Approach Summary
Find length n and tail. Make it circular. New tail is at position n - (k % n) - 1; new head is its next. Break the ring.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Rotate List optimally, we apply the linked list algorithmic strategy. Find length n and tail. Make it circular. New tail is at position n - (k % n) - 1; new head is its next. Break the ring. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n) time, utilizing O(1) auxiliary memory.
Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
class Solution {
// Time: O(n), Space: O(1)
public ListNode rotateList(ListNode head) {
if (head == null || head.next == null) return head;
ListNode dummy = new ListNode(0, head);
ListNode slow = dummy, fast = dummy;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return dummy.next;
}
}class Solution {
public:
// Time: O(n), Space: O(1)
ListNode* rotateList(ListNode* head) {
if (!head || !head->next) return head;
ListNode dummy(0, head);
ListNode* slow = &dummy;
ListNode* fast = &dummy;
while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
}
return dummy.next;
}
};# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def rotate_list(self, head):
# Time: O(n), Space: O(1)
dummy = ListNode(0)
dummy.next = head
slow, fast = dummy, dummy
# Fast/slow pointer traversal or reversal logic
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
return dummy.next/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
var rotateList = function(head) {
if (!head || !head.next) return head;
let dummy = new ListNode(0, head);
let slow = dummy, fast = dummy;
while (fast.next && fast.next.next) {
slow = slow.next;
fast = fast.next.next;
}
return dummy.next;
};Edge Cases to Watch
- Key recognition trigger: k can exceed list length — k % n is the effective rotation.
- Constraint condition: Connect tail to head, then cut at the right point.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.
Master the Linked List Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Linked List pattern.
How to Recognize This Pattern
- k can exceed list length — k % n is the effective rotation
- Connect tail to head, then cut at the right point
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(1)