Why Insert at Middle Specific Position in DSA Go?

The Big Idea

What if you could add anything anywhere in your list instantly without mistakes?

The Scenario

Imagine you have a list of names on paper, and you want to add a new name exactly in the middle. You have to rewrite all the names after that spot to make space. This is slow and tiring.

The Problem

Manually moving each name down one by one is slow and easy to mess up. You might lose names or write them in the wrong order. It takes a lot of time and effort.

The Solution

Using a data structure that supports inserting at any position lets you add the new name right where you want without rewriting everything. It handles the shifting automatically and safely.

Before vs After

✗ Before

names := []string{"Alice", "Bob", "Charlie"}
// Insert "David" at position 1 manually
newNames := make([]string, len(names)+1)
copy(newNames[:1], names[:1])
newNames[1] = "David"
copy(newNames[2:], names[1:])

✓ After

names := []string{"Alice", "Bob", "Charlie"}
pos := 1
names = append(names[:pos], append([]string{"David"}, names[pos:]...)...)

What It Enables

You can quickly and safely add items anywhere in your list, making your programs flexible and efficient.

Real Life Example

Think about a music playlist where you want to add a new song exactly after the second song without disturbing the rest manually.

Key Takeaways

Manual insertion is slow and error-prone.

Data structures help insert at any position easily.

This makes managing lists faster and safer.

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