Detect Cycle in Linked List Floyd's Algorithm in DSA C - 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 C

struct Node* slow = [1];

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

Aslow

BNULL

Chead

Dfast

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing slow to NULL will cause the loop to never start.

Using fast instead of head confuses the pointers.

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

2fill in blank

medium

Complete the condition to continue the loop while both fast and fast->next are not NULL.

DSA C

while (fast != NULL && [1] != NULL) {

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

Aslow->next

Bslow

Chead->next

Dfast->next

Attempts:

3 left

💡 Hint

Common Mistakes

Checking slow->next instead of fast->next causes incorrect loop conditions.

Not checking fast->next leads to segmentation faults.

The loop continues only if fast and fast->next are not NULL to avoid accessing invalid memory.

3fill in blank

hard

Fix the error in moving the fast pointer two steps ahead.

DSA C

fast = [1]->next;

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

Afast->next

Bslow->next

Cfast

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using slow->next moves the wrong pointer.

Assigning fast to fast without moving causes infinite loop.

The fast pointer moves two steps ahead by assigning fast to fast->next->next. Here, fast->next is the first step.

4fill in blank

hard

Fill both blanks to correctly move slow and fast pointers inside the loop.

DSA C

slow = slow[1];
fast = fast[2]->next;

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A->next

B->prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using ->prev moves backward, which is incorrect for singly linked list.

Moving fast only one step causes incorrect cycle detection.

Slow pointer moves one step: slow = slow->next;
Fast pointer moves two steps: fast = fast->next->next;

5fill in blank

hard

Fill all three blanks to complete the cycle detection condition inside the loop.

DSA C

if (slow [1] fast) {
    return [2];
}
slow = slow[3];

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

A==

B1

C->next

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == misses the cycle detection.

Returning 0 means no cycle, which is wrong here.

Not moving slow pointer causes infinite loop.

If slow equals fast, a cycle is detected, so return 1 (true). Then move slow one step forward.

Practice

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

easy

A. To find the middle element of the linked list

B. To detect if there is a cycle (loop) in the linked list

C. To reverse the linked list

D. To sort the linked list in ascending order

Solution

1. Step 1: Understand Floyd's Algorithm goal

   Floyd's Cycle Detection Algorithm is designed to find if a linked list has a cycle or loop.

2. Step 2: Compare with other linked list operations

   Sorting, reversing, or finding the middle element are different operations and not related to cycle detection.

3. Final Answer:

   To detect if there is a cycle (loop) in the linked list -> Option B

4. Quick Check:

   Cycle detection = Detect cycle [OK]

Hint: Floyd's algorithm is famous for cycle detection [OK]

Common Mistakes:

• Confusing cycle detection with sorting
• Thinking it finds middle element
• Assuming it reverses the list

2. Which of the following is the correct way to initialize the two pointers used in Floyd's Cycle Detection Algorithm in C?

easy

A. Node *slow = head; Node *fast = head->next;

B. Node *slow = NULL; Node *fast = head;

C. Node *slow = head->next; Node *fast = head->next->next;

D. Node *slow = head; Node *fast = head;

Solution

1. Step 1: Understand pointer initialization in Floyd's algorithm

   Both slow and fast pointers start at the head node to begin traversal.

2. Step 2: Check options for correct initialization

   Node *slow = head; Node *fast = head; initializes both pointers at head, which is the standard approach.

3. Final Answer:

   Node *slow = head; Node *fast = head; -> Option D

4. Quick Check:

   Both pointers start at head [OK]

Hint: Both pointers start at head node in Floyd's algorithm [OK]

Common Mistakes:

• Starting fast pointer ahead of slow incorrectly
• Initializing slow as NULL
• Skipping initialization of one pointer

3. What will be the output of the following C code snippet using Floyd's algorithm on a linked list with a cycle?

typedef struct Node {
  int data;
  struct Node *next;
} Node;

int detectCycle(Node *head) {
  Node *slow = head, *fast = head;
  while (fast && fast->next) {
    slow = slow->next;
    fast = fast->next->next;
    if (slow == fast) return 1;
  }
  return 0;
}

// Assume linked list has a cycle
printf("%d", detectCycle(head));

medium

A. 1

B. 0

C. Compilation error

D. Infinite loop

Solution

1. Step 1: Analyze the detectCycle function logic

   The function moves slow pointer by one and fast pointer by two nodes. If they meet, it returns 1 indicating a cycle.

2. Step 2: Given the list has a cycle, pointers will meet

   Since the list has a cycle, slow and fast pointers will eventually meet, so function returns 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   Cycle present = return 1 [OK]

Hint: Cycle means slow and fast pointers meet, return 1 [OK]

Common Mistakes:

• Assuming function returns 0 for cycle
• Thinking code causes infinite loop
• Confusing return values

4. Identify the bug in this Floyd's cycle detection code snippet:

int detectCycle(Node *head) {
  Node *slow = head, *fast = head;
  while (fast != NULL && fast->next != NULL) {
    slow = slow->next;
    fast = fast->next;
    if (slow == fast) return 1;
    fast = fast->next;
  }
  return 0;
}

medium

A. fast pointer moves two steps but not checked for NULL before second move

B. slow pointer moves two steps instead of one

C. fast pointer moves only one step instead of two

D. slow pointer is not initialized properly

Solution

1. Step 1: Analyze pointer movements in the loop

   Slow moves one step, fast moves one step, then checks if slow == fast, before the second fast move.

2. Step 2: Problem with check timing

   In Floyd's algorithm, fast should move two steps before the cycle check. Here, the check occurs after fast has moved only one step, and the second fast move is done without checking if fast->next is NULL, which can cause a NULL pointer dereference.

3. Final Answer:

   fast pointer moves two steps but not checked for NULL before second move -> Option A

4. Quick Check:

   Second fast move lacks NULL check [OK]

Hint: Fast pointer must complete two steps with NULL checks before cycle check [OK]

Common Mistakes:

• Checking cycle after fast's first step only
• Confusing slow and fast pointer steps
• Not initializing pointers correctly

5. You have a linked list where a cycle exists. After detecting the cycle using Floyd's algorithm, how can you find the starting node of the cycle?

hard

A. Move fast pointer two steps and slow pointer one step until they meet again

B. Count the number of nodes in the list and divide by two

C. Reset slow pointer to head, then move slow and fast one step at a time until they meet

D. Reverse the linked list and check for cycle again

Solution

1. Step 1: Detect cycle and pointers meet inside the loop

   Floyd's algorithm detects cycle when slow and fast pointers meet inside the loop.

2. Step 2: Reset slow to head and move both pointers one step at a time

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

3. Final Answer:

   Reset slow pointer to head, then move slow and fast one step at a time until they meet -> Option C

4. Quick Check:

   Reset slow + move both one step = cycle start [OK]

Hint: Reset slow to head, move both pointers stepwise to find cycle start [OK]

Common Mistakes:

• Moving fast pointer two steps again
• Trying to count nodes to find cycle start
• Reversing list to detect cycle start