What is the brute force approach and complexity for Reverse Nodes in k-Group?

Put the node pointers in a vector, reverse each full block there, then relink in array order with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for Reverse Nodes in k-Group?

Count k nodes ahead; if there are k, reverse that block and connect it to the previous group's tail with optimal time complexity O(N) and space complexity O(1).

Reverse Nodes in k-Group

Q: How do you identify the pattern for Reverse Nodes in k-Group?

Look for key signals in the prompt: "reverse the nodes in groups of k" indicates Each block of k is reversed, one after another, and the blocks keep their order → Groups of k, reversing every full block.; "if fewer than k nodes are left, leave them" indicates Count k ahead before reversing. A short last block is never touched.; "O(1) extra memory" indicates No vector of nodes. Reverse each block in place and join it through groupPrev..

Hard·LeetCode #25 Pro OnlyInterview-Level

LeetCode

Problem Statement

You are given the head of a linked list and a positive integer k. Split the list into blocks of k nodes, starting from the front. Reverse each full block, and keep the blocks in their original order. Return the head of the new list.

A block must have exactly k nodes. If fewer than k nodes are left at the end, leave that last block as it is.

Only the next arrows may change. Each node keeps its own value.

Input

head, the first node of a singly linked list, and k.

Output

The head of the list after every full block of k nodes is reversed.

Example 1

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

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

The blocks (1, 2) and (3, 4) are full, so they are reversed. Node 5 is a block of 1, which is less than k, so it stays.

Example 2

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

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

The block (1, 2, 3) is reversed. The short block (4, 5) is left as it is.

Example 3

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

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

A block of 1 node is the same when reversed, so nothing changes.

Constraints

• The list has n nodes, with 1 ≤ k ≤ n ≤ 5000.
• 0 ≤ Node.val ≤ 1000
• Follow-up: can you solve it with O(1) extra memory?

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“reverse the nodes in groups of k”

→ Each block of k is reversed, one after another, and the blocks keep their order → Groups of k, reversing every full block.

“if fewer than k nodes are left, leave them”

→ Count k ahead before reversing. A short last block is never touched.

“O(1) extra memory”

→ No vector of nodes. Reverse each block in place and join it through groupPrev.

Approach Complexity Ladder

Brute Force — collect the nodes, reverse full blocks in an array, relinkO(N)O(N)

Put the node pointers in a vector, reverse each full block there, then relink in array order

Optimal — count k ahead, reverse the block in place, link the groupsO(N)O(1)

Count k nodes ahead; if there are k, reverse that block and connect it to the previous group's tail

Brute Force Intuition

Store a pointer to every node in a vector. For each full block of k in the vector, reverse that range. A short last block is left alone. Then point every node at the entry after it in the vector. The nodes are reordered, never copied, and their values never change.

Recognition rule for this sub-pattern:

Repeat sublist reversal on each block of k nodes; leave a short last block alone (or not, as asked).

Remember this · Linked List — Reversal

Trigger

Reverse whole / part / every k nodes, reorder, palindrome check, numbers stored as lists.

Think

Which piece do I reverse, and which nodes must I remember to reconnect afterwards?

Key question

Before reversing, who is the node just before the piece and who is just after?

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