Reorder Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle node of the linked list using slow and fast pointers.

DSA Python

slow = head
fast = head
while fast and fast.next:
    slow = slow.[1]
    fast = fast.next.next

Drag options to blanks, or click blank then click option'

Anext

Bprev

Crandom

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using fast pointer's next instead of slow's next

Using prev instead of next

The slow pointer moves one step at a time using next to find the middle node.

2fill in blank

medium

Complete the code to reverse the second half of the linked list starting from the middle node.

DSA Python

prev = None
current = slow
while current:
    next_node = current.[1]
    current.next = prev
    prev = current
    current = next_node

Drag options to blanks, or click blank then click option'

Aprev

Bnext

Cchild

Drandom

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next

Not saving the next node before reversing

We store the next node using current.next before reversing the pointer.

3fill in blank

hard

Fix the error in merging two halves of the linked list alternately.

DSA Python

first = head
second = prev
while second:
    temp1 = first.next
    temp2 = second.next
    first.next = second
    second.next = [1]
    first = temp1
    second = temp2

Drag options to blanks, or click blank then click option'

Afirst

Btemp2

Csecond

Dtemp1

Attempts:

3 left

💡 Hint

Common Mistakes

Linking second.next to temp2 causes cycle

Linking second.next to first causes incorrect order

After linking second node, we link it to the next node of the first half, which is temp1.

4fill in blank

hard

Fill both blanks to complete the function that reorders the linked list.

DSA Python

def reorderList(head):
    if not head or not head.next:
        return
    slow = head
    fast = head
    while fast and fast.next:
        slow = slow.[1]
        fast = fast.next.next
    prev = None
    current = slow.[2]
    slow.next = None

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cchild

Drandom

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next

Using different pointers inconsistently

Both blanks use next to move forward in the list.

5fill in blank

hard

Fill all three blanks to merge the two halves of the list correctly.

DSA Python

first = head
second = prev
while second:
    temp1 = first.next
    temp2 = second.next
    first.next = second
    second.next = [1]
    first = [2]
    second = [3]

Drag options to blanks, or click blank then click option'

Atemp1

Btemp2

Cfirst

Dsecond

Attempts:

3 left

💡 Hint

Common Mistakes

Linking second.next incorrectly

Not updating pointers properly

We link second.next to temp1, then move first and second pointers to temp1 and temp2 respectively.

Practice

1. What is the main goal of the Reorder Linked List problem?

easy

A. Remove duplicate nodes from the linked list

B. Sort the linked list in ascending order

C. Reverse the entire linked list

D. Rearrange the list so nodes alternate from start and end towards the center

Solution

1. Step 1: Understand the problem statement

   The problem asks to reorder the list so that nodes come alternately from the start and end, like L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2 ...

2. Step 2: Differentiate from other operations

   This is not sorting or reversing the entire list, but a specific pattern of rearrangement.

3. Final Answer:

   Rearrange the list so nodes alternate from start and end towards the center -> Option D

4. Quick Check:

   Reorder means alternate start/end nodes = A [OK]

Hint: Remember: reorder means alternate front and back nodes [OK]

Common Mistakes:

• Confusing reorder with sorting
• Thinking reorder means full reversal
• Assuming duplicates are removed

2. Which of the following Python code snippets correctly splits a linked list into two halves for reordering?

easy

A. slow, fast = head, head.next
while fast and fast.next:
slow = slow.next
fast = fast.next.next
second = slow.next
slow.next = null

B. slow, fast = head, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
second = slow.next
slow.next = null

C. slow, fast = head.next, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
second = slow.next
slow.next = null

D. slow, fast = head, head
while fast.next and fast.next.next:
slow = slow.next
fast = fast.next.next
second = slow.next
slow.next = null

Solution

1. Step 1: Understand slow and fast pointer technique

   Using slow and fast pointers starting at head, fast moves twice as fast to find middle.

2. Step 2: Check code correctness

   slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next second = slow.next slow.next = null correctly initializes both at head and uses while fast and fast.next to move pointers safely.

3. Final Answer:

   Option B code snippet correctly splits the list into two halves -> Option B

4. Quick Check:

   Slow and fast start at head, fast moves twice = B [OK]

Hint: Start slow and fast at head; fast moves twice as fast [OK]

Common Mistakes:

• Starting fast at head.next causes off-by-one errors
• Incorrect while loop conditions causing crashes
• Not breaking the list by setting slow.next = null

3. What is the output linked list after reordering the list 1 -> 2 -> 3 -> 4 -> 5 -> null?

medium

A. 1 -> 5 -> 2 -> 4 -> 3 -> null

B. 5 -> 4 -> 3 -> 2 -> 1 -> null

C. 1 -> 2 -> 3 -> 4 -> 5 -> null

D. 1 -> 3 -> 5 -> 2 -> 4 -> null

Solution

1. Step 1: Identify the reorder pattern

   Reorder means first node, last node, second node, second last node, and so on.

2. Step 2: Apply pattern to given list

   For 1->2->3->4->5, reordered is 1->5->2->4->3.

3. Final Answer:

   1 -> 5 -> 2 -> 4 -> 3 -> null -> Option A

4. Quick Check:

   Alternate front and back nodes = C [OK]

Hint: Pick nodes alternately from front and back [OK]

Common Mistakes:

• Reversing entire list instead of reordering
• Keeping original order unchanged
• Skipping nodes or mixing order

4. Given the following code snippet for reordering a linked list, what is the bug?

def reorderList(head):
    if not head or not head.next:
        return
    slow, fast = head, head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
    second = slow.next
    slow.next = null
    prev = null
    while second:
        temp = second.next
        second.next = prev
        prev = second
        second = temp
    first, second = head, prev
    while second:
        temp1 = first.next
        temp2 = second.next
        first.next = second
        second.next = temp1
        first = temp1
        second = temp2

medium

A. The function does not return the head after reordering

B. The slow pointer should start at head.next to find middle correctly

C. The while loop merging two halves should check first and second, not just second

D. The code does not handle the case when the list has only one node

Solution

1. Step 1: Analyze function return behavior

   The function modifies the list in place but does not return anything, which is acceptable if list nodes are modified directly.

2. Step 2: Check if returning head is necessary

   Since the list is modified in place, returning head is optional; however, if used in some contexts, returning head is safer.

3. Step 3: Identify actual bug

   All other parts correctly find middle, reverse second half, and merge. The main issue is no return statement if function expected to return head.

4. Final Answer:

   The function does not return the head after reordering -> Option A

5. Quick Check:

   Missing return statement = A [OK]

Hint: Check if function returns the modified list head [OK]

Common Mistakes:

• Starting slow pointer at head.next causing wrong middle
• Merging loop condition incorrect causing infinite loop
• Ignoring single node edge case

5. You have a linked list 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null. After applying the reorder linked list algorithm, what will be the final list? Also, explain why the middle node is chosen as it is.

hard

A. 1 -> 6 -> 2 -> 5 -> 3 -> 4 -> null; middle chosen to split list evenly

B. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null; no reordering done for even length

C. 1 -> 6 -> 2 -> 5 -> 3 -> 4 -> null; middle chosen as the first of the two middle nodes

D. 1 -> 5 -> 2 -> 4 -> 3 -> 6 -> null; middle chosen as the last node

Solution

1. Step 1: Apply reorder pattern to even length list

   For 1->2->3->4->5->6, reordered list is 1->6->2->5->3->4.

2. Step 2: Explain middle node choice

   Using slow and fast pointers, slow stops at the first middle node (3), splitting list into two halves evenly.

3. Final Answer:

   1 -> 6 -> 2 -> 5 -> 3 -> 4 -> null; middle chosen as the first of the two middle nodes -> Option C

4. Quick Check:

   Even length split uses first middle node = D [OK]

Hint: For even length, slow stops at first middle node [OK]

Common Mistakes:

• Choosing second middle node as split point
• Not reordering even length lists
• Mixing node order in final list