Is Reverse Nodes in k-Group easy or hard?

Reverse Nodes in k-Group is categorized as a hard problem because it requires careful pointer manipulation, handling partial groups, and maintaining correct connections between segments. The algorithm itself is linear, but implementing it without mistakes can be challenging.

Reverse Nodes in k-Group Python/Java solution

Both Python and Java solutions follow the same algorithm: iterate through the list, confirm k nodes exist, reverse the block using pointer updates, and reconnect it to the rest of the list. The logic remains identical across languages, with differences only in syntax and class definitions for the linked list node.

How to solve Reverse Nodes in k-Group in O(n)?

Process the linked list in groups of size k. For each group, first verify that k nodes exist, then reverse those k nodes using in-place pointer reversal. After reversing, connect the previous group's tail to the new head of the reversed block and continue scanning the list. This ensures each node is processed only a few times, giving O(n) time complexity.

What is the best approach for Reverse Nodes in k-Group?

The iterative group reversal approach is considered the best solution. It scans the list, reverses nodes in blocks of k using pointer manipulation, and reconnects the segments. This method runs in O(n) time and uses O(1) extra space, which is typically the optimal solution expected in interviews.

Is Reverse Nodes in k-Group asked at Google/Amazon/Meta?

Reverse Nodes in k-Group is a well-known hard linked list problem frequently discussed in interviews at large tech companies such as Amazon, Google, and Meta. Interviewers use it to evaluate pointer manipulation, edge case handling, and understanding of linked list operations.

What data structure is used in Reverse Nodes in k-Group?

The problem operates on a singly linked list. The solution relies on pointer manipulation to reverse segments of nodes and reconnect them correctly. No additional data structures are required in the optimal implementation.

What is the time complexity of Reverse Nodes in k-Group?

The optimal time complexity is O(n), where n is the number of nodes in the linked list. Each node is visited a constant number of times during group validation and reversal. The iterative approach achieves this with O(1) extra space, while the recursive version uses O(n/k) stack space.

Description Solution Video 24:31

25. Reverse Nodes in k-Group

Hard67.4% acceptance27 companies

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.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Constraints:

The number of nodes in the list is n.
1 <= k <= n <= 5000
0 <= Node.val <= 1000

Follow-up: Can you solve the problem in O(1) extra memory space?

Asked at top 20 of 27 companies

Topics

Linked List Recursion

24:31

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25. Reverse Nodes in k-Group

Asked at top 20 of 27 companies

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