Clone Linked List with Random Pointer in DSA Java - Time & Space Complexity

Time Complexity: Clone Linked List with Random Pointer

O(n)

Understanding Time Complexity

We want to understand how the time needed to clone a linked list with random pointers changes as the list grows.

How does the work increase when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node {
    int val;
    Node next;
    Node random;
    Node(int val) { this.val = val; }
}

public Node cloneList(Node head) {
    if (head == null) return null;
    Node curr = head;
    while (curr != null) {
        Node copy = new Node(curr.val);
        copy.next = curr.next;
        curr.next = copy;
        curr = copy.next;
    }
    // ... (rest of cloning steps)
    return head.next;
}
    

This code creates a copy of each node and inserts it right after the original node in the list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing the linked list multiple times.
• How many times: The list is traversed about three times: once to insert copies, once to assign random pointers, and once to separate the lists.

How Execution Grows With Input

Each traversal goes through all nodes once, so the total work grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 30 steps (3 passes x 10 nodes)
100	About 300 steps (3 passes x 100 nodes)
1000	About 3000 steps (3 passes x 1000 nodes)

Pattern observation: The total steps increase linearly as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to clone the list grows in direct proportion to the number of nodes.

Common Mistake

[X] Wrong: "Because there are multiple loops, the time complexity is quadratic O(n²)."

[OK] Correct: Each loop runs separately over the list, not nested inside each other, so the total work adds up linearly, not multiplies.

Interview Connect

Understanding this helps you explain how to handle complex linked list problems efficiently, a skill valued in many coding challenges.

Self-Check

"What if the list had an additional pointer that required another traversal? How would the time complexity change?"

Practice

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

easy

A. Copying both next and random pointers correctly

B. Only copying the next pointers

C. Ignoring the random pointers

D. Reversing the linked list during cloning

Solution

1. Step 1: Understand linked list structure

   A linked list with random pointers has two references per node: next and random.

2. Step 2: Identify cloning challenge

   Cloning requires copying both next and random pointers so the new list matches the original structure.

3. Final Answer:

   Copying both next and random pointers correctly -> Option A

4. Quick Check:

   Both pointers must be copied = B [OK]

Hint: Remember to copy both next and random pointers [OK]

Common Mistakes:

• Copying only next pointers
• Ignoring random pointers
• Assuming random pointers are null

2. Which Java code snippet correctly declares a node class for a linked list with a random pointer?

easy

A. class Node { int val; Node next; Node random; Node(int val) { val = val; } }

B. class Node { int val; Node next; Node random; Node() { val = 0; } }

C. class Node { int val; Node next; Node random; void Node(int val) { this.val = val; } }

D. class Node { int val; Node next; Node random; Node(int val) { this.val = val; } }

Solution

1. Step 1: Check constructor syntax

   Constructor must have the same name as class and assign value with this.val = val.

2. Step 2: Validate field declarations

   Fields val, next, and random must be declared correctly as int and Node types.

3. Final Answer:

   class Node { int val; Node next; Node random; Node(int val) { this.val = val; } } -> Option D

4. Quick Check:

   Constructor with this.val = val is correct = A [OK]

Hint: Constructor name must match class and use this keyword [OK]

Common Mistakes:

• Using void return type for constructor
• Not using this keyword for assignment
• Incorrect constructor name

3. What will be the output of the following code snippet that clones a linked list with random pointers?

class Node {
  int val;
  Node next, random;
  Node(int val) { this.val = val; }
}

// Original list: 1 -> 2 -> null
// random pointers: 1.random = 2, 2.random = 1

Node head = new Node(1);
head.next = new Node(2);
head.random = head.next;
head.next.random = head;

Node clonedHead = cloneList(head);
System.out.println(clonedHead.val + " -> " + clonedHead.next.val + " -> null");
System.out.println(clonedHead.random.val + " -> " + clonedHead.next.random.val + " -> null");

// Assume cloneList correctly clones the list with random pointers

medium

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

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

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

D. 1 -> 1 -> null
2 -> 2 -> null

Solution

1. Step 1: Understand original list and random pointers

   Original list nodes: 1 -> 2 -> null; random: 1.random = 2, 2.random = 1.

2. Step 2: Cloned list should replicate values and random links

   Cloned list nodes: 1 -> 2 -> null; cloned random pointers: cloned 1.random = cloned 2, cloned 2.random = cloned 1.

3. Final Answer:

   1 -> 2 -> null 2 -> 1 -> null -> Option B

4. Quick Check:

   Random pointers cycle = C [OK]

Hint: Random pointers in clone match original's random targets [OK]

Common Mistakes:

• Printing original random values instead of cloned
• Mixing up random pointer values
• Assuming random pointers are null

4. Identify the bug in this cloning function for a linked list with random pointers:

Node cloneList(Node head) {
  if (head == null) return null;
  Node curr = head;
  while (curr != null) {
    Node copy = new Node(curr.val);
    copy.next = curr.next;
    curr.next = copy;
    curr = curr.next.next;
  }
  // Set random pointers
  curr = head;
  while (curr != null) {
    curr.next.random = curr.random != null ? curr.random.next : null;
    curr = curr.next.next;
  }
  // Separate lists
  curr = head;
  Node copyHead = head.next;
  while (curr != null) {
    Node copy = curr.next;
    curr.next = copy.next;
    copy.next = (copy.next != null) ? copy.next.next : null;
    curr = curr.next;
  }
  return copyHead;
}

medium

A. In the last while loop, copy.next = copy.next.next causes error when copy.next is null

B. In the second while loop, curr.next.random = curr.random.next may cause NullPointerException

C. In the first while loop, curr = curr.next should be curr = curr.next.next

D. The function does not clone the random pointers at all

Solution

1. Step 1: Analyze last while loop for separating lists

   copy.next = copy.next.next assumes copy.next is not null, but at list end copy.next is null causing error.

2. Step 2: Identify fix for null check

   Must check if copy.next is null before accessing copy.next.next to avoid NullPointerException.

3. Final Answer:

   In the last while loop, copy.next = copy.next.next causes error when copy.next is null -> Option A

4. Quick Check:

   Null check needed before copy.next.next = A [OK]

Hint: Check for null before accessing next of next pointers [OK]

Common Mistakes:

• Not checking null before copy.next.next
• Incorrect iteration step in first loop
• Ignoring random pointer cloning

5. Given a linked list with random pointers, how can you clone it in O(n) time and O(1) extra space? Choose the correct approach.

hard

A. Use a hash map to store original to clone node mapping

B. Copy only next pointers and ignore random pointers

C. Interleave cloned nodes with original nodes, set random pointers, then separate lists

D. Create new nodes and assign random pointers by traversing multiple times

Solution

1. Step 1: Understand space constraints

   Using a hash map uses O(n) extra space, so not allowed for O(1) space.

2. Step 2: Use interleaving method

   Interleave cloned nodes between original nodes, assign random pointers using original nodes, then separate lists to get cloned list.

3. Final Answer:

   Interleave cloned nodes with original nodes, set random pointers, then separate lists -> Option C

4. Quick Check:

   Interleaving nodes = O(1) space cloning [OK]

Hint: Interleave clones to avoid extra space for mapping [OK]

Common Mistakes:

• Using hash map increases space complexity
• Multiple traversals without interleaving
• Ignoring random pointers