Reverse Linked List Groups
Company: Bytedance
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Technical Screen
Quick Answer: This question evaluates linked list manipulation, pointer/reference management, in-place iterative algorithms, and the ability to enforce O(1) extra space constraints.
Part 1: Reverse Linked List in Groups Including the Final Partial Group
Constraints
- 1 <= len(values) <= 10^5
- 1 <= k <= len(values)
- -10^9 <= values[i] <= 10^9
- The core linked-list reversal must be iterative.
- Use O(1) auxiliary space for pointer manipulation, excluding the input/output serialization used by this problem.
Examples
Input: ([1, 2, 3, 4, 5], 2)
Expected Output: [2, 1, 4, 3, 5]
Explanation: Groups are [1,2], [3,4], and [5]. Each group is reversed.
Input: ([1, 2, 3, 4, 5], 3)
Expected Output: [3, 2, 1, 5, 4]
Explanation: The final partial group [4,5] is reversed to [5,4].
Input: ([7], 1)
Expected Output: [7]
Explanation: Single-node list with k = 1 remains unchanged.
Input: ([1, 2, 3, 4], 4)
Expected Output: [4, 3, 2, 1]
Explanation: When k equals the list length, the whole list is one reversed group.
Hints
- Reverse one group at a time, keeping track of the tail of the previously processed group.
- When a group ends early because the list runs out, reverse the nodes you have anyway instead of restoring them.
Part 2: Behavior of K-Group Reversal When k = 1
Constraints
- 0 <= len(values) <= 10^5
- -10^9 <= values[i] <= 10^9
- k is fixed to 1.
- The relative order of all nodes must remain unchanged.
Examples
Input: ([1, 2, 3, 4],)
Expected Output: [1, 2, 3, 4]
Explanation: Each group has one node, so the list is unchanged.
Input: ([],)
Expected Output: []
Explanation: An empty list remains empty.
Input: ([42],)
Expected Output: [42]
Explanation: A single node remains unchanged.
Input: ([-1, 0, -1, 5],)
Expected Output: [-1, 0, -1, 5]
Explanation: Values and order are preserved exactly.
Hints
- A one-element group is identical before and after reversal.
- Think about whether any pointer rewiring is necessary for a single-node group.
Part 3: Reverse Linked List Groups When k May Exceed the List Length
Constraints
- 0 <= len(values) <= 10^5
- 1 <= k <= 10^9
- -10^9 <= values[i] <= 10^9
- The core linked-list reversal must be iterative.
- Use O(1) auxiliary space for pointer manipulation, excluding input/output serialization.
Examples
Input: ([1, 2, 3], 5)
Expected Output: [3, 2, 1]
Explanation: k is larger than the list length, so the whole list is one partial group and is reversed.
Input: ([1, 2, 3, 4, 5], 2)
Expected Output: [2, 1, 4, 3, 5]
Explanation: Regular groups of size 2 are reversed, and the final single-node group remains itself.
Input: ([], 3)
Expected Output: []
Explanation: The empty list remains empty.
Input: ([9], 10)
Expected Output: [9]
Explanation: A single-node list is unchanged even though k is greater than the length.
Input: ([1, 2, 3, 4], 4)
Expected Output: [4, 3, 2, 1]
Explanation: When k equals the length, the entire list is reversed.
Hints
- You do not need to precompute the length to handle k > length.
- Start reversing a group and stop either after k nodes or when you reach the end of the list.
Part 4: Determine Complexity of Iterative K-Group Linked List Reversal
Constraints
- 0 <= n <= 10^18
- 1 <= k <= 10^18
- Use integer arithmetic; do not simulate the linked list.
- The reported asymptotic complexities are for the iterative linked-list reversal algorithm.
Examples
Input: (5, 2)
Expected Output: ['groups=3', 'pointer_writes=8', 'time=O(n)', 'space=O(1)']
Explanation: The groups are sizes 2, 2, and 1. Pointer writes are 5 node reversals plus 3 group connections.
Input: (5, 3)
Expected Output: ['groups=2', 'pointer_writes=7', 'time=O(n)', 'space=O(1)']
Explanation: The groups are sizes 3 and 2.
Input: (1, 1)
Expected Output: ['groups=1', 'pointer_writes=2', 'time=O(n)', 'space=O(1)']
Explanation: One node forms one group.
Input: (0, 4)
Expected Output: ['groups=0', 'pointer_writes=0', 'time=O(n)', 'space=O(1)']
Explanation: An empty list has no groups and no pointer writes.
Input: (10, 20)
Expected Output: ['groups=1', 'pointer_writes=11', 'time=O(n)', 'space=O(1)']
Explanation: When k > n, the whole non-empty list is one partial group.
Hints
- Every node is visited and reversed once, regardless of k.
- The number of groups is ceil(n / k), except that an empty list has zero groups.