How do you solve Reverse Nodes in Even Length Groups on LeetCode?

Reverse Nodes in Even Length Groups (LeetCode #2074) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding group lengths is crucial for determining when to reverse nodes.

What is the brute force solution for Reverse Nodes in Even Length Groups?

The brute force approach for Reverse Nodes in Even Length Groups is Brute Force, with O(n²) time complexity and O(1) space complexity. We can traverse the linked list to identify groups based on their lengths and reverse the nodes in even-length groups. This approach is straightforward but inefficient as it requires multiple passes over the list. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Nodes in Even Length Groups?

Reverse Nodes in Even Length Groups uses the following concepts: Linked List. The recommended patterns to study are: Two Pointers, Linked List Manipulation.

What are the interview tips for Reverse Nodes in Even Length Groups?

Always clarify the problem statement and edge cases before diving into coding. Think about the structure of the linked list and how groups are formed. Practice reversing linked lists in place, as this is a common operation in many problems.

What are common mistakes when solving Reverse Nodes in Even Length Groups?

Not correctly managing pointers when reversing nodes, leading to broken lists. Failing to account for the last group, which may not be fully populated.

#2074

Reverse Nodes in Even Length Groups

Medium

Linked ListTwo PointersLinked List Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can achieve a more efficient solution by processing the linked list in a single pass while keeping track of the group length and reversing nodes in place when needed. This reduces the number of traversals and uses less memory.

⚙️

Algorithm

4 steps

1Step 1: Initialize pointers to traverse the linked list and keep track of the current group length.
2Step 2: For each group, determine its length and check if it is even.
3Step 3: If the group length is even, reverse the nodes in place by adjusting pointers.
4Step 4: Move to the next group and repeat until the end of the list.

solution.py38 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def reverseEvenLengthGroups(head):
8    if not head:
9        return head
10    current = head
11    group_length = 1
12    prev_tail = None
13    while current:
14        group_start = current
15        group_nodes = []
16        for _ in range(group_length):
17            if current:
18                group_nodes.append(current)
19                current = current.next
20        if group_length % 2 == 0:
21            # Reverse in place
22            prev = None
23            for node in group_nodes:
24                node.next = prev
25                prev = node
26            if prev_tail:
27                prev_tail.next = group_nodes[-1]
28            else:
29                head = group_nodes[-1]
30            group_nodes[0].next = current
31            prev_tail = group_nodes[0]
32        else:
33            if prev_tail:
34                prev_tail.next = group_start
35            prev_tail = group_start
36        group_length += 1
37    return head
38

ℹ

Complexity note: The time complexity is O(n) because we traverse the linked list only once, and the space complexity is O(1) since we do not use any additional data structures that grow with input size.

1Understanding group lengths is crucial for determining when to reverse nodes.
2Reversing nodes in place is more efficient than using additional data structures.

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