Reverse Nodes in k-Group — DSA Visualizer

Problem

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 is. You may not alter the values in the list's nodes, only the actual 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: 1 ≤ k ≤ n ≤ 5000 | 0 ≤ Node.val ≤ 1000

k-Group Bracket Visualization

Reversed (done)

Active group

Keep as-is

Pending

State Variables

Group #

—

Reversal Step

—

Nodes Reversed

Step Logic

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Ready

0 / 0

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Algorithm

Create dummy → head. Set groupPrev = dummy.

Find kth node from groupPrev. If null → fewer than k remain, stop.

Save groupNext = kth.next. Reverse the k-group using insert-at-front: pull curr toward prev.

Reconnect: groupPrev.next = kth. Advance groupPrev to old group head (now tail).

Repeat until no full group remains. Return dummy.next.

Time

O(n)

Space

O(1)

Why It Works

The insert-at-front technique avoids tracking six pointers simultaneously. We set prev = groupNext (the node just past the k-group), then repeatedly detach curr and splice it before prev. After k iterations, what was the last node (kth) is now at the front — and groupPrev.next = kth stitches the reversed group into the list. The original group head (tmp) becomes the new groupPrev (it is now the tail).