Clone Linked List with Random Pointer in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of cloning a linked list with random pointers

What is the printed state of the cloned linked list after running the given Go code?

DSA Go

package main
import "fmt"
type Node struct {
    Val int
    Next *Node
    Random *Node
}
func cloneList(head *Node) *Node {
    if head == nil {
        return nil
    }
    curr := head
    // Step 1: Insert cloned nodes
    for curr != nil {
        newNode := &Node{Val: curr.Val}
        newNode.Next = curr.Next
        curr.Next = newNode
        curr = newNode.Next
    }
    // Step 2: Assign random pointers
    curr = head
    for curr != nil {
        if curr.Random != nil {
            curr.Next.Random = curr.Random.Next
        }
        curr = curr.Next.Next
    }
    // Step 3: Separate lists
    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
}
func printList(head *Node) {
    curr := head
    for curr != nil {
        randomVal := "null"
        if curr.Random != nil {
            randomVal = fmt.Sprint(curr.Random.Val)
        }
        fmt.Printf("%d(random:%s) -> ", curr.Val, randomVal)
        curr = curr.Next
    }
    fmt.Println("null")
}
func main() {
    // Create list: 1->2->3->null
    n1 := &Node{Val: 1}
    n2 := &Node{Val: 2}
    n3 := &Node{Val: 3}
    n1.Next = n2
    n2.Next = n3
    // Random pointers
    n1.Random = n3
    n2.Random = n1
    n3.Random = n2
    clone := cloneList(n1)
    printList(clone)
}

A 1(random:1) -> 2(random:2) -> 3(random:3) -> null

B 1(random:null) -> 2(random:null) -> 3(random:null) -> null

C 1(random:2) -> 2(random:3) -> 3(random:1) -> null

D 1(random:3) -> 2(random:1) -> 3(random:2) -> null

💡 Hint

Focus on how the random pointers are assigned in the cloned list.

The cloned list preserves the original random pointer connections. Node 1's random points to Node 3, Node 2's random points to Node 1, and Node 3's random points to Node 2.

🧠 Conceptual

intermediate

Understanding the purpose of interleaving cloned nodes

Why does the cloning algorithm insert cloned nodes between original nodes before assigning random pointers?

A To easily assign random pointers by using original node's random's next as the clone's random

B To reduce the time complexity from O(n^2) to O(n^3)

C To avoid using extra space for a hash map by linking clones directly after originals

D To sort the linked list nodes in ascending order

💡 Hint

Think about how the random pointer of the clone is assigned using the original node's random pointer.

By placing cloned nodes right after original nodes, the algorithm can assign the clone's random pointer by accessing original.Random.Next, which is the clone of the original's random node.

🔧 Debug

advanced

Identify the error in this cloning code snippet

What error will this Go code produce when cloning a linked list with random pointers?

DSA Go

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
    }
    return head.Next
}

A Nil pointer dereference when accessing curr.Next.Next

B Correct cloning with no errors

C Infinite loop because curr never advances properly

D Compilation error due to missing return statement

💡 Hint

Check how curr is updated inside the loop.

The line curr = curr.Next moves curr to the newly inserted clone node, causing the loop to never reach the original next node, resulting in an infinite loop.

❓ Predict Output

advanced

Output after separating cloned list from original

Given the interleaved list after cloning, what is the state of the original list after separation?

DSA Go

package main
import "fmt"
type Node struct {
    Val int
    Next *Node
    Random *Node
}
func separateLists(head *Node) *Node {
    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
}
func printList(head *Node) {
    curr := head
    for curr != nil {
        fmt.Printf("%d -> ", curr.Val)
        curr = curr.Next
    }
    fmt.Println("null")
}
func main() {
    // Original list: 1->2->3->null
    n1 := &Node{Val: 1}
    n2 := &Node{Val: 2}
    n3 := &Node{Val: 3}
    n1.Next = n2
    n2.Next = n3
    // Interleaved cloned nodes: 1->1'->2->2'->3->3'->null
    n1.Next = &Node{Val: 1, Next: n2}
    n2.Next = &Node{Val: 2, Next: n3}
    n3.Next = &Node{Val: 3, Next: nil}
    clone := separateLists(n1)
    printList(n1)
}

A 1 -> 2 -> null

B 1 -> 2 -> 3 -> null

C null

D 1 -> 1 -> 2 -> 2 -> 3 -> 3 -> null

💡 Hint

After separation, the original list should be restored to its initial form.

The separation loop restores the original list by skipping cloned nodes, so the original list remains 1 -> 2 -> 3 -> null.

🧠 Conceptual

expert

Time and space complexity of cloning a linked list with random pointers

What are the time and space complexities of the optimal algorithm that clones a linked list with random pointers without using extra hash maps?

A Time: O(n), Space: O(1)

B Time: O(n log n), Space: O(1)

C Time: O(n), Space: O(n)

D Time: O(n^2), Space: O(n)

💡 Hint

Consider the interleaving technique that avoids extra data structures.

The interleaving method clones nodes in O(n) time and uses no extra space besides the cloned nodes themselves, achieving O(1) auxiliary space.

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