Delete by Value in Doubly Linked List in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the list is empty before deletion.

DSA Go

if dll.head == [1] {
    return
}

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

Atrue

B0

Cnil

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if head equals 0 instead of nil.

Using 'true' or 'head' which are incorrect types.

In Go, the head pointer is nil when the list is empty, so we check if dll.head == nil.

2fill in blank

medium

Complete the code to traverse the list to find the node with the target value.

DSA Go

current := dll.head
for current != [1] && current.value != value {
    current = current.next
}

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

Anil

Bdll.tail

Cdll.head

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Using dll.tail instead of nil to check end of list.

Using current instead of nil causing infinite loop.

We traverse until current is nil (end of list) or we find the value.

3fill in blank

hard

Fix the error in updating the previous node's next pointer when deleting the current node.

DSA Go

if current.prev != [1] {
    current.prev.next = current.next
}

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

Adll.tail

Bnil

Ccurrent

Ddll.head

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against dll.head instead of nil.

Using current or dll.tail which are incorrect.

We check if current.prev is not nil before updating its next pointer.

4fill in blank

hard

Fill both blanks to correctly update the next node's previous pointer and handle tail update.

DSA Go

if current.next != [1] {
    current.next.prev = current.prev
} else {
    dll.tail = [2]
}

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

Anil

Bcurrent.prev

Cdll.head

Dcurrent.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using dll.head instead of nil in first blank.

Setting tail to current.next instead of current.prev.

If current.next is not nil, update its prev pointer; else update tail to current.prev.

5fill in blank

hard

Fill all three blanks to complete the DeleteByValue function in Go.

DSA Go

func (dll *DoublyLinkedList) DeleteByValue(value int) {
    if dll.head == [1] {
        return
    }
    current := dll.head
    for current != [2] && current.value != value {
        current = current.next
    }
    if current == [3] {
        return
    }
    if current.prev != nil {
        current.prev.next = current.next
    } else {
        dll.head = current.next
    }
    if current.next != nil {
        current.next.prev = current.prev
    } else {
        dll.tail = current.prev
    }
}

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

Anil

Bdll.tail

Ddll.head

Attempts:

3 left

💡 Hint

Common Mistakes

Using dll.tail or dll.head instead of nil for empty checks.

Not returning when value not found.

Check if head is nil, traverse until current is nil or value found, return if not found (current == nil).

Practice

1. What happens when you delete a node by value in a doubly linked list?

easy

A. The node with the matching value is removed and links are updated

B. All nodes are deleted regardless of value

C. Only the first node is deleted no matter the value

D. The list is reversed instead of deleting

Solution

1. Step 1: Locate the node with the given value

   Traverse the doubly linked list from head to tail to find the node whose data matches the value to delete.

2. Step 2: Update the links to remove the node

   Adjust the previous node's next pointer and the next node's previous pointer to bypass the node to be deleted, effectively removing it from the list.

3. Final Answer:

   The node with the matching value is removed and links are updated -> Option A

4. Quick Check:

   Delete by value means remove matching node and fix links [OK]

Hint: Find node by value, then fix previous and next links [OK]

Common Mistakes:

• Not updating previous node's next pointer
• Not updating next node's previous pointer
• Deleting wrong node when multiple have same value

2. Which of the following Go code snippets correctly deletes a node by value from a doubly linked list?

easy

A. if current.data == value { current.next.next = current.prev.prev }

B. if current.data == value { current.next = current.prev; current.prev = current.next }

C. if current.data == value { current = nil }

D. if current.data == value { current.prev.next = current.next; current.next.prev = current.prev }

Solution

1. Step 1: Check node data against value

   The condition current.data == value correctly identifies the node to delete.

2. Step 2: Update links correctly

   Setting current.prev.next = current.next and current.next.prev = current.prev properly removes the node by linking its neighbors.

3. Final Answer:

   if current.data == value { current.prev.next = current.next; current.next.prev = current.prev } -> Option D

4. Quick Check:

   Correct link updates remove node [OK]

Hint: Update prev.next and next.prev to remove node [OK]

Common Mistakes:

• Swapping prev and next pointers incorrectly
• Setting current to nil without fixing links
• Accessing next or prev without nil checks causing panic

3. Given the doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5, what is the list after deleting the node with value 3?

medium

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

B. 1 <-> 2 <-> 4 <-> 5

C. 2 <-> 3 <-> 4 <-> 5

D. 1 <-> 2 <-> 3 <-> 5

Solution

1. Step 1: Identify node with value 3

   Traverse the list to find the node containing 3, which is between nodes 2 and 4.

2. Step 2: Remove node 3 and update links

   Set node 2's next pointer to node 4 and node 4's previous pointer to node 2, skipping node 3.

3. Final Answer:

   1 <-> 2 <-> 4 <-> 5 -> Option B

4. Quick Check:

   Node 3 removed, neighbors linked [OK]

Hint: Remove node and link neighbors directly [OK]

Common Mistakes:

• Leaving node 3 linked causing broken list
• Deleting wrong node by value
• Not updating both prev and next pointers

4. Consider this Go code snippet to delete a node by value in a doubly linked list:

func deleteByValue(head *Node, value int) *Node {
    current := head
    for current != nil {
        if current.data == value {
            if current.prev != nil {
                current.prev.next = current.next
            } else {
                head = current.next
            }
            if current.next != nil {
                current.next.prev = current.prev
            }
            break
        }
        current = current.next
    }
    return head
}

What is the bug in this code?

medium

A. It does not handle deleting the last node correctly

B. It does not update head when deleting the first node

C. It breaks after deleting only the first matching node

D. It does not check for nil pointers before accessing prev or next

Solution

1. Step 1: Analyze loop and break behavior

   The loop breaks immediately after deleting the first node with matching value, so only one node is deleted even if multiple nodes have the same value.

2. Step 2: Understand implications for multiple matches

   If the list has multiple nodes with the same value, only the first is removed, which may not be the intended behavior if all matches should be deleted.

3. Final Answer:

   It breaks after deleting only the first matching node -> Option C

4. Quick Check:

   Break stops after first delete [OK]

Hint: Check if loop breaks too early when deleting multiple nodes [OK]

Common Mistakes:

• Ignoring multiple nodes with same value
• Not updating head when first node deleted
• Not checking nil before pointer access

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 20 <-> 40. You want to delete all nodes with value 20. Which approach correctly deletes all such nodes?

hard

A. Traverse the list and delete nodes with value 20, continuing without breaking the loop

B. Delete the first node with value 20 and then stop traversal

C. Delete nodes with value 20 only if they are at the head or tail

D. Replace nodes with value 20 with a new node containing 0

Solution

1. Step 1: Traverse entire list without breaking

   To delete all nodes with value 20, you must check every node and not stop after the first deletion.

2. Step 2: Delete matching nodes and update links carefully

   For each node with value 20, update the previous and next pointers to remove it, then continue traversal to find others.

3. Final Answer:

   Traverse the list and delete nodes with value 20, continuing without breaking the loop -> Option A

4. Quick Check:

   Delete all matches by full traversal [OK]

Hint: Do not break loop; delete all matching nodes during traversal [OK]

Common Mistakes:

• Breaking loop after first deletion
• Not updating links properly causing broken list
• Only deleting head or tail nodes