How hard is LeetCode 2487. Remove Nodes From Linked List?

LeetCode 2487. Remove Nodes From Linked List is rated Medium on LeetCode.

What is the time complexity of LeetCode 2487. Remove Nodes From Linked List?

The Python solution on this page runs in O(n).

What is the space complexity of LeetCode 2487. Remove Nodes From Linked List?

The Python solution on this page uses O(n), where n is the length of the linked list auxiliary space.

What topics does LeetCode 2487. Remove Nodes From Linked List cover?

LeetCode 2487. Remove Nodes From Linked List is tagged Stack, Recursion, Linked List and Monotonic Stack on LeetCode.

MediumNo. 2487

Remove Nodes From Linked List — LeetCode 2487 Python Solution

Pattern: Linked List
Topics: Stack, Recursion, Linked List, Monotonic Stack
Reading time: 4 min
Source: leetcode.com

The problem

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.

Example

Input: head = [5,2,13,3,8]
Output: [13,8]
Explanation: The nodes that should be removed are 5, 2 and 3.

Python solution

PythonSolution

# 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

Time: O(n)
Space · auxiliary: O(n)where n is the length of the linked list

Pattern: Linked List

Rewire pointers in place, with a dummy head and a saved next to keep it safe. LeetCode 2487. Remove Nodes From Linked List is filed here because LeetCode tags it Linked List, which is the vocabulary this hub collects.

The linked list guide has the Python template for the pattern and the 75 LeetCode problems that use it.

Related problems
No.	Problem	Difficulty
1019	Next Greater Node In Linked List	Medium
2289	Steps to Make Array Non-decreasing	Medium
0084	Largest Rectangle in Histogram	Hard
0456	132 Pattern	Medium
0503	Next Greater Element II	Medium
0654	Maximum Binary Tree	Medium

Frequently asked questions

How hard is LeetCode 2487. Remove Nodes From Linked List?: LeetCode 2487. Remove Nodes From Linked List is rated Medium on LeetCode.
What is the time complexity of LeetCode 2487. Remove Nodes From Linked List?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 2487. Remove Nodes From Linked List?: The Python solution on this page uses O(n), where n is the length of the linked list auxiliary space.
What topics does LeetCode 2487. Remove Nodes From Linked List cover?: LeetCode 2487. Remove Nodes From Linked List is tagged Stack, Recursion, Linked List and Monotonic Stack on LeetCode.

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 Terra

Transcribing...

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

The problem

Example

Python solution

Complexity

Pattern: Linked List

Related problems

Frequently asked questions

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Approach

Approach

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