The Question

Reverse Nodes in k-Group

Given the head of a singly linked list, reverse the nodes of the list $k$ at a time and return the modified list. $k$ is a positive integer and is guaranteed to be less than or equal to the total number of nodes. If the number of nodes is not a multiple of $k$, the left-out nodes at the end should remain in their original order. You must perform the reversal in-place without modifying the values within the nodes; only the node pointers themselves may be changed. Example: Input: head = [1,2,3,4,5], k = 2 Output: [2,1,4,3,5] Constraints: - The number of nodes in the list is $n$. - $1 \le k \le n \le 5000$. - $0 \le Node.val \le 1000$.

Python

Linked List

Iterative Algorithm

Questions & Insights

Clarifying Questions

What are the constraints on the number of nodes $n$ and the value $k$?

Assumption:

1 \le k \le n \le 5000

. The list will not be empty based on

k \le n

, but we should handle the

k=1

case efficiently.

Is $O(1)$ extra space complexity a requirement?

Assumption: Yes, we should perform the reversal in-place by manipulating pointers, avoiding recursion (which would use

O(n/k)

stack space) or auxiliary arrays.

How should we handle cases where $k$ is larger than the total number of nodes?

Assumption: The problem states

k \le

length of the list, so we don't need to worry about

k > n

initially, but we must stop reversing when the remaining nodes are fewer than

k

Can we modify the node values?

Assumption: No, the requirement explicitly states only pointers should be changed.

Thinking Process

Identify Group Boundaries: Before reversing a segment, we must traverse

k

steps forward to ensure a full group exists. If we reach the end of the list before

k

steps, we leave the remaining nodes as they are.

Iterative Reversal Logic: Use a "dummy" node to handle the case where the head of the list changes. Maintain a groupPrev pointer to the node immediately before the segment to be reversed.

Pointer Re-wiring: For each group of

k

Identify the

k

-th node (groupEnd) and the node following it (nextGroupStart).

Disconnect the group temporarily or simply track the boundaries.

Reverse the

k

nodes.

Connect groupPrev.next to the new head of the reversed segment (which was the old groupEnd).

Connect the tail of the reversed segment (the old groupStart) to nextGroupStart.

Progress: Move the groupPrev pointer to the tail of the newly reversed segment and repeat until the remaining nodes are fewer than

k

Implementation Breakdown

Problem Set

Functional Requirements: Reverse every

k

nodes in a linked list; leave trailing nodes

(< k)

as is.

Constraints:

1 \le k \le n \le 5000

; Node values

0 \le val \le 1000

;

O(1)

auxiliary space.

Approach

Algorithm: Iterative Segment Reversal.

Data Structure: Singly Linked List.

Complexity:

Time:

O(n)

where

n

is the number of nodes (each node is visited at most twice).

Space:

O(1)

as we only use a few pointer variables.

Implementation

Wrap Up

Advanced Topics

Recursion vs. Iteration: While a recursive solution is often cleaner to write for linked lists, it consumes

O(n/k)

space on the call stack. In a production environment with very long lists, the iterative approach is preferred to prevent StackOverflowError.

Memory Locality: Linked lists have poor cache locality compared to arrays. If performance is critical and the list is modified frequently, one might consider if the data structure choice is optimal for the access patterns.

Readability: Using a helper function getKth improves the separation of concerns, making the main loop focus on the "re-wiring" logic rather than the traversal logic.