Detect Cycle in Linked List Floyd's Algorithm in DSA Rust - Time & Space Complexity

Time Complexity: Detect Cycle in Linked List Floyd's Algorithm

O(n)

Understanding Time Complexity

We want to understand how long it takes to check if a linked list has a cycle using Floyd's Algorithm.

The question is: how does the time grow as the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn has_cycle(head: Option<Box<Node>>) -> bool {
    let mut slow = head.as_ref();
    let mut fast = head.as_ref();
    while let (Some(s), Some(f)) = (slow, fast) {
        slow = s.next.as_ref();
        fast = f.next.as_ref().and_then(|n| n.next.as_ref());
        if let (Some(ref ss), Some(ref ff)) = (slow, fast) {
            if std::ptr::eq(ss.as_ref(), ff.as_ref()) {
                return true;
            }
        }
    }
    false
}
    

This code uses two pointers moving at different speeds to detect if the linked list loops back on itself.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves slow and fast pointers through the list.
• How many times: At most, the loop runs once per node until pointers meet or reach the end.

How Execution Grows With Input

As the list grows, the number of steps to detect a cycle grows roughly in proportion to the number of nodes.

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

Pattern observation: The steps increase linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to detect a cycle grows in a straight line with the number of nodes in the list.

Common Mistake

[X] Wrong: "The algorithm checks every pair of nodes, so it must be O(n²)."

[OK] Correct: The two pointers move through the list independently, not checking all pairs, so the total steps grow only linearly.

Interview Connect

Understanding this linear time check shows you can analyze pointer-based algorithms efficiently, a key skill in many coding challenges.

Self-Check

"What if we used only one pointer moving one step at a time? How would the time complexity change?"

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