Create a Circular Singly Linked List in DSA Go - Time & Space Complexity

Time Complexity: Create a Circular Singly Linked List

O(n)

Understanding Time Complexity

We want to understand how the time needed to create a circular singly linked list changes as we add more items.

How does the work grow 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
}

func createCircularList(values []int) *Node {
    if len(values) == 0 {
        return nil
    }
    head := &Node{value: values[0]}
    current := head
    for i := 1; i < len(values); i++ {
        current.next = &Node{value: values[i]}
        current = current.next
    }
    current.next = head // make it circular
    return head
}

func main() {
    list := createCircularList([]int{1, 2, 3, 4})
    fmt.Println(list.value) // Output: 1
}

This code creates a circular singly linked list from a slice of integers by linking nodes and connecting the last node back to the first.

Identify Repeating Operations

• Primary operation: Looping through the input slice to create and link nodes.
• How many times: The loop runs once for each element after the first, so (n - 1) times where n is the number of values.

How Execution Grows With Input

As the number of values increases, the number of steps to create nodes and link them grows roughly the same amount.

Input Size (n)	Approx. Operations
10	About 9 loops to link nodes
100	About 99 loops to link nodes
1000	About 999 loops to link nodes

Pattern observation: The work grows directly with the number of items; doubling items roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to create the list grows in a straight line with the number of items you add.

Common Mistake

[X] Wrong: "Creating a circular list takes more time than a normal linked list because of the extra link back to the head."

[OK] Correct: Adding the link from the last node back to the head is just one extra step, which does not grow with input size, so it does not change the overall time complexity.

Interview Connect

Understanding how creating linked lists scales helps you explain your code clearly and shows you know how data structures behave as they grow.

Self-Check

"What if we changed the code to insert each new node at the head instead of the tail? How would the time complexity change?"

Practice

1. What is a key characteristic of a circular singly linked list compared to a regular singly linked list?

easy

A. The last node points back to the first node, forming a circle.

B. Each node has two pointers, one to the next and one to the previous node.

C. It uses an array internally to store nodes.

D. Nodes are stored in sorted order automatically.

Solution

1. Step 1: Understand the structure of a circular singly linked list

   In a circular singly linked list, the last node's next pointer points back to the first node, creating a loop.

2. Step 2: Compare with other linked lists

   Unlike doubly linked lists, nodes have only one pointer. Unlike arrays, nodes are linked by pointers, not indices.

3. Final Answer:

   The last node points back to the first node, forming a circle. -> Option A

4. Quick Check:

   Circular list last node points to head = D [OK]

Hint: Last node points to head in circular singly linked list [OK]

Common Mistakes:

• Confusing circular singly linked list with doubly linked list
• Thinking nodes have two pointers
• Assuming nodes are stored in arrays

2. Which of the following Go code snippets correctly initializes a circular singly linked list with one node containing value 10?

easy

A. node := &Node{data: 10}; node.next = node

B. node := Node{data: 10}; node.next = nil

C. node := &Node{data: 10, next: nil}

D. node := &Node{data: 10}; node.next = nil

Solution

1. Step 1: Create a node pointer and assign data

   Using &Node creates a pointer to a Node struct with data 10.

2. Step 2: Make the node point to itself

   Setting node.next = node creates a circular link for a single node list.

3. Final Answer:

   node := &Node{data: 10}; node.next = node -> Option A

4. Quick Check:

   Single node points to itself = C [OK]

Hint: Single node's next points to itself for circular list [OK]

Common Mistakes:

• Setting next to nil breaks circular property
• Not using pointer for node
• Confusing struct initialization syntax

3. Given the following Go code, what will be printed?

package main
import "fmt"

type Node struct {
    data int
    next *Node
}

func main() {
    head := &Node{data: 1}
    second := &Node{data: 2}
    third := &Node{data: 3}

    head.next = second
    second.next = third
    third.next = head

    current := head
    for i := 0; i < 5; i++ {
        fmt.Print(current.data, " ")
        current = current.next
    }
}

medium

A. 1 2 3 1 1

B. 1 2 3 3 3

C. 1 2 3 1 2

D. 1 2 3 2 3

Solution

1. Step 1: Understand the circular list structure

   Nodes 1 -> 2 -> 3 -> back to 1, forming a circle.

2. Step 2: Trace the loop printing 5 nodes

   Prints: 1, then 2, then 3, then back to 1, then 2.

3. Final Answer:

   1 2 3 1 2 -> Option C

4. Quick Check:

   Loop prints nodes in circle = A [OK]

Hint: Follow next pointers in circle for 5 steps [OK]

Common Mistakes:

• Assuming list ends after 3 nodes
• Not looping back to head
• Miscounting loop iterations

4. Identify the error in this Go code that tries to create a circular singly linked list with two nodes:

package main

type Node struct {
    data int
    next *Node
}

func main() {
    first := &Node{data: 1}
    second := &Node{data: 2}

    first.next = second
    second.next = nil

    // Make it circular
    second.next = first
}

medium

A. No error, code correctly creates circular list

B. second.next is set to nil then overwritten, unnecessary step

C. second.next is set to nil before setting to first, redundant but no error

D. Missing initialization of first.next before assigning second

Solution

1. Step 1: Analyze the assignments to second.next

   Initially second.next is set to nil, then immediately overwritten to first.

2. Step 2: Identify redundancy

   Setting second.next to nil is unnecessary since it is overwritten right after.

3. Final Answer:

   second.next is set to nil then overwritten, unnecessary step -> Option B

4. Quick Check:

   Redundant assignment detected = B [OK]

Hint: Avoid redundant pointer assignments in circular list [OK]

Common Mistakes:

• Thinking nil assignment causes error
• Ignoring redundant code
• Confusing order of pointer assignments

5. You want to insert a new node with value 4 at the end of an existing circular singly linked list with nodes 1 -> 2 -> 3 -> back to 1. Which of the following Go code snippets correctly performs this insertion?

hard

A. newNode := &Node{data: 4}
current := head
for current.next != nil {
current = current.next
}
current.next = newNode
newNode.next = head

B. newNode := &Node{data: 4, next: head}
head.next = newNode

C. newNode := &Node{data: 4}
newNode.next = head
head.next = newNode

D. newNode := &Node{data: 4}
current := head
for current.next != head {
current = current.next
}
current.next = newNode
newNode.next = head

Solution

1. Step 1: Traverse to last node before head

   Loop until current.next == head to find last node in circular list.

2. Step 2: Insert new node after last node and link back to head

   Set current.next = newNode and newNode.next = head to maintain circularity.

3. Final Answer:

   Traverse to last node, link new node, then link new node back to head -> Option D

4. Quick Check:

   Proper insertion at end keeps circle intact = A [OK]

Hint: Find last node where next is head, insert new node there [OK]

Common Mistakes:

• Using nil instead of head to find last node
• Not linking new node back to head
• Incorrectly changing head pointer