Reverse a Doubly Linked List in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to swap the next and prev pointers of the current node.

DSA Go

current.[1], current.prev = current.prev, current.next

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

Anext

Bprev

Cvalue

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping with the wrong fields like value or head.

Not swapping both pointers at once.

Swapping current.next and current.prev reverses the direction of the links for the current node.

2fill in blank

medium

Complete the code to move to the next node after swapping pointers.

DSA Go

current = current.[1]

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

Atail

Bnext

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using current.next instead of current.prev after swapping.

Using unrelated fields like head or tail.

After swapping, current.prev points to the next node in the original list, so we move to current.prev.

3fill in blank

hard

Fix the error in updating the head pointer after reversing the list.

DSA Go

if temp != nil {
    head = temp.[1]
}

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

Anext

Bprev

Cvalue

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head to temp.next which is nil or incorrect.

Using value or tail fields incorrectly.

After the loop, temp.prev points to the new head, so update head to temp.prev.

4fill in blank

hard

Fill both blanks to correctly initialize and traverse the doubly linked list for reversal.

DSA Go

temp := [1]
for current != nil {
    current.[2], current.prev = current.prev, current.next
    current = current.prev
}

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

Anil

Bhead

Cnext

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from nil or wrong node.

Swapping wrong pointers.

We start from head and swap current.next with current.prev to reverse links.

5fill in blank

hard

Fill all three blanks to complete the function that reverses a doubly linked list and returns the new head.

DSA Go

func reverseDoublyLinkedList(head *Node) *Node {
    var current = head
    var temp *Node = [1]
    for current != nil {
        temp = current.[2]
        current.[3], current.prev = current.prev, current.next
        current = current.prev
    }
    if temp != nil {
        head = temp.prev
    }
    return head
}

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

Anil

Bprev

Cnext

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing temp incorrectly.

Swapping or updating pointers in wrong order.

Initialize temp as nil, update temp to current.prev before swapping, then swap current.next and current.prev.

Practice

1. What is the main purpose of reversing a doubly linked list?

easy

A. To sort the list in ascending order

B. To delete all nodes from the list

C. To make the last node the first and the first node the last

D. To add a new node at the end

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes with pointers to both previous and next nodes.

2. Step 2: Define reversing operation

   Reversing means swapping the previous and next pointers for all nodes, so the last node becomes the first.

3. Final Answer:

   To make the last node the first and the first node the last -> Option C

4. Quick Check:

   Reverse = swap previous and next pointers [OK]

Hint: Reversing swaps next and previous pointers [OK]

Common Mistakes:

• Confusing reversing with sorting
• Thinking reversing deletes nodes
• Assuming reversing adds nodes

2. Which of the following Go code snippets correctly swaps the next and prev pointers of a node curr in a doubly linked list?

easy

A. curr.next, curr.prev = curr.prev, curr.next

B. curr.next = curr.prev; curr.prev = curr.next

C. curr.prev = curr.next; curr.next = curr.prev

D. curr.next = curr.next; curr.prev = curr.prev

Solution

1. Step 1: Understand multiple assignment in Go

   Go allows swapping two variables in one line using multiple assignment.

2. Step 2: Identify correct swap syntax

   The correct way to swap is curr.next, curr.prev = curr.prev, curr.next which exchanges the values.

3. Final Answer:

   curr.next, curr.prev = curr.prev, curr.next -> Option A

4. Quick Check:

   Swap pointers with multiple assignment [OK]

Hint: Use Go's multiple assignment to swap pointers [OK]

Common Mistakes:

• Assigning one pointer twice without swapping
• Using separate lines causing overwrite
• Not swapping pointers at all

3. Given the following Go code to reverse a doubly linked list, what will be the printed output after reversal?

type Node struct {
    data int
    prev *Node
    next *Node
}

func reverse(head *Node) *Node {
    var temp *Node
    curr := head
    for curr != nil {
        temp = curr.prev
        curr.prev = curr.next
        curr.next = temp
        curr = curr.prev
    }
    if temp != nil {
        head = temp.prev
    }
    return head
}

// List before reversal: 1 <-> 2 <-> 3 <-> nil
// After reversal, print from new head
func printList(head *Node) {
    curr := head
    for curr != nil {
        fmt.Print(curr.data, " <-> ")
        curr = curr.next
    }
    fmt.Println("nil")
}

medium

A. 1 <-> 2 <-> 3 <-> nil

B. 3 <-> 2 <-> 1 <-> nil

C. nil

D. 1 <-> nil

Solution

1. Step 1: Trace pointer swaps in the loop

   Each node swaps its prev and next pointers, moving curr to original next node via curr.prev after swap.

2. Step 2: Update head to last processed node

   After loop, head is set to temp.prev which points to the new first node (original last node).

3. Final Answer:

   3 <-> 2 <-> 1 <-> nil -> Option B

4. Quick Check:

   Reversed list prints nodes in reverse order [OK]

Hint: Reversed list prints nodes backward [OK]

Common Mistakes:

• Not updating head after reversal
• Confusing prev and next pointers
• Printing original list without reversal

4. Identify the bug in this Go function that attempts to reverse a doubly linked list:

func reverse(head *Node) *Node {
    curr := head
    for curr != nil {
        curr.prev, curr.next = curr.next, curr.prev
        curr = curr.next
    }
    return head
}

medium

A. The loop moves curr to curr.next instead of curr.prev after swapping

B. The pointers are not swapped correctly

C. The function does not return the new head

D. The function deletes nodes accidentally

Solution

1. Step 1: Analyze pointer swapping and traversal

   The pointers are swapped correctly, but after swapping, curr.next points to the previous node, so moving curr to curr.next moves backward.

2. Step 2: Identify correct traversal direction

   After swapping, curr should move to curr.prev (original next) to continue forward traversal.

3. Final Answer:

   The loop moves curr to curr.next instead of curr.prev after swapping -> Option A

4. Quick Check:

   Traverse using curr.prev after swap [OK]

Hint: After swap, move curr to curr.prev, not curr.next [OK]

Common Mistakes:

• Moving curr in wrong direction after swap
• Not updating head to new first node
• Swapping pointers incorrectly

5. You have a doubly linked list with nodes containing integers. After reversing it using the standard pointer swap method, you want to find the sum of all nodes at even positions (2nd, 4th, etc.) in the reversed list. Which approach correctly combines reversal and summation?

hard

A. Sum all nodes without reversal

B. Sum nodes at even positions before reversal, then reverse the list

C. Reverse the list and sum nodes at odd positions instead

D. First reverse the list, then traverse from new head summing nodes at even positions

Solution

1. Step 1: Reverse the doubly linked list

   Use pointer swapping to reverse the list so the order is changed.

2. Step 2: Traverse from new head and sum nodes at even positions

   After reversal, positions change, so summing even positions requires traversal from new head counting nodes.

3. Final Answer:

   First reverse the list, then traverse from new head summing nodes at even positions -> Option D

4. Quick Check:

   Reverse first, then sum even positions [OK]

Hint: Reverse first, then sum nodes by position [OK]

Common Mistakes:

• Summing before reversal causing wrong positions
• Summing odd positions instead of even
• Ignoring reversal effect on node order