Reverse Linked List in Groups of K (Generalization)

📋

Reverse Linked List in Groups of K (Generalization)

Edge cases:

</>

IDE

def reverseKGroup(head: ListNode, k: int) -> ListNode:public ListNode reverseKGroup(ListNode head, int k)ListNode* reverseKGroup(ListNode* head, int k)function reverseKGroup(head, k)

def reverseKGroup(head, k):
    # Write your solution here
    pass

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        // Write your solution here
        return null;
    }
}

#include <vector>
using namespace std;

ListNode* reverseKGroup(ListNode* head, int k) {
    // Write your solution here
    return nullptr;
}

function reverseKGroup(head, k) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [1, 2, 3, 4, 5]Code never reverses any group, returning original list always.✅ Add logic to check for k nodes and reverse groups accordingly.

Wrong: [2, 1, 4, 3, 5, 6]Code reverses only first group and leaves rest unchanged.✅ Implement loop or recursion to process all groups until end of list.

Wrong: [3, 2, 1, 5, 4]Code reverses last partial group incorrectly instead of leaving it as is.✅ Check if k nodes are available before reversal; skip reversal if fewer than k.

Wrong: [1, 2, 3, 4, 5]Code reverses nodes even when k=1, causing no change but extra operations or errors.✅ Return head immediately if k=1 to avoid unnecessary reversal.

Wrong: Timeout or stack overflowRecursive solution with large input causing stack overflow or TLE.✅ Use iterative approach with O(1) space to handle large inputs efficiently.

✓

Test Cases

Focus on handling empty and minimal input cases correctly before moving on.

Pay attention to partial groups and pointer reconnections to avoid subtle bugs.

Optimize your solution to run in linear time and constant space for large inputs.

t1_01basic

Input{"head":[1,2,3,4,5],"k":2}

Expected[2,1,4,3,5]

⏱ Performance - must finish in 2000ms

The list is reversed in groups of 2. First two nodes (1,2) reversed to (2,1), next two (3,4) reversed to (4,3), last node (5) remains as is.

💡 Think about reversing nodes in chunks of size k rather than the entire list.

💡 Check if there are at least k nodes before reversing each group to avoid partial reversal.

💡 Reverse exactly k nodes at a time and reconnect the reversed groups properly.

Why it failed: Output differs because the code either reverses fewer than k nodes or reverses the last partial group incorrectly. Fix by checking if k nodes exist before reversing and leave the last partial group as is.

✓ Correctly reverses nodes in groups of k and handles leftover nodes properly.

t1_02basic

Input{"head":[1,2,3,4,5,6],"k":3}

Expected[3,2,1,4,5,6]

⏱ Performance - must finish in 2000ms

The list is reversed in groups of 3. First three nodes (1,2,3) reversed to (3,2,1), next three (4,5,6) reversed to (6,5,4).

💡 Try dividing the list into groups of size k and reverse each group separately.

💡 Use a pointer to check if there are k nodes available before reversing a group.

💡 After reversing a group, connect the tail of the reversed group to the head of the next group.

Why it failed: Incorrect output due to reversing partial groups or incorrect reconnection of reversed groups. Fix by verifying k nodes exist before reversal and properly linking groups.

✓ Successfully reverses all full groups of k nodes and reconnects the list correctly.

t2_01edge

Input{"head":[],"k":1}

Expected[]

⏱ Performance - must finish in 2000ms

Empty list input should return empty list regardless of k.

💡 Consider what happens when the input list is empty.

💡 Check if your code handles null or empty head pointers gracefully.

💡 Return the head immediately if it is null to avoid errors.

Why it failed: Code crashes or returns non-empty output on empty input. Fix by adding a base case to return head if it is null.

✓ Handles empty list input correctly by returning null.

t2_02edge

Input{"head":[1],"k":1}

Expected[1]

⏱ Performance - must finish in 2000ms

Single element list with k=1 should remain unchanged.

💡 Check behavior when list length equals 1 and k equals 1.

💡 Reversing groups of size 1 means no change to the list.

💡 Return the head as is when k=1.

Why it failed: Code incorrectly modifies or reverses single node when k=1. Fix by returning head immediately if k=1.

✓ Correctly returns the original list when k=1.

t2_03edge

Input{"head":[1,2,3,4],"k":4}

Expected[4,3,2,1]

⏱ Performance - must finish in 2000ms

k equals list length, so entire list is reversed.

💡 Test the case where k equals the length of the list.

💡 The entire list should be reversed as one group.

💡 Ensure your reversal logic handles full list reversal correctly.

Why it failed: Code fails to reverse entire list when k equals list length, possibly reversing partial groups only. Fix by allowing reversal when k equals list length.

✓ Successfully reverses the entire list when k equals list length.

t2_04edge

Input{"head":[7,7,7,7,7],"k":3}

Expected[7,7,7,7,7]

⏱ Performance - must finish in 2000ms

All elements identical; reversal should maintain values but change node order in groups.

💡 Check if your code reverses nodes by pointers, not by swapping values.

💡 Reversing identical values should still change node order.

💡 Verify that node connections are updated, not just values.

Why it failed: Code swaps values instead of nodes or fails to reverse nodes properly. Fix by reversing node pointers, not values.

✓ Correctly reverses nodes even when values are identical.

t3_01corner

Input{"head":[1,2,3,4,5],"k":3}

Expected[3,2,1,4,5]

⏱ Performance - must finish in 2000ms

Last group has fewer than k nodes and should remain unchanged.

💡 Check if your code reverses the last group even if it has fewer than k nodes.

💡 You must verify the availability of k nodes before reversal.

💡 Skip reversal for the last group if it contains fewer than k nodes.

Why it failed: Code incorrectly reverses last partial group with fewer than k nodes. Fix by adding a check to reverse only if k nodes are available.

✓ Correctly leaves last partial group unchanged.

t3_02corner

Input{"head":[1,2,3,4,5,6],"k":2}

Expected[2,1,4,3,6,5]

⏱ Performance - must finish in 2000ms

Tests correct reconnection of reversed groups; last group reversed properly.

💡 Ensure that after reversing each group, the next pointer of the group's tail points to the head of the next reversed group.

💡 Check pointer updates carefully to avoid broken links.

💡 Use a dummy node to simplify reconnection logic.

Why it failed: Code fails to reconnect reversed groups properly, causing broken or cyclic lists. Fix by updating next pointers after each reversal group.

✓ Successfully reconnects all reversed groups maintaining list integrity.

t3_03corner

Input{"head":[1,2,3,4,5],"k":1}

Expected[1,2,3,4,5]

⏱ Performance - must finish in 2000ms

Tests that no reversal occurs when k=1, a common greedy trap is to reverse anyway.

💡 Reversing groups of size 1 should not change the list.

💡 Check if your code unnecessarily reverses when k=1.

💡 Add a condition to return head immediately if k=1.

Why it failed: Code reverses nodes even when k=1, causing unnecessary operations or errors. Fix by returning head immediately if k=1.

✓ Correctly handles k=1 by returning the original list.

t4_01performance

Input{"head":[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100],"k":10}

Expectednull

⏱ Performance - must finish in 2000ms

Large input with n=100 and k=10 to test O(n) time complexity and O(1) space. Must complete within 2 seconds.

💡 Use iterative approach with constant extra space to handle large inputs efficiently.

💡 Avoid recursion to prevent stack overflow and extra space usage.

💡 Reverse nodes in-place and update pointers carefully to maintain O(n) time.

Why it failed: Code exceeds time limit due to inefficient recursion or repeated traversals. Fix by implementing iterative O(n) solution with O(1) space.

✓ Solution runs within time limit confirming O(n) time and O(1) space complexity.

Practice

1. What is the time complexity of the in-place iterative flattening algorithm for a multilevel doubly linked list with total n nodes, considering the worst-case scenario where each node has a child list?

medium

A. O(n) amortized, since the total number of pointer updates and traversals sums to linear over all nodes.

B. O(n) because each node is visited a constant number of times during the flattening process.

C. O(n^2) because for each node with a child, we traverse its entire child list to find the tail.

D. O(n log n) due to repeated tail searches in nested child lists.

Solution

1. Step 1: Analyze tail-finding loop

   Although tail is found by traversing child lists, each node is visited at most once as tail or curr.

2. Step 2: Amortized analysis

   All next pointers are traversed linearly; tail searches do not overlap nodes multiple times, so total work is linear.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Each node processed once, tail searches amortize to O(n) [OK]

Hint: Tail searches look nested but total visits are linear [OK]

Common Mistakes:

• Assuming tail search is repeated fully for each child causing O(n^2)
• Confusing amortized with worst-case per iteration
• Ignoring that nodes are not revisited multiple times

2. What is the time and space complexity of the optimal in-place odd-even linked list rearrangement algorithm for a list of length n?

medium

A. Time: O(n), Space: O(1) because each node is visited once and rearranged in-place

B. Time: O(n log n), Space: O(1) due to sorting nodes by index parity

C. Time: O(n), Space: O(n) due to auxiliary lists for odd and even nodes

D. Time: O(n^2), Space: O(1) because of nested pointer updates

Solution

1. Step 1: Identify loop iterations

   The while loop advances even and odd pointers through the list, each node visited once -> O(n) time.

2. Step 2: Analyze space usage

   No extra data structures are created; rearrangement is done by pointer manipulation -> O(1) space.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Linear time and constant space is standard for in-place linked list rearrangement [OK]

Hint: One pass with two pointers -> O(n) time, no extra lists -> O(1) space [OK]

Common Mistakes:

• Assuming nested loops cause O(n^2) time
• Confusing auxiliary space with recursion stack

3. Suppose the linked list can be extremely long (millions of nodes), causing recursion stack overflow in the recursive solution. Which modification is the best to handle this scenario efficiently?

hard

A. Increase the recursion limit to accommodate the large linked list depth.

B. Switch to an iterative approach that accumulates the decimal value using bit shifts.

C. Use a divide-and-conquer recursive approach to reduce recursion depth.

D. Convert the linked list to a string first, then parse the binary string to integer.

Solution

1. Step 1: Identify the problem with recursion

   Deep recursion causes stack overflow for very long linked lists.

2. Step 2: Evaluate alternatives

   Increasing recursion limit is risky and platform-dependent; divide-and-conquer adds complexity without reducing total depth; string conversion is inefficient.

3. Step 3: Choose iterative approach

   Iterative bitwise accumulation uses constant stack space and linear time, handling large inputs safely.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   Iterative approach avoids recursion stack overflow [OK]

Hint: Iterative avoids recursion stack overflow on large inputs [OK]

Common Mistakes:

• Relying on recursion limit increase which is unsafe
• Assuming divide-and-conquer reduces recursion depth effectively
• Using string conversion which is inefficient for large inputs

4. Suppose the multilevel doubly linked list can contain cycles via child pointers (i.e., a child pointer may point to an ancestor node). Which modification to the in-place iterative flattening algorithm correctly handles this scenario without infinite loops?

hard

A. Ignore cycles since the algorithm naturally terminates when curr is null.

B. Remove all child pointers before flattening to break cycles.

C. Use recursion with a depth limit to avoid infinite recursion.

D. Add a visited set to track nodes already processed and skip nodes if revisited.

Solution

1. Step 1: Understand cycle problem

   Cycles via child pointers cause infinite loops if nodes are revisited during flattening.

2. Step 2: Evaluate solutions

   Ignoring cycles (A) causes infinite loops. Removing child pointers (B) loses data. Recursion with depth limit (C) is unreliable. Tracking visited nodes (D) prevents revisiting and infinite loops.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Visited set prevents infinite loops in cyclic structures [OK]

Hint: Cycle detection requires tracking visited nodes explicitly [OK]

Common Mistakes:

• Assuming no cycles exist in input
• Relying on recursion depth limits
• Ignoring child pointers that create cycles

5. Suppose the linked list nodes can be reused multiple times (i.e., the list is circular or nodes can appear multiple times). Which modification to the odd-even rearrangement algorithm is necessary to handle this scenario correctly?

hard

A. Add a visited set to track nodes already processed to avoid infinite loops or duplicates

B. No modification needed; the original in-place algorithm works correctly even with reused nodes

C. Convert the list to an array first, then reorder and reconstruct the list to handle duplicates

D. Use recursion to process nodes and detect cycles automatically

Solution

1. Step 1: Identify problem with reused nodes

   If nodes are reused or list is circular, naive pointer traversal causes infinite loops or duplicates.

2. Step 2: Use a visited set

   Tracking visited nodes prevents revisiting and infinite loops, ensuring correct rearrangement.

3. Step 3: Why other options fail

   No modification ignores cycles; array conversion adds overhead; recursion risks stack overflow without cycle detection.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Visited set is standard for cycle detection in linked lists [OK]

Hint: Detect cycles with visited set to avoid infinite loops [OK]

Common Mistakes:

• Assuming original algorithm handles cycles or reused nodes
• Using recursion without cycle detection