Detect Cycle in Linked List Floyd's Algorithm in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the slow pointer to the head of the linked list.

DSA Rust

let mut slow = [1];

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

Ahead

BNone

Ctail

Dfast

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing slow to None or tail instead of head.

The slow pointer starts at the head of the linked list to begin traversal.

2fill in blank

medium

Complete the code to move the fast pointer two steps ahead safely.

DSA Rust

[1]

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

Afast = fast.next.next;

Bfast = fast.unwrap().next.unwrap().next;

Cfast = fast.next;

Dfast = fast.and_then(|node| node.next.as_ref()).and_then(|node| node.next.as_ref()).map(|node| &**node);

Attempts:

3 left

💡 Hint

Common Mistakes

Using unwrap which can cause panic if next is None.

This safely moves the fast pointer two steps ahead using Rust's Option chaining.

3fill in blank

hard

Fix the error in the loop condition to check if fast and fast.next are not None.

DSA Rust

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

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

Afast.is_none()

Bfast.next.is_some()

Cfast.unwrap().next.is_some()

Dfast.next.is_none()

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to access fast.next directly without unwrap.

We must check if fast's next node exists by unwrapping fast safely.

4fill in blank

hard

Fill both blanks to move slow by one step and fast by two steps safely.

DSA Rust

slow = [1];
fast = [2];

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

Aslow.and_then(|node| node.next.as_ref()).map(|node| &**node)

Bslow.next

Cfast.next

Dfast.and_then(|node| node.next.as_ref()).and_then(|node| node.next.as_ref()).map(|node| &**node)

Attempts:

3 left

💡 Hint

Common Mistakes

Using direct next without Option handling.

Slow moves one step using Option chaining; fast moves two steps similarly.

5fill in blank

hard

Fill all three blanks to return true if slow equals fast, else false after loop.

DSA Rust

if slow == [1] {
    return [2];
}

return [3];

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

Afast

Btrue

Cfalse

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true or false incorrectly.

If slow and fast meet, a cycle exists (return true). Otherwise, return false.

Practice

1. What is the main idea behind Floyd's Cycle Detection Algorithm in a linked list?

easy

A. Use two pointers moving at different speeds to detect a cycle

B. Count the number of nodes and check if it exceeds a limit

C. Reverse the linked list and check if it matches the original

D. Use a hash set to store visited nodes and check for repeats

Solution

1. Step 1: Understand Floyd's algorithm concept

   It uses two pointers, one moving faster (fast) and one slower (slow) through the list.

2. Step 2: Detect cycle by pointer meeting

   If the fast pointer meets the slow pointer, a cycle exists; otherwise, no cycle.

3. Final Answer:

   Use two pointers moving at different speeds to detect a cycle -> Option A

4. Quick Check:

   Two pointers at different speeds = cycle detection [OK]

Hint: Two pointers at different speeds detect cycle fast [OK]

Common Mistakes:

• Thinking counting nodes detects cycle
• Assuming reversing list helps detect cycle
• Using extra memory instead of pointers

2. Which Rust code snippet correctly initializes the slow and fast pointers for Floyd's cycle detection in a linked list named head?

easy

A. let mut slow = head.as_ref(); let mut fast = head.as_ref();

B. let mut slow = head.clone(); let mut fast = head.clone();

C. let mut slow = &head.next; let mut fast = &head.next.next;

D. let mut slow = &head; let mut fast = &head.next;

Solution

1. Step 1: Initialize both pointers at head

   Floyd's algorithm starts both slow and fast pointers at the head node.

2. Step 2: Correct Rust reference usage

   Using head.as_ref() for both slow and fast is correct; fast moves faster later.

3. Final Answer:

   let mut slow = head.as_ref(); let mut fast = head.as_ref(); -> Option A

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head node [OK]

Common Mistakes:

• Starting fast pointer at head.next before loop
• Cloning nodes unnecessarily
• Using wrong references causing borrow errors

3. What will be the output of the following Rust code snippet detecting a cycle in a linked list?

struct Node { val: i32, next: Option }

fn has_cycle(head: &Option) -> bool {
    let mut slow = head.as_ref();
    let mut fast = head.as_ref();
    while fast.is_some() && fast.unwrap().next.is_some() {
        slow = slow.unwrap().next.as_ref();
        fast = fast.unwrap().next.as_ref().unwrap().next.as_ref();
        if slow.map(|n| n as *const ()) == fast.map(|n| n as *const ()) {
            return true;
        }
    }
    false
}

// List: 1 -> 2 -> 3 -> 4 -> 2 (cycle back to node with value 2)

medium

A. Infinite loop

B. false

C. true

D. Compilation error

Solution

1. Step 1: Understand the list structure

   The list has a cycle starting at node with value 2, so cycle exists.

2. Step 2: Trace Floyd's pointers

   Slow and fast pointers will eventually meet inside the cycle, returning true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Cycle present = true [OK]

Hint: Cycle in list means pointers meet = true [OK]

Common Mistakes:

• Assuming no cycle returns true
• Confusing Option with references
• Missing unwrap() causing panic

4. Identify the bug in this Rust code snippet for detecting a cycle using Floyd's algorithm:

fn has_cycle(head: &Option) -> bool {
    let mut slow = head.as_ref();
    let mut fast = head.as_ref();
    while fast.is_some() && fast.unwrap().next.is_some() {
        slow = slow.unwrap().next.as_ref();
        fast = fast.unwrap().next.as_ref(); // Bug here
        if slow.map(|n| n as *const ()) == fast.map(|n| n as *const ()) {
            return true;
        }
    }
    false
}

medium

A. slow pointer moves two steps instead of one

B. fast pointer moves only one step instead of two

C. Missing check for slow being None

D. Incorrect comparison of slow and fast pointers

Solution

1. Step 1: Analyze fast pointer movement

   Fast pointer should move two steps each loop, but here it moves only one step.

2. Step 2: Understand impact on cycle detection

   Moving fast only one step breaks Floyd's logic, so cycle detection fails or slows.

3. Final Answer:

   fast pointer moves only one step instead of two -> Option B

4. Quick Check:

   Fast pointer must move two steps [OK]

Hint: Fast pointer must jump two nodes each iteration [OK]

Common Mistakes:

• Moving fast pointer one step only
• Not checking fast pointer's next before moving
• Comparing pointers incorrectly

5. Given a linked list where the last node points back to the head (forming a cycle), how can you modify Floyd's algorithm to find the exact node where the cycle begins?

hard

A. Use a hash set to store nodes and find first repeated node

B. Count nodes until slow and fast meet, then return that count as cycle start

C. Reverse the list and check where reversal breaks

D. After detecting cycle, reset slow to head and move both pointers one step until they meet

Solution

1. Step 1: Detect cycle using Floyd's algorithm

   First, detect if cycle exists by moving slow and fast pointers.

2. Step 2: Find cycle start node

   Reset slow to head, then move slow and fast one step each until they meet; meeting point is cycle start.

3. Final Answer:

   After detecting cycle, reset slow to head and move both pointers one step until they meet -> Option D

4. Quick Check:

   Reset slow and move both pointers to find cycle start [OK]

Hint: Reset slow to head after cycle detection to find start [OK]

Common Mistakes:

• Trying to count nodes instead of pointer moves
• Using extra memory instead of pointer reset
• Reversing list which breaks cycle detection