Intersection Point of Two Linked Lists in DSA Rust - Time & Space Complexity

Time Complexity: Intersection Point of Two Linked Lists

O(m + n)

Understanding Time Complexity

We want to know how long it takes to find where two linked lists meet.

How does the time grow as the lists get longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn get_intersection_node(head_a: Option<&Box>, head_b: Option<&Box>) -> Option<&Box> {
    let mut a = head_a;
    let mut b = head_b;

    while a != b {
        a = if a.is_none() { head_b } else { a.unwrap().next.as_ref() };
        b = if b.is_none() { head_a } else { b.unwrap().next.as_ref() };
    }
    a
}
    

This code tries to find the node where two linked lists join by moving pointers through both lists.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving two pointers step-by-step through the lists.
• How many times: Each pointer moves at most twice the length of its list.

How Execution Grows With Input

As the lists get longer, the pointers take more steps, but only up to the sum of both list lengths.

Input Size (n)	Approx. Operations
10	About 20 steps
100	About 200 steps
1000	About 2000 steps

Pattern observation: The steps grow roughly in a straight line with the total length of both lists combined.

Final Time Complexity

Time Complexity: O(m + n)

This means the time grows linearly with the sum of the lengths of the two lists.

Common Mistake

[X] Wrong: "The time is O(m * n) because we compare every node of one list with every node of the other."

[OK] Correct: The code uses two pointers moving through the lists, not nested loops, so it only takes steps proportional to the total length, not their product.

Interview Connect

Understanding this time complexity helps you explain how efficient pointer techniques can solve problems without extra memory or slow nested loops.

Self-Check

"What if we changed the code to use nested loops comparing each node of one list to every node of the other? How would the time complexity change?"

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