Insert at Middle Specific Position in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a new node at the given position in the linked list.

DSA Go

func InsertAtPosition(head *Node, data int, pos int) *Node {
    newNode := &Node{data: data}
    if pos == 1 {
        newNode.next = head
        return newNode
    }
    current := head
    for i := 1; i < [1]-1 && current != nil; i++ {
        current = current.next
    }
    if current == nil {
        return head
    }
    newNode.next = current.next
    current.next = newNode
    return head
}

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

Apos

Bdata

CnewNode

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using data or newNode instead of pos in the loop condition.

Looping until pos instead of pos-1 causing off-by-one error.

The loop runs until the node before the position where the new node will be inserted, so it should run until pos-1.

2fill in blank

medium

Complete the code to check if the position is valid before insertion.

DSA Go

func InsertAtPosition(head *Node, data int, pos int) *Node {
    if pos <= 0 {
        return head
    }
    length := 0
    current := head
    for current != nil {
        length++
        current = current.next
    }
    if pos > [1]+1 {
        return head
    }
    // insertion code continues...
    return head
}

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

Apos

Bdata

Clength

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing position with pos or data instead of length.

Allowing position greater than length+1 causing invalid insertion.

The position should not be greater than length+1 because you can insert at the end (length+1).

3fill in blank

hard

Fix the error in the loop that traverses to the node before insertion.

DSA Go

func InsertAtPosition(head *Node, data int, pos int) *Node {
    newNode := &Node{data: data}
    if pos == 1 {
        newNode.next = head
        return newNode
    }
    current := head
    for i := 1; i < [1]-1 && current != nil; i++ {
        current = current.next
    }
    if current == nil {
        return head
    }
    newNode.next = current.next
    current.next = newNode
    return head
}

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

Adata

Bpos

Chead

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= instead of < causing the loop to move one node too far.

Using data or newNode instead of pos in the loop condition.

The loop should run while i < pos-1, not i <= pos-1, to stop at the node before insertion.

4fill in blank

hard

Fill both blanks to create a map of word lengths for words longer than 3 characters.

DSA Go

lengths := map[string]int{}
for _, word := range words {
    if len(word) [1] 3 {
        lengths[word] = [2]
    }
}

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

A>

Blen(word)

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > causing wrong filtering.

Assigning word instead of len(word) as value.

We want words longer than 3, so len(word) > 3, and the value is the length of the word.

5fill in blank

hard

Fill all three blanks to create a filtered map with values greater than 0.

DSA Go

result := map[string]int{}
for k, v := range data {
    if v [3] 0 {
        result[[1]] = [2]
    }
}

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

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.upper() which is not valid Go syntax.

Using < instead of > causing wrong filtering.

We check if v > 0, then assign result[k] = v. The key is k, value is v, and condition is > 0.

Practice

1. What does inserting a node at a specific middle position in a linked list mean?

easy

A. Adding a new node only at the start of the list

B. Adding a new node only at the end of the list

C. Adding a new node exactly between two existing nodes at a given position

D. Removing a node from the middle of the list

Solution

1. Step 1: Understand linked list positions

   Positions in a linked list start from 1 at the head and increase by 1 for each node.

2. Step 2: Define middle insertion

   Inserting at a middle position means placing a new node between two existing nodes at that position.

3. Final Answer:

   Adding a new node exactly between two existing nodes at a given position -> Option C

4. Quick Check:

   Middle insertion = Insert between nodes [OK]

Hint: Middle insertion means placing node between two nodes [OK]

Common Mistakes:

• Confusing insertion with deletion
• Thinking insertion only happens at start or end
• Ignoring position indexing starting at 1

2. Which Go code snippet correctly inserts a new node at position 3 in a singly linked list?

easy

A. temp := head
for i := 1; i < 2; i++ {
temp = temp.next
}
newNode.next = temp.next
temp.next = newNode

B. temp := head
for i := 0; i < 3; i++ {
temp = temp.next
}
newNode.next = temp.next
temp.next = newNode

C. temp := head
for i := 1; i <= 3; i++ {
temp = temp.next
}
newNode.next = temp.next
temp.next = newNode

D. temp := head
for i := 1; i < 3; i++ {
temp = temp.next
}
newNode.next = temp.next
temp.next = newNode

Solution

1. Step 1: Loop to node before position 3

   To insert at position 3, loop until position 2 (i < 2) to get node before insertion point.

2. Step 2: Adjust pointers correctly

   Set newNode.next to temp.next, then temp.next to newNode to insert newNode at position 3.

3. Final Answer:

   Code snippet in option A correctly inserts at position 3 -> Option A

4. Quick Check:

   Loop to position-1, then insert [OK]

Hint: Loop to position-1 node before inserting [OK]

Common Mistakes:

• Looping too many times and skipping nodes
• Setting pointers in wrong order causing lost nodes
• Using wrong loop conditions (<= instead of <)

3. What will be the linked list after inserting value 99 at position 2 in the list 10 -> 20 -> 30 -> null using this Go code?

type Node struct {
  val  int
  next *Node
}

head := &Node{10, &Node{20, &Node{30, nil}}}
newNode := &Node{99, nil}
temp := head
for i := 1; i < 1; i++ {
  temp = temp.next
}
newNode.next = temp.next
temp.next = newNode

medium

A. 10 -> 99 -> 20 -> 30 -> null

B. 99 -> 10 -> 20 -> 30 -> null

C. 10 -> 20 -> 99 -> 30 -> null

D. 10 -> 20 -> 30 -> 99 -> null

Solution

1. Step 1: Loop runs zero times to position 1 node

   The loop condition i < 1 means no iteration; temp stays at head (value 10).

2. Step 2: Insert newNode after temp

   newNode.next points to temp.next (node with 20), then temp.next points to newNode, inserting 99 after 10.

3. Final Answer:

   10 -> 99 -> 20 -> 30 -> null -> Option A

4. Quick Check:

   Insert after position 1 = position 2 [OK]

Hint: Loop to node before position, then insert new node [OK]

Common Mistakes:

• Misunderstanding loop range causing wrong insertion
• Confusing head node position as 0 instead of 1
• Incorrect pointer assignments losing nodes

4. Identify the error in this Go code that tries to insert a node at position 4 in a linked list of length 3:

temp := head
for i := 1; i <= 4; i++ {
  temp = temp.next
}
newNode.next = temp.next
temp.next = newNode

medium

A. New node is not created before insertion

B. Pointers are assigned in wrong order causing lost nodes

C. Position 4 is invalid for a list of length 3

D. Loop runs too many times causing temp to be nil and panic

Solution

1. Step 1: Analyze loop iterations

   The loop runs from i=1 to i=4 inclusive, moving temp 4 times forward.

2. Step 2: Check list length and temp value

   List length is 3, so after 3 moves temp becomes nil; moving one more causes panic on temp.next.

3. Final Answer:

   Loop runs too many times causing temp to be nil and panic -> Option D

4. Quick Check:

   Loop beyond list length = nil pointer error [OK]

Hint: Loop only to position-1 within list length [OK]

Common Mistakes:

• Looping beyond list length causing runtime errors
• Not checking for nil before accessing next
• Confusing <= with < in loop condition

5. Given a singly linked list and a position k, write a Go function to insert a new node with value v at position k. If k is greater than the list length + 1, insert at the end. What will be the list after inserting 50 at position 5 in 10 -> 20 -> 30 -> 40 -> null?

hard

A. 10 -> 20 -> 50 -> 30 -> 40 -> null

B. 10 -> 20 -> 30 -> 40 -> 50 -> null

C. 50 -> 10 -> 20 -> 30 -> 40 -> null

D. 10 -> 20 -> 30 -> 50 -> 40 -> null

Solution

1. Step 1: Check position against list length

   List length is 4; position 5 is length+1, so insert at end.

2. Step 2: Insert new node at end

   Traverse to last node (40), set its next to new node with value 50, new node next is nil.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> 50 -> null -> Option B

4. Quick Check:

   Position > length = insert at end [OK]

Hint: If position too big, insert at list end [OK]

Common Mistakes:

• Trying to insert beyond list length without appending
• Misplacing new node in middle instead of end
• Not handling nil next pointer at list end