To solve the Next Greater Node In Linked List problem, immediately map the linked list into an array to access indices efficiently. Use a monotonic decreasing stack to track nodes without a known next greater value. When encountering a larger node, pop from the stack and assign the current value as the next greater for all applicable nodes, ensuring linear time complexity.
Problem Statement
Given the head of a singly linked list, determine for each node the first subsequent node that has a strictly larger value. Return the results in an array corresponding to the original node order, using 0 if no such node exists.
You must handle the linked list efficiently without unnecessary repeated traversal. Employ techniques that combine linked list pointer manipulation with array and stack structures to achieve optimal performance for up to 10,000 nodes.
Examples
Example 1
Input: head = [2,1,5]
Output: [5,5,0]
Example details omitted.
Example 2
Input: head = [2,7,4,3,5]
Output: [7,0,5,5,0]
Example details omitted.
Constraints
- The number of nodes in the list is n.
- 1 <= n <= 104
- 1 <= Node.val <= 109
Solution Approach
Convert Linked List to Array
Iterate through the linked list to create an array of node values, preserving original order. This allows index-based access for easier application of stack-based logic.
Use a Monotonic Stack
Maintain a stack of indices whose next greater value hasn't been found yet. Traverse the array, and for each value, pop indices from the stack when the current value is larger, assigning it as their next greater value.
Populate Remaining Nodes
After traversal, any indices left in the stack do not have a next greater node. Set their corresponding array entries to 0, finalizing the output array.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(n) |
| Space | O(n) |
Time complexity is O(n) because each node is pushed and popped at most once from the stack. Space complexity is O(n) for storing the array of node values and the stack.
What Interviewers Usually Probe
- Expecting use of a monotonic stack to efficiently track next greater elements in a linked list context.
- Looking for correct handling of node indexing when mapping linked list to array.
- Assessing understanding of linear-time solutions versus naive O(n^2) traversal approaches.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to convert the linked list into an array first, which complicates index-based stack management.
- Incorrectly popping from the stack too early or late, leading to wrong next greater assignments.
- Failing to set remaining nodes to 0 after stack processing, resulting in incomplete output.
Follow-up variants
- Finding next smaller node instead of next greater, adjusting stack condition accordingly.
- Applying the pattern to a circular linked list where next greater may wrap around to the start.
- Handling multiple linked lists simultaneously, requiring separate stacks or merged tracking.
How GhostInterview Helps
- Automatically suggests the monotonic stack pattern tied to this exact linked-list pointer manipulation problem.
- Highlights failure modes like missing array conversion or leftover stack nodes, preventing common mistakes.
- Provides example-driven reasoning showing how to efficiently populate the next greater values in linear time.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the most efficient approach for Next Greater Node In Linked List?
Convert the linked list into an array and use a monotonic decreasing stack to track nodes without known next greater values, achieving O(n) time.
Can this pattern be applied to arrays directly?
Yes, the monotonic stack approach works directly on arrays, but mapping from a linked list requires initial conversion to preserve order.
Why do we need a stack for this problem?
The stack tracks nodes whose next greater value hasn’t been determined yet, allowing efficient updates as larger values are encountered.
What happens if a node has no next greater node?
Any node remaining in the stack after full traversal has no next greater node, and its corresponding array entry is set to 0.
How does this solution avoid O(n^2) traversal?
Each node is pushed and popped from the stack at most once, ensuring linear traversal instead of checking every subsequent node for each element.
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