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;

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

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.