Clone Linked List with Random Pointer in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a 'random pointer' in a linked list?

A random pointer is an extra pointer in each node of a linked list that can point to any node in the list or be null, not necessarily the next node.

beginner

Why can't we just copy nodes one by one to clone a linked list with random pointers?

Because random pointers can point anywhere, copying nodes one by one without tracking original-to-copy mapping will lose the correct random pointer connections.

intermediate

What is the main idea behind the three-step approach to clone a linked list with random pointers?

1. Insert cloned nodes next to original nodes.
2. Assign random pointers for cloned nodes using original nodes.
3. Separate the cloned list from the original list.

beginner

In Go, how do you define a node struct for a linked list with a random pointer?

type Node struct { Val int Next *Node Random *Node }

intermediate

What is the time complexity of cloning a linked list with random pointers using the three-step approach?

The time complexity is O(n), where n is the number of nodes, because we traverse the list a constant number of times.

What does the random pointer in a linked list node point to?

A Any node in the list or null

B Only the next node

C Only the previous node

D The head node always

The random pointer can point to any node in the list or be null.

Which step is NOT part of the three-step cloning approach?

A Reverse the original linked list

B Assign random pointers for cloned nodes

C Insert cloned nodes next to original nodes

D Separate the cloned list from the original

Reversing the original linked list is not part of the cloning process.

How do you assign the random pointer of a cloned node in the three-step approach?

A Set clonedNode.Random = originalNode.Next.Random

B Set clonedNode.Random = originalNode.Random

C Set clonedNode.Random = originalNode.Random.Next

D Set clonedNode.Random = nil

Because cloned nodes are inserted next to original nodes, originalNode.Random.Next points to the cloned node corresponding to originalNode.Random.

What is the space complexity of the three-step cloning method?

A O(n) extra space

B O(log n) extra space

C O(n^2) extra space

D O(1) extra space

The method uses no extra data structures, so extra space is O(1).

In Go, what type should the random pointer field have in the node struct?

A int

B *Node

C string

D bool

The random pointer points to another node, so its type is *Node.

Explain the three-step approach to clone a linked list with random pointers.

Think about how to keep track of original and cloned nodes together.

Describe why a simple node-by-node copy fails for linked lists with random pointers.

Consider what happens if you copy nodes without remembering their original connections.

Practice

1. What is the main challenge when cloning a linked list with a random pointer?

easy

A. Reversing the linked list during cloning

B. Only copying the next pointers

C. Ignoring the random pointers

D. Copying both the next and random pointers correctly

Solution

1. Step 1: Understand the structure of the linked list

   The linked list has two pointers per node: next and random. Both need to be copied to clone the list correctly.

2. Step 2: Identify the cloning challenge

   Copying next pointers is straightforward, but random pointers can point anywhere, so they must be handled carefully to maintain correct references.

3. Final Answer:

   Copying both the next and random pointers correctly -> Option D

4. Quick Check:

   Cloning requires both pointers copied correctly [OK]

Hint: Remember: random pointers can point anywhere, copy carefully [OK]

Common Mistakes:

• Copying only next pointers and ignoring random
• Assuming random pointers follow next pointers
• Not handling null random pointers

2. Which Go syntax correctly declares a struct node with next and random pointers for cloning a linked list?

easy

A. type Node struct { Val int; Next Node; Random Node }

B. type Node struct { Val int; Next *Node; Random *Node }

C. type Node struct { Val int; Next *int; Random *int }

D. type Node struct { Val int; Next *Node; Random int }

Solution

1. Step 1: Understand pointer usage in Go structs

   Next and Random should be pointers to Node to link nodes properly, so use *Node type.

2. Step 2: Check each option for correct pointer types

   type Node struct { Val int; Next *Node; Random *Node } uses *Node for both Next and Random, which is correct. Others use wrong types or non-pointers.

3. Final Answer:

   type Node struct { Val int; Next *Node; Random *Node } -> Option B

4. Quick Check:

   Use pointers for linked nodes in Go structs [OK]

Hint: Use *Node type for next and random pointers in struct [OK]

Common Mistakes:

• Using value type instead of pointer for next/random
• Using wrong pointer types like *int
• Mixing pointer and non-pointer types incorrectly

3. Given the following Go code snippet for cloning a linked list with random pointers, what will be the output of printing the cloned list values in order?

type Node struct {
    Val    int
    Next   *Node
    Random *Node
}

func cloneList(head *Node) *Node {
    if head == nil {
        return nil
    }
    // Step 1: Insert cloned nodes
    curr := head
    for curr != nil {
        newNode := &Node{Val: curr.Val}
        newNode.Next = curr.Next
        curr.Next = newNode
        curr = newNode.Next
    }
    // Step 2: Copy random pointers
    curr = head
    for curr != nil {
        if curr.Random != nil {
            curr.Next.Random = curr.Random.Next
        }
        curr = curr.Next.Next
    }
    // Step 3: Separate cloned list
    curr = head
    cloneHead := head.Next
    cloneCurr := cloneHead
    for curr != nil {
        curr.Next = curr.Next.Next
        if cloneCurr.Next != nil {
            cloneCurr.Next = cloneCurr.Next.Next
        }
        curr = curr.Next
        cloneCurr = cloneCurr.Next
    }
    return cloneHead
}

// Original list: 1 -> 2 -> 3 -> nil
// Random pointers: 1.random = 3, 2.random = 1, 3.random = 2

func main() {
    n3 := &Node{Val: 3}
    n2 := &Node{Val: 2, Next: n3}
    n1 := &Node{Val: 1, Next: n2}
    n1.Random = n3
    n2.Random = n1
    n3.Random = n2

    cloned := cloneList(n1)
    curr := cloned
    for curr != nil {
        fmt.Print(curr.Val)
        if curr.Next != nil {
            fmt.Print(" -> ")
        }
        curr = curr.Next
    }
    fmt.Println(" -> nil")
}

medium

A. 1 -> 2 -> 3 -> nil

B. 3 -> 2 -> 1 -> nil

C. 1 -> 3 -> 2 -> nil

D. nil

Solution

1. Step 1: Understand cloning steps

   The code clones nodes by inserting new nodes after originals, copies random pointers, then separates cloned list.

2. Step 2: Trace the cloned list values

   Original list is 1->2->3. Cloned list copies values in same order: 1->2->3.

3. Final Answer:

   1 -> 2 -> 3 -> nil -> Option A

4. Quick Check:

   Cloned list values match original order [OK]

Hint: Cloned list keeps original order of nodes [OK]

Common Mistakes:

• Assuming cloned list reverses order
• Confusing random pointer order with next pointer order
• Not separating cloned list properly

4. Identify the bug in this Go function that attempts to clone a linked list with random pointers:

func cloneList(head *Node) *Node {
    if head == nil {
        return nil
    }
    curr := head
    for curr != nil {
        newNode := &Node{Val: curr.Val}
        newNode.Next = curr.Next
        curr.Next = newNode
        curr = curr.Next
    }
    curr = head
    for curr != nil {
        if curr.Random != nil {
            curr.Next.Random = curr.Random.Next
        }
        curr = curr.Next.Next
    }
    curr = head
    cloneHead := head.Next
    cloneCurr := cloneHead
    for curr != nil {
        curr.Next = curr.Next.Next
        if cloneCurr.Next != nil {
            cloneCurr.Next = cloneCurr.Next.Next
        }
        curr = curr.Next
        cloneCurr = cloneCurr.Next
    }
    return cloneHead
}

medium

A. In the first loop, curr = curr.Next should be curr = curr.Next.Next

B. Random pointers are not copied correctly

C. The cloned list is not separated from the original

D. The function does not handle empty lists

Solution

1. Step 1: Analyze the first loop for inserting cloned nodes

   After inserting newNode, curr should move two steps to skip the new node and reach next original node.

2. Step 2: Identify incorrect pointer update

   The code uses curr = curr.Next, which moves to the new node again, causing an infinite loop or incorrect behavior.

3. Final Answer:

   In the first loop, curr = curr.Next should be curr = curr.Next.Next -> Option A

4. Quick Check:

   Move curr two steps after inserting clone [OK]

Hint: After inserting clone, skip it by moving curr two steps [OK]

Common Mistakes:

• Moving curr only one step causes infinite loop
• Not updating random pointers properly
• Forgetting to separate cloned list

5. You want to clone a linked list with random pointers but also reverse the cloned list. Which approach correctly combines cloning and reversing in Go? A) Clone the list first, then reverse the cloned list by changing next pointers. B) Reverse the original list first, then clone it. C) Clone the list and reverse random pointers instead of next pointers. D) Clone the list and keep the order same, reversing is not possible with random pointers.

hard

A. Clone the list and reverse random pointers instead of next pointers.

B. Reverse the original list first, then clone it.

C. Clone the list first, then reverse the cloned list by changing next pointers.

D. Clone the list and keep the order same, reversing is not possible with random pointers.

Solution

1. Step 1: Understand cloning and reversing separately

   Cloning creates a copy with correct next and random pointers. Reversing changes next pointers order.

2. Step 2: Combine cloning and reversing correctly

   Clone first to preserve random pointers, then reverse cloned list's next pointers without affecting random pointers.

3. Final Answer:

   Clone the list first, then reverse the cloned list by changing next pointers. -> Option C

4. Quick Check:

   Clone then reverse next pointers to keep random intact [OK]

Hint: Clone first, then reverse next pointers only [OK]

Common Mistakes:

• Reversing original list breaks random pointer references
• Reversing random pointers instead of next pointers
• Assuming reversing is impossible with random pointers