#25

Hard

auto_awesomeLinked-list pointer manipulation

LeetCode Problem Workspace

Reverse Nodes in k-Group

Reverse Nodes in k-Group challenges you to reverse segments of a linked list in groups of size k.

Problem Statement

Given a linked list, you need to reverse the nodes in groups of size k. You cannot alter the values of the nodes, only the order of the nodes themselves. If there are fewer than k nodes left at the end of the list, they should remain unchanged.

You must return the modified linked list with nodes reversed in k-sized groups. Note that k is a positive integer and the number of nodes in the list is at least k, ensuring that full k-sized groups can be reversed.

Examples

Example 1

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

Output: [2,1,4,3,5]

Example details omitted.

Example 2

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

Output: [3,2,1,4,5]

Example details omitted.

Constraints

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

Solution Approach

Linked List Pointer Manipulation

The key to this problem is to reverse nodes in-place using linked list pointer manipulation. We iterate through the list in chunks of k, reversing each chunk while keeping track of the head and tail of each segment.

Recursion and Base Case

Recursion helps break the problem into manageable chunks. The base case handles when there are fewer than k nodes left, in which case we return the list as is, avoiding any unnecessary changes.

Maintaining List Integrity

While reversing each group, we ensure that pointers are correctly adjusted so that the modified list remains intact, avoiding issues such as circular references or lost nodes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the number of nodes in the list. This is because we traverse the list once, reversing each k-sized segment. The space complexity is O(1) since we only use constant extra space during the reversal process.

What Interviewers Usually Probe

Candidates who focus on pointer manipulation and recursion will likely show a strong understanding of the problem's core pattern.
Look for candidates who efficiently reverse nodes without excessive memory allocation or recursion depth.
Candidates should be able to identify edge cases, such as when the number of nodes is less than k, and handle them appropriately.

Common Pitfalls or Variants

Common pitfalls

Failing to properly reverse nodes and inadvertently breaking the list structure can lead to lost or corrupted data.
Mismanaging the list when fewer than k nodes remain will result in incorrect answers or infinite loops.
Candidates may struggle with recursion depth limitations in cases with long lists, especially when not using iteration.

Follow-up variants

Handling different k values, including small values (k = 1) and larger values relative to list size.
Implementing the solution iteratively instead of recursively for performance optimization in cases with large n.
Reversing nodes in different patterns, such as reversing every other k nodes, instead of all k nodes consecutively.

FAQ

What is the core pattern of the Reverse Nodes in k-Group problem?

The core pattern is linked-list pointer manipulation, where you reverse segments of the list in-place in groups of k nodes, using recursion or iteration.

Can I alter the values of the nodes in this problem?

No, you can only alter the order of the nodes, not the values within the nodes.

How do I handle the case when there are fewer than k nodes left?

When fewer than k nodes remain, you simply leave them unchanged and return the rest of the list as is.

What are the time and space complexities for this problem?

The time complexity is O(n), where n is the number of nodes, and the space complexity is O(1) as no additional memory is used for node storage.

How can GhostInterview assist with this problem?

GhostInterview helps by providing simulated test cases and feedback on pointer manipulation techniques, helping you improve both recursion and iteration approaches.

terminal

Solution

Solution 1: Simulation

We can simulate the entire reversal process according to the problem description.

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseKGroup(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        def reverse(head: Optional[ListNode]) -> Optional[ListNode]:
            dummy = ListNode()
            cur = head
            while cur:
                nxt = cur.next
                cur.next = dummy.next
                dummy.next = cur
                cur = nxt
            return dummy.next

        dummy = pre = ListNode(next=head)
        while pre:
            cur = pre
            for _ in range(k):
                cur = cur.next
                if cur is None:
                    return dummy.next
            node = pre.next
            nxt = cur.next
            cur.next = None
            pre.next = reverse(node)
            node.next = nxt
            pre = node
        return dummy.next

Continue Practicing

#24 swap nodes in pairs #21 merge two sorted lists #2 add two numbers