Clone Linked List with Random Pointer in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

DSA C

struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
newNode->data = [1];
newNode->next = NULL;
newNode->random = NULL;

Drag options to blanks, or click blank then click option'

Avalue

Bdata

Cnode

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using an undefined variable instead of the input value.

Assigning the pointer instead of the data value.

The new node's data field should be set to the given value, which is stored in the variable 'value'.

2fill in blank

medium

Complete the code to insert the cloned node right after the original node.

DSA C

clonedNode->next = original->next;
original->next = [1];

Drag options to blanks, or click blank then click option'

AclonedNode

Boriginal

Ctemp

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning original->next to original instead of clonedNode.

Using an unrelated pointer variable.

To insert the cloned node after the original, set original->next to clonedNode.

3fill in blank

hard

Fix the error in setting the random pointer of the cloned node.

DSA C

if (original->random != NULL) {
    original->next->random = original->random[1]2;
}

Drag options to blanks, or click blank then click option'

A->random->

B->next->

C->next

D->random

Attempts:

3 left

💡 Hint

Common Mistakes

Using '->random' instead of '->next' to get the cloned node.

Not checking if original->random is NULL.

The cloned node is right after the original, so to set its random pointer, use original->random->next.

4fill in blank

hard

Fill both blanks to separate the cloned list from the original list.

DSA C

original->next = original->next[1];
cloned->next = cloned->next[2];

Drag options to blanks, or click blank then click option'

A->next

B->random

C->next->next

D->random->next

Attempts:

3 left

💡 Hint

Common Mistakes

Using '->next' only, which does not skip nodes.

Using '->random' which is unrelated here.

To separate lists, skip one node ahead by using '->next->next' for both original and cloned nodes.

5fill in blank

hard

Fill all three blanks to complete the function that clones the linked list with random pointers.

DSA C

struct Node* cloneList(struct Node* head) {
    struct Node* curr = head;
    while (curr != NULL) {
        struct Node* cloned = (struct Node*)malloc(sizeof(struct Node));
        cloned->data = curr->data;
        cloned->next = curr->next;
        curr->next = [1];
        curr = cloned->next;
    }

    curr = head;
    while (curr != NULL) {
        if (curr->random != NULL) {
            curr->next->random = curr->random[2];
        }
        curr = curr->next->next;
    }

    struct Node* original = head;
    struct Node* cloneHead = head ? head->next : NULL;
    struct Node* cloned = cloneHead;
    while (original != NULL && cloned != NULL) {
        original->next = original->next[3];
        cloned->next = cloned->next[3];
        original = original->next;
        cloned = cloned->next;
    }
    return cloneHead;
}

Drag options to blanks, or click blank then click option'

Acloned

B->next

C->next->next

D->random->next

Attempts:

3 left

💡 Hint

Common Mistakes

Not inserting cloned node correctly.

Incorrectly setting random pointers.

Not separating lists properly causing cycles.

Insert cloned node after current (curr->next = cloned), set random pointer using original's random's next, and separate lists by skipping nodes with '->next->next'.

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