Memory Layout Comparison Array vs Linked List in DSA C++ - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an array of 5 integers.

DSA C++

int arr[[1]];

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

A5

B1

C0

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 as size will create too small arrays.

The array size must be 5 to hold 5 integers.

2fill in blank

medium

Complete the code to define a node struct for a singly linked list.

DSA C++

struct Node {
    int data;
    Node* [1];
};

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

Anext

Bprev

Cchild

Dlink

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which is for doubly linked lists.

The pointer to the next node in a singly linked list is usually named next.

3fill in blank

hard

Fix the error in the code to correctly allocate memory for a new node.

DSA C++

Node* newNode = (Node*) malloc(sizeof([1]));

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

Aint

BNode

Cvoid

Dchar

Attempts:

3 left

💡 Hint

Common Mistakes

Using sizeof(int) allocates too little memory.

We must allocate memory for a Node, not for other types.

4fill in blank

hard

Fill both blanks to traverse an array and print its elements.

DSA C++

for (int [1] = 0; [2] < size; ++[1]) {
    std::cout << arr[[1]] << " ";
}

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

Ai

Bj

Csize

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in initialization and condition.

The loop variable is 'i' and must be used consistently in the condition.

5fill in blank

hard

Fill all three blanks to insert a new node at the beginning of a linked list.

DSA C++

Node* newNode = new Node();
newNode->data = value;
newNode->[1] = [2];
head = [3];

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

Anext

Bhead

CnewNode

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' pointer which does not exist in singly linked list.

To insert at the start, set newNode->next to head, then update head to newNode.

Practice

1. Which of the following best describes the memory layout of an array compared to a linked list?

easy

A. Both array and linked list nodes are scattered randomly in memory.

B. Array elements are scattered in memory, while linked list nodes are stored contiguously.

C. Both array and linked list nodes are stored contiguously in memory.

D. Array elements are stored in contiguous memory locations, while linked list nodes are scattered in memory.

Solution

1. Step 1: Understand array memory layout

   Arrays store elements in contiguous memory blocks, meaning elements are placed one after another in memory.

2. Step 2: Understand linked list memory layout

   Linked list nodes are allocated dynamically and can be anywhere in memory, connected by pointers.

3. Final Answer:

   Array elements are stored in contiguous memory locations, while linked list nodes are scattered in memory. -> Option D

4. Quick Check:

   Array = contiguous, Linked List = scattered [OK]

Hint: Arrays are contiguous; linked lists use pointers to scattered nodes [OK]

Common Mistakes:

• Thinking linked list nodes are contiguous
• Confusing array with linked list memory layout
• Assuming both have same memory layout

2. Which C++ declaration correctly defines a linked list node struct with an integer data and a pointer to the next node?

easy

A. struct Node { int data; Node* next; };

B. struct Node { int data; Node next; };

C. struct Node { int data; Node& next; };

D. struct Node { int data; int* next; };

Solution

1. Step 1: Understand pointer usage in struct

   The next node pointer must be a pointer type to allow linking nodes dynamically.

2. Step 2: Analyze each option

   struct Node { int data; Node next; }; uses Node next (not pointer), invalid. struct Node { int data; Node* next; }; uses Node* next (correct pointer). struct Node { int data; Node& next; }; uses reference Node& next (not suitable for linked list). struct Node { int data; int* next; }; uses int* next (wrong type).

3. Final Answer:

   struct Node { int data; Node* next; }; -> Option A

4. Quick Check:

   Pointer to next node = Node* next [OK]

Hint: Use Node* for next pointer in linked list node struct [OK]

Common Mistakes:

• Using object instead of pointer for next
• Using reference instead of pointer
• Using wrong pointer type

3. Consider the following C++ code snippet that creates an array and a linked list with 3 elements each. What is the output of the memory addresses printed for the array and linked list nodes?

int arr[3] = {10, 20, 30};
struct Node { int data; Node* next; };
Node* n1 = new Node{10, nullptr};
Node* n2 = new Node{20, nullptr};
Node* n3 = new Node{30, nullptr};
n1->next = n2; n2->next = n3;

std::cout << &arr[0] << " " << &arr[1] << " " << &arr[2] << std::endl;
std::cout << n1 << " " << n2 << " " << n3 << std::endl;

medium

A. Both array and linked list addresses are contiguous

B. Array addresses are contiguous; linked list addresses are non-contiguous

C. Array addresses are non-contiguous; linked list addresses are contiguous

D. Both array and linked list addresses are non-contiguous

Solution

1. Step 1: Analyze array memory addresses

   Array elements are stored contiguously, so &arr[0], &arr[1], &arr[2] will be adjacent addresses.

2. Step 2: Analyze linked list node addresses

   The nodes are separately allocated dynamically with new; their addresses (printed as n1, n2, n3) are non-contiguous in memory.

3. Final Answer:

   Array addresses are contiguous; linked list addresses are non-contiguous -> Option B

4. Quick Check:

   Array contiguous, linked list scattered [OK]

Hint: Array addresses are adjacent; linked list nodes scattered [OK]

Common Mistakes:

• Assuming linked list nodes are contiguous
• Confusing variable addresses with pointer values
• Ignoring dynamic allocation behavior

4. The following code attempts to create a linked list of 3 nodes but has a bug related to memory layout. Identify the bug.

struct Node {
  int data;
  Node* next;
};

Node* head = new Node{1, nullptr};
Node* second = new Node{2, nullptr};
Node* third = new Node{3, nullptr};

head->next = second;
second = third;
third->next = nullptr;

medium

A. The line 'second = third;' breaks the link between first and second nodes.

B. The nodes are not allocated dynamically.

C. The next pointers are not set correctly for all nodes.

D. The head pointer is not initialized.

Solution

1. Step 1: Trace pointer assignments

   head->next = second links first node to second node correctly. Then 'second = third;' changes the local second pointer to point to third node, but node2->next remains nullptr. third->next = nullptr sets nothing new.

2. Step 2: Understand impact on linked list

   The next pointer of the second node is never set to the third node, so the list is 1 -> 2 -> null with third node unreachable from head.

3. Final Answer:

   The next pointers are not set correctly for all nodes. -> Option C

4. Quick Check:

   Missing second->next = third [OK]

Hint: Changing pointer variable breaks links; update next pointers instead [OK]

Common Mistakes:

• Confusing pointer variable with node's next pointer
• Ignoring pointer reassignment effect
• Assuming nodes are not dynamically allocated

5. You want to store 1 million integers and frequently access elements by index. Considering memory layout and access speed, which data structure is better and why?

1. Array
2. Linked List

hard

A. Array, because contiguous memory allows fast index access and better cache performance.

B. Linked List, because scattered nodes reduce memory fragmentation and improve access speed.

C. Linked List, because dynamic allocation uses less memory than arrays.

D. Array, because linked lists require extra memory for pointers making them slower.

Solution

1. Step 1: Consider access by index

   Arrays allow direct access by index in O(1) time due to contiguous memory layout.

2. Step 2: Consider memory and cache effects

   Contiguous memory improves cache locality, speeding up access. Linked lists have scattered nodes, causing cache misses and slower access.

3. Final Answer:

   Array, because contiguous memory allows fast index access and better cache performance. -> Option A

4. Quick Check:

   Fast index access = Array with contiguous memory [OK]

Hint: For fast index access, choose arrays due to contiguous memory [OK]

Common Mistakes:

• Assuming linked list is faster for index access
• Ignoring cache locality effects
• Confusing memory fragmentation with access speed