How do you solve Remove Nodes From Linked List on LeetCode?

Remove Nodes From Linked List (LeetCode #2487) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Iterating from the end allows us to keep track of the maximum value efficiently.

What is the time complexity of Remove Nodes From Linked List?

The optimal solution for Remove Nodes From Linked List runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the linked list once. The space complexity is O(n) due to the stack storing nodes.

What is the brute force solution for Remove Nodes From Linked List?

The brute force approach for Remove Nodes From Linked List is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check each node and see if there is any node with a greater value to its right. If there is, we remove the node. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Remove Nodes From Linked List?

Remove Nodes From Linked List uses the following concepts: Linked List, Stack, Recursion, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Two Pointers.

What are the interview tips for Remove Nodes From Linked List?

Always clarify the problem constraints and edge cases with the interviewer. Think about the order of operations and how it affects the outcome. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Remove Nodes From Linked List?

Not considering the order of nodes when reconstructing the linked list. Confusing the conditions for removing nodes, leading to incorrect results.

#2487

Remove Nodes From Linked List

Medium

Linked ListStackRecursionMonotonic StackMonotonic StackTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This approach uses a stack to keep track of nodes that should remain in the list. By processing the list from the end to the beginning, we can efficiently determine which nodes to keep based on the maximum value seen so far.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack and a variable to track the maximum value seen so far.
2Step 2: Traverse the linked list from the end to the start.
3Step 3: For each node, if its value is greater than the maximum value seen, push it onto the stack and update the maximum value.
4Step 4: After processing all nodes, pop from the stack to reconstruct the linked list in the correct order.

solution.py22 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def removeNodes(head):
8    stack = []
9    max_val = float('-inf')
10    current = head
11    while current:
12        if current.val > max_val:
13            stack.append(current)
14            max_val = current.val
15        current = current.next
16    dummy = ListNode(0)
17    tail = dummy
18    while stack:
19        tail.next = stack.pop()
20        tail = tail.next
21    tail.next = None
22    return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse the linked list once. The space complexity is O(n) due to the stack storing nodes.

1Iterating from the end allows us to keep track of the maximum value efficiently.
2Using a stack helps in maintaining the order of nodes that need to be kept.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.