Leetcode #1474: Delete N Nodes After M Nodes of a Linked List
In this guide, we solve Leetcode #1474 Delete N Nodes After M Nodes of a 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 and two integers m and n. Traverse the linked list and remove some nodes in the following way: Start with the head as the current node.
Quick Facts
- Difficulty: Easy
- Premium: Yes
- Tags: Linked List
Intuition
Linked list problems often require pointer manipulation rather than extra memory.
Two-pointer techniques expose cycles, midpoints, or reordering patterns.
Approach
Traverse with fast/slow pointers or reverse sublists when needed.
Maintain invariants carefully to avoid losing nodes.
Steps:
- Traverse with pointers.
- Reverse or split if required.
- Reconnect nodes correctly.
Example
Input: head = [1,2,3,4,5,6,7,8,9,10,11,12,13], m = 2, n = 3
Output: [1,2,6,7,11,12]
Explanation: Keep the first (m = 2) nodes starting from the head of the linked List (1 ->2) show in black nodes.
Delete the next (n = 3) nodes (3 -> 4 -> 5) show in read nodes.
Continue with the same procedure until reaching the tail of the Linked List.
Head of the linked list after removing nodes is returned.
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 deleteNodes(self, head: ListNode, m: int, n: int) -> ListNode:
pre = head
while pre:
for _ in range(m - 1):
if pre:
pre = pre.next
if pre is None:
return head
cur = pre
for _ in range(n):
if cur:
cur = cur.next
pre.next = None if cur is None else cur.next
pre = pre.next
return head
Complexity
The time complexity is , where is the number of nodes in the linked list. The space complexity is .
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.