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

Time Complexity: Clone Linked List with Random Pointer

O(n)

Understanding Time Complexity

We want to understand how long it takes to copy a linked list where each node has a random pointer.

How does the time grow as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Clone linked list with random pointer
Node* cloneList(Node* head) {
    if (!head) return NULL;
    Node* curr = head;
    // Step 1: Insert cloned nodes
    while (curr) {
        Node* newNode = malloc(sizeof(Node));
        newNode->val = curr->val;
        newNode->next = curr->next;
        curr->next = newNode;
        curr = newNode->next;
    }
    // Step 2: Set random pointers
    curr = head;
    while (curr) {
        if (curr->random) curr->next->random = curr->random->next;
        curr = curr->next->next;
    }
    // Step 3: Separate lists
    curr = head;
    Node* cloneHead = head->next;
    Node* cloneCurr = cloneHead;
    while (curr) {
        curr->next = curr->next->next;
        if (cloneCurr->next) cloneCurr->next = cloneCurr->next->next;
        curr = curr->next;
        cloneCurr = cloneCurr->next;
    }
    return cloneHead;
}
    

This code creates a copy of each node, sets the random pointers for the copies, and then separates the original and cloned lists.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Three separate while loops each traversing the entire list once.
• How many times: Each loop runs once over all n nodes, so about 3n steps total.

How Execution Grows With Input

As the list size grows, the number of steps grows roughly in direct proportion.

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

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to clone the list grows in a straight line as the list gets longer.

Common Mistake

[X] Wrong: "Because there are nested steps, the time must be quadratic O(n²)."

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

Interview Connect

Understanding this linear time cloning method shows you can handle complex pointer structures efficiently, a valuable skill in many coding challenges.

Self-Check

"What if we used a hash map to store original-to-clone node mappings instead of inserting cloned nodes inline? How would the time complexity change?"

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