How do you solve Reverse Nodes in k-Group on LeetCode?

Reverse Nodes in k-Group (LeetCode #25) can be solved using Brute Force or Optimal Solution (Iterative). The optimal approach is Optimal Solution (Iterative) with O(n) time complexity and O(1) space complexity. Key insight: Reversing a linked list in groups requires careful management of pointers to maintain the structure of the list.

What is the brute force solution for Reverse Nodes in k-Group?

The brute force approach for Reverse Nodes in k-Group is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves reversing each group of k nodes one at a time. We traverse the linked list, and for each k nodes, we reverse their order and then link them back to the rest of the list. The optimal Optimal Solution (Iterative) approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Nodes in k-Group?

Reverse Nodes in k-Group uses the following concepts: Linked List, Recursion. The recommended patterns to study are: Linked List, Two Pointers, Recursion.

What are the interview tips for Reverse Nodes in k-Group?

When you first see this problem, clarify the constraints and edge cases, such as what happens when the list length is not a multiple of k. Communicate your thought process clearly, explaining how you plan to reverse the nodes and link the groups together. Mention edge cases like when k is 1 (no change) or when k equals the length of the list (reverse entire list).

What are common mistakes when solving Reverse Nodes in k-Group?

Students often forget to handle the connection between the end of one reversed group and the start of the next group. Not checking if there are enough nodes left to reverse can lead to incorrect outputs.

#25

Reverse Nodes in k-Group

Hard

Linked ListRecursionLinked ListTwo PointersRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Iterative)★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses an iterative approach to reverse the nodes in groups of k without using extra space. This method efficiently links the reversed groups together while maintaining the overall structure of the list.

⚙️

Algorithm

3 steps

1Step 1: Count the total number of nodes in the linked list.
2Step 2: Use a loop to reverse nodes in groups of k until there are fewer than k nodes left.
3Step 3: For each group, reverse the nodes and link them back to the previous and next groups.

solution.py37 lines

1# Full working Python code with comments
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7class Solution:
8    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
9        # Count the number of nodes in the linked list
10        count = 0
11        current = head
12        while current:
13            count += 1
14            current = current.next
15
16        # Dummy node to help with the head of the new list
17        dummy = ListNode(0)
18        dummy.next = head
19        prev_group_end = dummy
20
21        # Reverse nodes in k-group
22        while count >= k:
23            current = prev_group_end.next
24            next_group_start = current
25            prev = None
26            for _ in range(k):
27                next_node = current.next
28                current.next = prev
29                prev = current
30                current = next_node
31            # Connect the end of the previous group to the start of the new group
32            prev_group_end.next = prev
33            next_group_start.next = current
34            prev_group_end = next_group_start
35            count -= k
36
37        return dummy.next

ℹ

Complexity note: The time complexity is O(n) because we traverse the list a constant number of times (once to count nodes and once to reverse). The space complexity is O(1) since we are not using any extra space that grows with input size.

1Reversing a linked list in groups requires careful management of pointers to maintain the structure of the list.
2Understanding how to manipulate pointers in linked lists is crucial for solving problems involving node rearrangement.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.