Leetcode #2487: Remove Nodes From Linked List
In this guide, we solve Leetcode #2487 Remove Nodes From Linked List in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given the head of a linked list. Remove every node which has a node with a greater value anywhere to the right side of it.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Stack, Recursion, Linked List, Monotonic Stack
Intuition
We need the next greater or smaller element efficiently, which is exactly what a monotonic stack offers.
Each element is pushed and popped at most once, yielding a linear-time scan.
Approach
Maintain a stack that is either increasing or decreasing, depending on the query.
When the invariant is broken, pop and resolve answers for those indices.
Steps:
- Scan elements once.
- Pop while the monotonic condition is violated.
- Use stack indices to update answers.
Example
Input: head = [5,2,13,3,8]
Output: [13,8]
Explanation: The nodes that should be removed are 5, 2 and 3.
- Node 13 is to the right of node 5.
- Node 13 is to the right of node 2.
- Node 8 is to the right of node 3.
Python Solution
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
nums = []
while head:
nums.append(head.val)
head = head.next
stk = []
for v in nums:
while stk and stk[-1] < v:
stk.pop()
stk.append(v)
dummy = ListNode()
head = dummy
for v in stk:
head.next = ListNode(v)
head = head.next
return dummy.next
Complexity
The time complexity is , and the space complexity is , where is the length of the linked list. The space complexity is , where is the length of the linked list.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.