Bird
Raised Fist0
DSA Rustprogramming~5 mins

Detect Cycle in Linked List Floyd's Algorithm in DSA Rust - Cheat Sheet & Quick Revision

Choose your learning style10 modes available
Recall & Review
beginner
What is Floyd's Cycle Detection Algorithm used for?
It is used to detect if a linked list has a cycle (loop) by using two pointers moving at different speeds.
Click to reveal answer
beginner
In Floyd's Algorithm, what are the two pointers called and how do they move?
They are called 'slow' and 'fast'. The slow pointer moves one step at a time, and the fast pointer moves two steps at a time.
Click to reveal answer
intermediate
Why does Floyd's Algorithm detect a cycle when slow and fast pointers meet?
If there is a cycle, the fast pointer will eventually catch up to the slow pointer inside the loop, meaning they point to the same node.
Click to reveal answer
intermediate
What is the time complexity of Floyd's Cycle Detection Algorithm?
The time complexity is O(n), where n is the number of nodes in the linked list.
Click to reveal answer
beginner
Show the Rust code snippet to initialize slow and fast pointers for Floyd's Algorithm on a linked list.
let mut slow = head.as_ref();
let mut fast = head.as_ref();
Click to reveal answer
What happens if the fast pointer reaches the end of the linked list in Floyd's Algorithm?
AThe algorithm restarts.
BA cycle is detected.
CThe slow pointer moves faster.
DThere is no cycle in the linked list.
How many steps does the slow pointer move in each iteration in Floyd's Algorithm?
AOne step
BTwo steps
CThree steps
DVariable steps
What is the main advantage of Floyd's Cycle Detection Algorithm?
AIt detects cycles without extra memory.
BIt sorts the linked list.
CIt uses extra memory to store visited nodes.
DIt reverses the linked list.
If slow and fast pointers meet, what does it indicate?
ANo cycle
BEnd of list
CCycle detected
DList is empty
What is the space complexity of Floyd's Cycle Detection Algorithm?
AO(n)
BO(1)
CO(log n)
DO(n^2)
Explain how Floyd's Cycle Detection Algorithm works step-by-step.
Think about two runners on a track moving at different speeds.
You got /4 concepts.
    Write the Rust code to detect a cycle in a linked list using Floyd's Algorithm.
    Use Option and as_ref() for safe pointer traversal.
    You got /6 concepts.

      Practice

      (1/5)
      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
      Recommended

      Ready to prove it?

      Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

      10 questions~5 mineasy hard