Insert at Middle Specific Position in DSA Go - Build from Scratch

Insert at Middle Specific Position in a Linked List

📖 Scenario: You are managing a waiting list for a popular event. People can join the list, and sometimes a new person wants to join at a specific position in the middle of the list.

🎯 Goal: Build a simple linked list and insert a new person at a specific middle position.

📋 What You'll Learn

Create a linked list with exactly these names in order: "Alice", "Bob", "Charlie", "David"

Create a variable called position and set it to 3

Insert a new node with the name "Eve" at the position in the linked list

Print the linked list names in order separated by -> and ending with null

💡 Why This Matters

🌍 Real World

Linked lists are used in many applications like managing waiting lists, undo functionality in apps, and dynamic memory management.

💼 Career

Understanding linked lists and pointer manipulation is essential for software developers working with low-level data structures, embedded systems, or performance-critical applications.

1

Create the initial linked list

Create a linked list with nodes containing the names "Alice", "Bob", "Charlie", and "David" in this exact order. Use a struct called Node with fields name string and next *Node. Create a variable called head that points to the first node.

DSA Go

// Define Node struct and create linked list with Alice, Bob, Charlie, David
// Your code here

Hint

Define a struct Node with name string and next *Node. Then link nodes in order.

2

Set the insertion position

Create an integer variable called position and set it to 3. This will be the position where you insert the new node.

DSA Go

package main

import "fmt"

type Node struct {
    name string
    next *Node
}

func main() {
    head := &Node{name: "Alice"}
    head.next = &Node{name: "Bob"}
    head.next.next = &Node{name: "Charlie"}
    head.next.next.next = &Node{name: "David"}

    // Create position variable and set to 3
    // Your code here
}

Hint

Just create a variable position and assign it the value 3.

3

Insert a new node at the specific position

Insert a new node with the name "Eve" at the position in the linked list. Use a variable called current to traverse the list until the node before the insertion point. Then adjust pointers to insert the new node.

DSA Go

package main

import "fmt"

type Node struct {
    name string
    next *Node
}

func main() {
    head := &Node{name: "Alice"}
    head.next = &Node{name: "Bob"}
    head.next.next = &Node{name: "Charlie"}
    head.next.next.next = &Node{name: "David"}

    position := 3

    // Insert new node "Eve" at position
    // Your code here
}

Hint

Use a loop to move current to the node before the insertion point. Then create newNode and adjust pointers.

4

Print the linked list

Print the names in the linked list starting from head. Print each name followed by -> and end with null. Use a variable called current to traverse the list.

DSA Go

package main

import "fmt"

type Node struct {
    name string
    next *Node
}

func main() {
    head := &Node{name: "Alice"}
    head.next = &Node{name: "Bob"}
    head.next.next = &Node{name: "Charlie"}
    head.next.next.next = &Node{name: "David"}

    position := 3

    current := head
    for i := 1; i < position-1; i++ {
        current = current.next
    }

    newNode := &Node{name: "Eve"}
    newNode.next = current.next
    current.next = newNode

    // Print linked list
    // Your code here
}

Hint

Use a loop to print each node's name followed by -> . After the loop, print null.

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