Reverse Nodes In K Group
Detailed guide and Python implementation for the 'Reverse Nodes In K Group' problem.
1. Concept Overview
The 'Reverse Nodes In K Group' problem is a key challenge in the Linked List section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Reverse Nodes In K Group.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Reverse Nodes In K Group carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given the head of a linked list, reverse the nodes of the list k at a time, and return the modified list.
k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.
You may not alter the values in the list's nodes, only nodes themselves may be changed.
The linked list is represented as a Python list. Implement a function reverseKGroup(head: list, k: int) -> list.
- •The number of nodes in the list is n
- •1 <= k <= n <= 5000
- •0 <= Node.val <= 1000
Examples
[1,2,3,4,5], 2
[2,1,4,3,5]
Reverse in groups of 2: [1,2] becomes [2,1], [3,4] becomes [4,3], and [5] stays as is.
[1,2,3,4,5], 3
[3,2,1,4,5]
Reverse in groups of 3: [1,2,3] becomes [3,2,1], and [4,5] has fewer than 3 nodes so it stays.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def reverse_k_group_opt(head, k: int):
if isinstance(head, list):
h = build_linked_list(head)
res = reverse_k_group_opt_helper(h, k)
return linked_list_to_list(res)
return reverse_k_group_opt_helper(head, k)
def reverse_k_group_opt_helper(head: ListNode, k: int) -> ListNode:
curr = head
for _ in range(k):
if not curr:
return head
curr = curr.next
prev = None
curr = head
for _ in range(k):
nxt = curr.next
curr.next = prev
prev = curr
curr = nxt
head.next = reverse_k_group_opt_helper(curr, k)
return prevBrute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def reverse_k_group_brute(head, k: int):
if isinstance(head, list):
h = build_linked_list(head)
res = reverse_k_group_brute_helper(h, k)
return linked_list_to_list(res)
return reverse_k_group_brute_helper(head, k)
def reverse_k_group_brute_helper(head: ListNode, k: int) -> ListNode:
# Recursively reverse k elements by extracting values
curr = head
vals = []
for _ in range(k):
if not curr: return head
vals.append(curr.val)
curr = curr.next
curr = head
for v in reversed(vals):
curr.val = v
curr = curr.next
reverse_k_group_brute_helper(curr, k)
return headAlgorithm Pattern Checklist
When dealing with Linked List data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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