Creating a Singly Linked List from Scratch in DSA Go - Complexity Walkthrough

Time Complexity: Creating a Singly Linked List from Scratch

O(n²)

Understanding Time Complexity

When we create a singly linked list by adding nodes one by one, it is important to understand how the time needed grows as we add more nodes.

We want to know how the work changes when the list gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

package main

import "fmt"

type Node struct {
  value int
  next  *Node
}

type LinkedList struct {
  head *Node
}

func (l *LinkedList) Append(val int) {
  newNode := &Node{value: val}
  if l.head == nil {
    l.head = newNode
    return
  }
  current := l.head
  for current.next != nil {
    current = current.next
  }
  current.next = newNode
}

func main() {
  list := LinkedList{}
  for i := 1; i <= 5; i++ {
    list.Append(i)
  }
  current := list.head
  for current != nil {
    fmt.Print(current.value, " -> ")
    current = current.next
  }
  fmt.Println("nil")
}

This code creates a singly linked list by adding new nodes at the end one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The loop inside the Append method that moves through the list to find the last node.
• How many times: For each new node added, this loop runs zero times for the first node (since the list is empty), once for the second, twice for the third, and so on, increasing with the list size.

How Execution Grows With Input

Explain the growth pattern intuitively.

Input Size (n)	Approx. Operations
10	About 45 steps (0+1+...+9)
100	About 4950 steps (sum of 0 to 99)
1000	About 499,500 steps (sum of 0 to 999)

Pattern observation: The total steps grow roughly like the square of the number of nodes added, because each new node requires walking through the existing list.

Final Time Complexity

Time Complexity: O(n²)

This means that as the list grows, the time to add all nodes grows very quickly, roughly like the square of the number of nodes.

Common Mistake

[X] Wrong: "Appending a node always takes the same amount of time, so adding n nodes is O(n)."

[OK] Correct: Each append must find the end by walking through the list, which takes longer as the list grows, so the total time adds up faster than just n.

Interview Connect

Understanding this helps you explain why adding nodes to a linked list can be slow if you don't keep track of the end, and shows your grasp of how data structure design affects performance.

Self-Check

"What if we kept a pointer to the last node (tail) and updated it on each append? How would the time complexity change?"

Practice

1. What is the main purpose of a singly linked list in data structures?

easy

A. To store elements in a hash table for fast lookup

B. To store elements in a tree-like structure with multiple branches

C. To store elements in a fixed-size array for quick access

D. To store elements in a linear order where each element points to the next

Solution

1. Step 1: Understand singly linked list structure

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike trees or arrays, singly linked lists have a linear sequence with pointers only to the next element.

3. Final Answer:

   To store elements in a linear order where each element points to the next -> Option D

4. Quick Check:

   Singly linked list = linear chain with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing singly linked list with trees or arrays
• Thinking nodes have pointers to previous nodes
• Assuming fixed size storage like arrays

2. Which of the following is the correct way to define a node struct for a singly linked list in Go?

easy

A. type Node struct { value int; next *Node }

B. type Node struct { value int; prev *Node }

C. type Node struct { value int; next Node }

D. type Node struct { value int; next interface{} }

Solution

1. Step 1: Identify correct pointer usage

   In Go, the next node must be a pointer to Node, so *Node is correct.

2. Step 2: Check struct fields

   Field 'value' stores data, 'next' points to the next node. 'prev' is for doubly linked lists, and 'interface{}' is too generic.

3. Final Answer:

   type Node struct { value int; next *Node } -> Option A

4. Quick Check:

   Node next pointer = *Node [OK]

Hint: Use *Node for next pointer in Go struct [OK]

Common Mistakes:

• Using value type instead of pointer for next
• Confusing singly linked list with doubly linked list (prev pointer)
• Using interface{} instead of *Node for next

3. What will be the output of the following Go code that creates a singly linked list with nodes 1 -> 2 -> 3?

package main
import "fmt"
type Node struct {
  value int
  next  *Node
}
func main() {
  head := &Node{value: 1}
  second := &Node{value: 2}
  third := &Node{value: 3}
  head.next = second
  second.next = third
  current := head
  for current != nil {
    fmt.Print(current.value)
    if current.next != nil {
      fmt.Print(" -> ")
    }
    current = current.next
  }
  fmt.Println(" -> null")
}

medium

A. 1 -> 2 -> null

B. 3 -> 2 -> 1 -> null

C. 1 -> 2 -> 3 -> null

D. Compilation error

Solution

1. Step 1: Trace node creation and linking

   Nodes with values 1, 2, 3 are created and linked in order: head.next = second, second.next = third.

2. Step 2: Trace printing loop

   Loop prints current.value, then ' -> ' if next exists, moving from head to third, then prints ' -> null'.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Print nodes in order 1,2,3 with arrows [OK]

Hint: Follow next pointers from head to end [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing last node in output
• Confusing pointer assignments causing nil errors

4. Identify the error in this Go code snippet that tries to append a new node with value 4 to the end of a singly linked list:

func appendNode(head *Node, val int) {
  newNode := &Node{value: val}
  current := head
  for current.next != nil {
    current = current.next
  }
  current = newNode
}

medium

A. The loop condition should be current != nil instead of current.next != nil

B. The new node is not linked to the list; current pointer is reassigned incorrectly

C. The new node should be created with next pointing to head

D. The function should return the new head pointer

Solution

1. Step 1: Analyze the loop and pointer update

   The loop correctly finds the last node where current.next is nil.

2. Step 2: Check how new node is linked

   Assigning current = newNode only changes the local variable, not the last node's next pointer. It should be current.next = newNode.

3. Final Answer:

   The new node is not linked to the list; current pointer is reassigned incorrectly -> Option B

4. Quick Check:

   Link new node by current.next = newNode [OK]

Hint: Assign new node to current.next, not current [OK]

Common Mistakes:

• Reassigning current instead of current.next
• Wrong loop condition causing nil pointer dereference
• Not returning updated head when needed

5. You want to create a function in Go that reverses a singly linked list. Which approach correctly reverses the list by changing the next pointers?

func reverseList(head *Node) *Node {
  var prev *Node = nil
  current := head
  for current != nil {
    nextTemp := current.next
    current.next = prev
    prev = current
    current = nextTemp
  }
  return prev
}

What will be the output if the original list is 1 -> 2 -> 3 -> null and you print the reversed list?

hard

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. null

D. Runtime error due to nil pointer

Solution

1. Step 1: Understand reversal logic

   The function iterates through the list, reversing each node's next pointer to point to the previous node.

2. Step 2: Trace reversal on list 1 -> 2 -> 3

   After reversal, the list becomes 3 -> 2 -> 1 -> null, and prev points to new head.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list head = prev = 3 [OK]

Hint: Reverse pointers by tracking prev, current, next [OK]

Common Mistakes:

• Not updating current.next correctly
• Losing track of next node causing infinite loop
• Returning old head instead of new head