Intersection Point of Two Linked Lists in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Rust

let new_node = ListNode::new([1]);

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

Anext

Bnode

Chead

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' or 'head' instead of the value.

Forgetting to pass any argument.

The new node is created with the given value 'val'.

2fill in blank

medium

Complete the code to advance the pointer to the next node.

DSA Rust

current = current[1];

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

A.next

B.prev

C.head

D.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.prev' which is for doubly linked lists.

Using '.value' which holds data, not the next node.

To move forward in the list, use '.next' to get the next node.

3fill in blank

hard

Fix the error in the loop condition to check if pointers are not null.

DSA Rust

while [1].is_some() && [2].is_some() {

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

Aptr1

Bptr2

Chead

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Checking the same pointer twice.

Using undefined variables.

Both pointers must be checked separately; here, the first pointer is checked twice by mistake.

4fill in blank

hard

Fill both blanks to compare node pointers and advance the correct pointer.

DSA Rust

if ptr1 == ptr2 {
    return ptr1;
} else if ptr1[1] < ptr2[2] {
    ptr1 = ptr1.unwrap().next.clone();
} else {
    ptr2 = ptr2.unwrap().next.clone();
}

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

A.as_ref()

B.unwrap()

C.clone()

D.next

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.unwrap()' which moves ownership and causes errors.

Using '.clone()' unnecessarily.

To compare the nodes, use '.as_ref()' to get references without moving ownership.

5fill in blank

hard

Fill all three blanks to return None if no intersection is found after traversal.

DSA Rust

while ptr1.is_some() && ptr2.is_some() {
    if ptr1 == ptr2 {
        return ptr1;
    }
    ptr1 = ptr1.unwrap().next.clone();
    ptr2 = ptr2.unwrap().next.clone();
}
return [1];

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

ASome(Box::new(ListNode::new(0)))

Bptr1

CNone

Dptr2

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a new node instead of None.

Returning one of the pointers without intersection.

If no intersection is found, the function returns 'None' to indicate no shared node.

Practice

1. What does the term intersection point mean in the context of two linked lists?

easy

A. The last node of the shorter linked list.

B. The node where the values of both linked lists are equal.

C. The node where both linked lists share the same reference (memory address).

D. The first node of the longer linked list.

Solution

1. Step 1: Understand linked list nodes

   Each node in a linked list has a value and a reference (memory address) to the next node.

2. Step 2: Define intersection point

   The intersection point is where two linked lists share the exact same node reference, meaning they merge at that node.

3. Final Answer:

   The node where both linked lists share the same reference (memory address). -> Option C

4. Quick Check:

   Intersection means shared node reference [OK]

Hint: Intersection means same node object, not just same value [OK]

Common Mistakes:

• Confusing same value with same node reference
• Thinking intersection is about list length
• Assuming intersection is always at the start

2. Which Rust code snippet correctly declares a singly linked list node struct for intersection detection?

easy

A. struct Node { value: i32, next: Option }

B. struct Node { value: i32, next: Node }

C. struct Node { value: i32, next: &Node }

D. struct Node { value: i32, next: Box<Node> }

Solution

1. Step 1: Understand Rust linked list node

   Rust needs Option<Box<Node>> for next to allow None and heap allocation.

2. Step 2: Check each option

   A uses raw Node causing infinite size error. C uses reference without lifetime. D lacks Option so cannot represent end of list. The correct declaration uses Option<Box<Node>>.

3. Final Answer:

   struct Node { value: i32, next: Option<Box<Node>> } -> Option A

4. Quick Check:

   Use Option<Box<Node>> for next pointer [OK]

Hint: Use Option<Box<Node>> for next to allow null and heap storage [OK]

Common Mistakes:

• Omitting Option causing infinite size error
• Using references without lifetimes
• Using raw Node type causing recursive size

3. Given two linked lists in Rust where list A is 1 -> 2 -> 3 -> 4 -> 5 and list B is 9 -> 4 -> 5, assuming nodes with values 4 and 5 are shared (intersection), what is the intersection point value?

medium

A. 4

B. 5

C. 3

D. 9

Solution

1. Step 1: Identify shared nodes

   Lists share nodes with values 4 and 5, meaning intersection starts at node with value 4.

2. Step 2: Determine intersection point

   The first shared node is the intersection point, which is value 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   First shared node value = 4 [OK]

Hint: Intersection is first shared node, not last [OK]

Common Mistakes:

• Choosing last shared node value
• Picking non-shared node values
• Confusing node value with node reference

4. Consider this Rust function snippet to find intersection point of two linked lists:

fn get_intersection_node(head_a: Option<&Box<Node>>, head_b: Option<&Box<Node>>) -> Option<&Box<Node>> {
    let mut a = head_a;
    let mut b = head_b;
    while a != b {
        a = if let Some(node) = a { node.next.as_ref() } else { head_b };
        b = if let Some(node) = b { node.next.as_ref() } else { head_a };
    }
    a
}

What is the main bug in this code?

medium

A. It does not handle None cases properly causing infinite loop.

B. It compares node references incorrectly using != instead of pointer equality.

C. It assigns head_b and head_a inside loop incorrectly causing logic error.

D. It uses references without proper lifetime annotations causing compile error.

Solution

1. Step 1: Analyze compilation

   The function uses bare references (&Box<Node>) in Option without lifetime parameters.

2. Step 2: Rust lifetime rules

   Explicit lifetime annotations are required (e.g., fn<'a>(head_a: Option<&'a Box<Node>>, ...) -> Option<&'a Box<Node>>), otherwise it fails to compile due to unresolved lifetimes.

3. Final Answer:

   It uses references without proper lifetime annotations causing compile error. -> Option D

4. Quick Check:

   Missing lifetime specifiers [OK]

Hint: Rust linked list functions need lifetime annotations for node references [OK]

Common Mistakes:

• Assuming != compares pointers correctly
• Ignoring infinite loop possibility
• Overlooking lifetime errors

5. You have two singly linked lists that intersect. List A length is 7, List B length is 5. To find the intersection point efficiently, which approach is best?

hard

A. Traverse both lists fully and store nodes in hash sets to find intersection.

B. Calculate length difference, advance longer list by difference, then move both pointers together.

C. Reverse both lists and compare nodes from start until mismatch.

D. Compare values of nodes one by one from heads until match found.

Solution

1. Step 1: Understand length difference method

   Calculate difference in lengths (7 - 5 = 2), advance longer list by 2 nodes to align both lists.

2. Step 2: Move both pointers together

   Then move both pointers one step at a time until they point to the same node (intersection point).

3. Final Answer:

   Calculate length difference, advance longer list by difference, then move both pointers together. -> Option B

4. Quick Check:

   Align lists by length difference for efficient intersection [OK]

Hint: Advance longer list by length difference before simultaneous traversal [OK]

Common Mistakes:

• Using extra memory unnecessarily
• Reversing lists which is inefficient
• Comparing node values instead of references