Clone Linked List with Random Pointer in DSA C - Execution Trace

Concept Flow - Clone Linked List with Random Pointer

Create new nodes interleaved with original nodes

↓

Assign random pointers for new nodes using original nodes

↓

Separate the new nodes to form the cloned list

↓

Return head of cloned list

The cloning process first creates new nodes inserted between original nodes, then sets random pointers for these new nodes, and finally separates the new nodes to form the cloned list.

Execution Sample

DSA C

Node* cloneList(Node* head) {
  Node* curr = head;
  while (curr) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->val = curr->val;
    newNode->next = curr->next;
    curr->next = newNode;
    curr = newNode->next;
  }
  // Set random pointers and separate lists
}

This code snippet clones nodes by inserting new nodes after each original node.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	Start with original list	1->2->3->null	head points to 1	1(val, random) -> 2(val, random) -> 3(val, random) -> null
2	Insert new node after 1	1->1'->2->3->null	1.next = 1', 1'.next = 2	1 -> 1' -> 2 -> 3 -> null
3	Insert new node after 2	1->1'->2->2'->3->null	2.next = 2', 2'.next = 3	1 -> 1' -> 2 -> 2' -> 3 -> null
4	Insert new node after 3	1->1'->2->2'->3->3'->null	3.next = 3', 3'.next = null	1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null
5	Assign random for 1'	Same nodes	1'.random = 1.random ? 1.random.next : NULL	1'.random points to clone of 1.random
6	Assign random for 2'	Same nodes	2'.random = 2.random ? 2.random.next : NULL	2'.random points to clone of 2.random
7	Assign random for 3'	Same nodes	3'.random = 3.random ? 3.random.next : NULL	3'.random points to clone of 3.random
8	Separate cloned list from original	Original: 1->2->3->null Clone: 1'->2'->3'->null	Restore original next pointers, set clone next pointers	Two separate lists: original and cloned
9	Return cloned list head	Clone list head is 1'	Return pointer to 1'	Cloned list ready
10	End	Process complete	No pointer changes	Original and cloned lists separated

💡 All nodes cloned, random pointers assigned, and lists separated

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 8	Final
curr	points to 1	points to 2	points to 3	null	null	null
head	points to 1	points to 1	points to 1	points to 1	points to 1	points to 1
newNode	null	points to 1'	points to 2'	points to 3'	null	null
cloneHead	null	null	null	null	points to 1'	points to 1'

Key Moments - 3 Insights

Why do we insert new nodes between original nodes instead of creating a separate list immediately?

Inserting new nodes between original nodes allows easy access to the corresponding cloned node when assigning random pointers, as shown in steps 2-4 and 5-7 in the execution_table.

How do we assign the random pointer for the cloned nodes?

We assign the cloned node's random pointer by using the original node's random pointer's next node, which is the cloned node, as shown in steps 5-7.

How do we separate the cloned list from the original list without losing connections?

We restore the original nodes' next pointers and set the cloned nodes' next pointers to form a separate list, as done in step 8.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the visual state of the list?

A 1 -> 1' -> 2 -> 2' -> 3 -> 3' -> null

B 1 -> 2 -> 3 -> null

C 1' -> 2' -> 3' -> null

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

💡 Hint

Refer to the Visual State column at step 4 in the execution_table.

At which step do we assign the random pointer for the cloned node of 2?

A Step 2

B Step 6

C Step 5

D Step 8

💡 Hint

Check the Operation column for assigning random pointers in the execution_table.

If we skip step 8 (separating the lists), what would be the state of the list?

A Only cloned nodes remain

B Only original nodes remain

C Original and cloned nodes interleaved

D Lists are already separated

💡 Hint

Look at the Pointer Changes and Visual State columns before step 8.

Concept Snapshot

Clone Linked List with Random Pointer:
1. Insert cloned nodes after each original node.
2. Assign cloned nodes' random pointers using original nodes' random.next.
3. Separate cloned nodes to form the cloned list.
4. Return cloned list head.
This method avoids extra space by interleaving nodes.

Full Transcript

To clone a linked list with random pointers, first insert new cloned nodes between original nodes. Then assign the random pointers of cloned nodes by referencing the original nodes' random pointers' next nodes. Finally, separate the cloned nodes from the original list to form a new cloned list. This process is efficient and uses no extra space for mapping. The execution table shows each step with the list's visual state and pointer changes, helping understand how the list transforms during cloning.

Practice

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

easy

A. Maintaining the correct random pointer connections in the cloned list

B. Copying the data values of each node

C. Reversing the linked list before cloning

D. Sorting the linked list nodes

Solution

1. Step 1: Understand the linked list structure

   A linked list with random pointers has two pointers per node: next and random. The random pointer can point to any node or null.

2. Step 2: Identify cloning challenge

   While copying nodes is straightforward, correctly assigning random pointers in the cloned list is tricky because random pointers may point to nodes not yet cloned.

3. Final Answer:

   Maintaining the correct random pointer connections in the cloned list -> Option A

4. Quick Check:

   Random pointer cloning = main challenge [OK]

Hint: Focus on copying random pointers correctly [OK]

Common Mistakes:

• Ignoring random pointers during cloning
• Assuming random pointers are always null
• Copying only next pointers

2. Which of the following code snippets correctly creates a new node for cloning a linked list with random pointers in C?

easy

A. Node* newNode = (Node*)malloc(sizeof(Node)); newNode->data = oldNode->data; newNode->next = NULL; newNode->random = NULL;

B. Node newNode = malloc(sizeof(Node)); newNode.data = oldNode.data; newNode.next = NULL; newNode.random = NULL;

C. Node* newNode = malloc(Node); newNode->data = oldNode->data; newNode->next = NULL; newNode->random = NULL;

D. Node* newNode = (Node*)malloc(sizeof(Node)); newNode->data = oldNode->data; newNode->next = oldNode->next; newNode->random = oldNode->random;

Solution

1. Step 1: Check correct malloc usage

   In C, malloc requires sizeof(Node) to allocate memory. Casting to (Node*) is good practice.

2. Step 2: Verify initialization

   New node's data is copied, next and random pointers are set to NULL initially to avoid dangling pointers.

3. Final Answer:

   Node* newNode = (Node*)malloc(sizeof(Node)); newNode->data = oldNode->data; newNode->next = NULL; newNode->random = NULL; -> Option A

4. Quick Check:

   Correct malloc and initialization = Node* newNode = (Node*)malloc(sizeof(Node)); newNode->data = oldNode->data; newNode->next = NULL; newNode->random = NULL; [OK]

Hint: Use malloc with sizeof and initialize pointers to NULL [OK]

Common Mistakes:

• Missing sizeof in malloc
• Not casting malloc result
• Copying next and random pointers directly

3. Given the following linked list with random pointers:

Node1(data=1) -> Node2(data=2) -> NULL
Node1.random = Node2
Node2.random = Node2

What will be the random pointer of the cloned Node1 after cloning?

medium

A. Points to NULL

B. Points to cloned Node2

C. Points to original Node2

D. Points to cloned Node1

Solution

1. Step 1: Understand original random pointers

   Original Node1's random points to original Node2.

2. Step 2: Cloning preserves structure

   Cloned Node1's random should point to cloned Node2, not original nodes, to maintain independence.

3. Final Answer:

   Points to cloned Node2 -> Option B

4. Quick Check:

   Random pointer in clone = cloned target node [OK]

Hint: Random pointers in clone point to cloned nodes [OK]

Common Mistakes:

• Random pointer points to original node
• Random pointer set to NULL by mistake
• Confusing next and random pointers

4. Consider this buggy code snippet for cloning a linked list with random pointers:

while (curr != NULL) {
  Node* copy = (Node*)malloc(sizeof(Node));
  copy->data = curr->data;
  copy->next = curr->next;
  curr->next = copy;
  curr = curr->next;
}

What is the main error causing an infinite loop?

medium

A. Copy node's next pointer is not set to NULL

B. Updating curr to curr->next skips the newly inserted node

C. Not updating curr to curr->next->next after insertion

D. malloc size is incorrect

Solution

1. Step 1: Analyze pointer updates in loop

   After inserting copy node after curr, curr->next points to copy. Moving curr = curr->next moves to copy again.

2. Step 2: Identify infinite loop cause

   Loop never advances past copy node because curr is updated to copy, not original next node. It should be curr = curr->next->next.

3. Final Answer:

   Not updating curr to curr->next->next after insertion -> Option C

4. Quick Check:

   Advance curr by two nodes after insertion [OK]

Hint: Move curr two steps after inserting copy [OK]

Common Mistakes:

• Advancing curr by one node causes infinite loop
• Forgetting to link copy nodes properly
• Incorrect malloc size

5. You want to clone a linked list with random pointers in O(n) time and O(1) extra space. Which approach correctly achieves this?

hard

A. Reverse the original list, clone it, then reverse back

B. Use a hash map to store original to copy node mapping, then assign pointers

C. Copy nodes one by one and assign random pointers by searching original list each time

D. Create copy nodes interleaved with original nodes, set random pointers using original nodes, then separate lists

Solution

1. Step 1: Understand space and time constraints

   O(n) time and O(1) space means no extra data structures like hash maps.

2. Step 2: Use interleaving method

   Insert cloned nodes between original nodes, assign random pointers using original nodes, then detach cloned list. This avoids extra space.

3. Final Answer:

   Create copy nodes interleaved with original nodes, set random pointers using original nodes, then separate lists -> Option D

4. Quick Check:

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

Hint: Interleave cloned nodes to avoid extra space [OK]

Common Mistakes:

• Using hash map increases space complexity
• Searching original list for random pointers is O(n^2)
• Reversing list does not help cloning random pointers