Insert at End Tail Insert in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

DSA Go

newNode := &Node{data: [1], next: nil}

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

Anode

Bval

Cdata

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undefined variable like 'val' or 'data' instead of 'value'.

The new node's data field should be assigned the variable value which holds the value to insert.

2fill in blank

medium

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

DSA Go

if head == [1] {
    head = newNode
    return head
}

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

Anull

Bnil

C0

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'null' which is not valid in Go.

Using 0 or 'empty' which are not pointer values.

In Go, the zero value for pointers is nil. Checking if head == nil means the list is empty.

3fill in blank

hard

Fix the error in the loop that finds the last node.

DSA Go

current := head
for current.next [1] nil {
    current = current.next
}

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

A!=

B==

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' which causes the loop to skip or never run.

The loop should continue while current.next is not nil, so use !=.

4fill in blank

hard

Fill both blanks to correctly link the new node at the end.

DSA Go

current.next = [1]
return [2]

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AnewNode

Bhead

Cnil

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the new node instead of the head.

Assigning nil or current instead of newNode.

Assign newNode to current.next to link it at the end, then return head as the list start.

5fill in blank

hard

Fill all three blanks to complete the tail insert function.

DSA Go

func InsertAtEnd(head *Node, value int) *Node {
    newNode := &Node{data: [1], next: nil}
    if head == [2] {
        head = newNode
        return head
    }
    current := head
    for current.next [3] nil {
        current = current.next
    }
    current.next = newNode
    return head
}

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

Avalue

Bnil

C!=

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names or comparison operators.

Not checking for empty list before looping.

The function creates a new node with value, checks if head is nil (empty list), then loops while current.next != nil to find the last node.

Practice

1. What does the "Insert at End (Tail Insert)" operation do in a singly linked list?

easy

A. Deletes the last node of the list

B. Adds a new node at the beginning of the list

C. Adds a new node at the end of the list

D. Deletes the first node of the list

Solution

1. Step 1: Understand the operation name

   "Insert at End" means adding a new element at the tail or last position of the list.

2. Step 2: Compare with other options

   Adding at the beginning is different, and deleting nodes is not insertion.

3. Final Answer:

   Adds a new node at the end of the list -> Option C

4. Quick Check:

   Insert at End = Add node at tail [OK]

Hint: Tail insert means add new node after last node [OK]

Common Mistakes:

• Confusing insert at end with insert at beginning
• Thinking it deletes nodes
• Assuming it inserts in the middle

2. Which of the following is the correct syntax to create a new node and insert it at the end in Go for a singly linked list?

easy

A. newNode := Node{value, nil}

B. newNode := &Node{data: value, next: nil}

C. newNode := &Node{value}

D. newNode := Node{data: value}

Solution

1. Step 1: Understand Go struct initialization

   In Go, to create a pointer to a struct with named fields, use &Node{field: value, ...}.

2. Step 2: Check each option

   newNode := &Node{data: value, next: nil} correctly uses &Node with named fields and sets next to nil for tail node.

3. Final Answer:

   newNode := &Node{data: value, next: nil} -> Option B

4. Quick Check:

   Pointer to struct with named fields = newNode := &Node{data: value, next: nil} [OK]

Hint: Use &Node{field: val, next: nil} to create tail node [OK]

Common Mistakes:

• Missing & to create pointer
• Not setting next to nil for tail
• Using unnamed fields incorrectly

3. What will be the printed linked list after inserting 4 at the end of the list 1 -> 2 -> 3 -> nil using tail insert?

medium

A. 1 -> 2 -> 3 -> 4 -> nil

B. 4 -> 1 -> 2 -> 3 -> nil

C. 1 -> 2 -> 4 -> 3 -> nil

D. 1 -> 2 -> 3 -> nil

Solution

1. Step 1: Understand tail insert effect

   Inserting 4 at the end adds it after 3, so list becomes 1 -> 2 -> 3 -> 4 -> nil.

2. Step 2: Compare options

   Only 1 -> 2 -> 3 -> 4 -> nil shows 4 at the end; others show wrong order or no change.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> nil -> Option A

4. Quick Check:

   Tail insert adds node at end [OK]

Hint: Tail insert appends new node after last node [OK]

Common Mistakes:

• Inserting at beginning instead of end
• Inserting in the middle
• Not updating last node's next pointer

4. Given the following Go code snippet for tail insertion, what is the error?

func InsertAtEnd(head *Node, value int) *Node {
    newNode := &Node{data: value, next: nil}
    if head == nil {
        head = newNode
    } else {
        temp := head
        for temp.next != nil {
            temp = temp.next
        }
        temp = newNode
    }
    return head
}

medium

A. temp = newNode does not link newNode to the list

B. head is not updated when list is empty

C. newNode is not created properly

D. Loop condition should be temp != nil

Solution

1. Step 1: Analyze the loop and insertion

   The loop finds the last node (temp.next == nil). Then temp = newNode only changes temp pointer, not the last node's next.

2. Step 2: Identify correct linking

   To insert, we must set temp.next = newNode, not temp = newNode.

3. Final Answer:

   temp = newNode does not link newNode to the list -> Option A

4. Quick Check:

   Link last node's next to newNode [OK]

Hint: Set lastNode.next = newNode, not lastNode = newNode [OK]

Common Mistakes:

• Assigning newNode to temp instead of temp.next
• Not handling empty list correctly
• Wrong loop condition

5. You have a singly linked list and want to insert multiple values [5, 10, 15] at the end one by one using tail insert. Which approach is most efficient?

hard

A. Traverse from head to end for each insertion

B. Create a new list for each value and merge later

C. Insert each new node at the beginning

D. Keep a tail pointer to last node and update it on each insert

Solution

1. Step 1: Understand repeated tail insert cost

   Traversing from head to end for each insert is costly (O(n) per insert), total O(n^2).

2. Step 2: Use tail pointer optimization

   Keeping a tail pointer allows direct access to last node, so insertion is O(1) each, total O(n).

3. Final Answer:

   Keep a tail pointer to last node and update it on each insert -> Option D

4. Quick Check:

   Tail pointer reduces insertion time [OK]

Hint: Use tail pointer to avoid full traversal each insert [OK]

Common Mistakes:

• Traversing full list every time
• Inserting at beginning instead of end
• Creating multiple lists unnecessarily