single-level list using an iterative approach with a stack. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing.

Understanding and solving such problems is crucial for improving problem-solving skills and preparing for technical interviews. We encourage readers to practice and explore further to deepen their understanding.

Additional Resources

For further reading and practice problems related to this topic, consider the following resources:

• LeetCode: Flatten a Multilevel Doubly Linked List
• GeeksforGeeks: Flatten a Linked List with Next and Child Pointers
• Coding Ninjas: Flatten a Multilevel Doubly Linked List