Rotate List
A medium Linked List problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Linked List
- Sheets
- 1
- Core for
- 3 roles
- Platform
- LeetCode
The problem
Rotate a linked list to the right by k places.
Example 1
- Input
- [1,2,3,4,5], k=2
- Output
- [4,5,1,2,3]
Example 2
- Input
- [0,1,2], k=4
- Output
- [2,0,1]
Example 3
- Input
- [1], k=0
- Output
- [1]
Constraints
- 1 <= n <= 500
- 0 <= k <= 2*10^9
How to think about it
Updated 2026-09-09Rotating a list shifts the cut point where the tail reconnects to the head. Connecting the tail to the original head forms a ring; breaking that ring at the right offset instantly turns the remainder into the new rotated list.
Approaches, worst first
Repeated single rotations
time O(k * n) · space O(1)
Pop the last element and prepend it to the head k times. Each rotation scans the entire list to find the second-to-last node, leading to quadratic running time that fails on large k.
Ring closure and cutWrite this one
time O(n) · space O(1)
Traverse to measure total length n and connect the tail back to head. Compute effective rotation `k % n`. Walk `n - (k % n) - 1` steps from head to reach the new tail, save the new head as `tail.next`, and break the ring by setting `tail.next = null`.
Where people lose marks · 3
- Failing to compute `k % n` causes timeouts or excessive loop iterations because k can reach 2 × 10^9 while n is at most 500.
- When `k % n == 0`, returning head directly avoids unnecessary ring-linking and unlinking operations.
- Off-by-one errors when counting steps to the new tail: you need the node whose next pointer becomes null, which sits at index `n - (k % n) - 1` from the start.
The theory behind it
Linked List — the ground this problem stands on. All Linked List problems
What Linked List is
A linked list is a chain of separate cargo cars connected by coupling hooks, scattered anywhere across memory rather than sitting in a tidy contiguous row. Each car, called a node, holds a single piece of data and a pointer directing traffic to the address of the next car in line. Because nodes connect only by directional links, jumping straight to the tenth car is impossible without walking past the first nine.
When to reach for it
Choose a linked list when a problem requires frequent insertions and deletions at known positions without shifting whole blocks of surrounding memory. Problems mentioning pointer splicing, reversing subsequences in place, merging sorted streams, or detecting cycles in linear chains strongly point here. It is ideal when total capacity is unpredictable and memory allocation must happen one individual node at a time.
How the pattern works
Think in terms of pointer rewiring before dereferencing. Keep a dummy head node pointing to the start of the list so modifications to the initial item do not require separate edge logic. Always save references to neighboring nodes into temporary variables before cutting or redirecting forward links. When diagnosing loops or locating middle nodes, advance two references simultaneously at differing velocities so traversal completes without supplementary storage.
What each operation costs
| Operation | Time |
|---|---|
| insert or delete at the head | O(1) |
| insert or delete after a known node | O(1) |
| find an element by value or position | O(n) |
What usually goes wrong with Linked List
- Losing access to the remainder of the chain by overwriting a next reference before caching the downstream node address in a temporary variable.
- Attempting to read properties of a null node reference after walking one step beyond the tail or advancing a fast runner without checking its next step.
- Creating an accidental infinite cycle by pointing a trailing node back into earlier segments of the chain without severing old outgoing links.
Which roles need this problem
Linked List is a core topic for these 3 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
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