Clone Linked List with Random Pointer in DSA C - 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 NULL, not just the next node.

intermediate

Why do we need to clone a linked list with random pointers carefully?

Because each node has a random pointer that can point anywhere, we must copy both the next and random pointers correctly to keep the structure intact.

intermediate

What is the first step in the common approach to clone a linked list with random pointers?

Insert a new cloned node right after each original node in the list, so the list looks like: original1 -> clone1 -> original2 -> clone2 -> ...

advanced

How do we set the random pointers of cloned nodes in the interleaved list?

For each original node, set the cloned node's random pointer to original node's random pointer's next node (which is the clone of the random target).

advanced

How do we separate the cloned list from the original after cloning?

We restore the original list by skipping cloned nodes and build the cloned list by linking cloned nodes together, effectively splitting the interleaved list into two.

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

A Only the next node

B Any node in the list or NULL

C Only the previous node

D The head of the list

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

In the cloning process, why do we insert cloned nodes between original nodes?

A To reverse the list

B To sort the list

C To delete original nodes later

D To easily set random pointers of clones using original nodes

Inserting clones between originals helps set random pointers by using the next pointers of original nodes.

How do we assign the random pointer of a cloned node?

A clone->random = original->random->next

B clone->random = original->next

C clone->random = NULL

D clone->random = original

Because the clone of the random target is right after the original random target node.

What is the final step after setting next and random pointers in cloning?

A Delete the original list

B Reverse the cloned list

C Separate the cloned list from the original list

D Sort the cloned list

We separate the interleaved list into original and cloned lists.

What is the time complexity of cloning a linked list with random pointers using the interleaving method?

A O(n)

B O(n^2)

C O(log n)

D O(1)

The method visits each node a constant number of times, so it runs in linear time.

Explain the three main steps to clone a linked list with random pointers using the interleaving method.

Think about how to keep track of clones and originals together.

Why can't we simply copy nodes and random pointers in one pass when cloning a linked list with random pointers?

Consider what happens if random points forward or backward.

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