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DSA Cprogramming~10 mins

Detect if a Linked List is Circular in DSA C - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

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

DSA C
struct Node* slow = [1];
Drag options to blanks, or click blank then click option'
ANULL
Bhead
Cslow
Dfast
Attempts:
3 left
💡 Hint
Common Mistakes
Initializing slow to NULL will cause no traversal.
Using fast instead of head mixes pointer roles.
2fill in blank
medium

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

DSA C
fast = fast->next->[1];
Drag options to blanks, or click blank then click option'
Anext
Bprev
Chead
Ddata
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'prev' is invalid in singly linked lists.
Accessing 'data' instead of 'next' causes errors.
3fill in blank
hard

Fix the error in the loop condition to check if fast and fast->next are not NULL.

DSA C
while (fast != NULL && fast->[1] != NULL) {
Drag options to blanks, or click blank then click option'
Aprev
Bdata
Cnext
Dhead
Attempts:
3 left
💡 Hint
Common Mistakes
Checking 'prev' causes compilation errors in singly linked lists.
Checking 'data' is meaningless for pointer safety.
4fill in blank
hard

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

DSA C
slow = slow->[1];
fast = fast->[2]->next;
Drag options to blanks, or click blank then click option'
Anext
Bprev
Chead
Ddata
Attempts:
3 left
💡 Hint
Common Mistakes
Using 'prev' in singly linked lists causes errors.
Using 'data' instead of 'next' breaks pointer logic.
5fill in blank
hard

Fill all three blanks to return true if the list is circular, else false.

DSA C
if (slow == [1]) {
    return [2];
} else {
    return [3];
}
Drag options to blanks, or click blank then click option'
Afast
Btrue
Cfalse
Dhead
Attempts:
3 left
💡 Hint
Common Mistakes
Returning false when pointers meet is wrong.
Comparing slow to head instead of fast misses the cycle.

Practice

(1/5)
1. What does it mean if a linked list is circular?
easy
A. The last node points back to one of the previous nodes, forming a loop
B. The linked list has no nodes
C. The linked list nodes are sorted in ascending order
D. The linked list has only one node

Solution

  1. Step 1: Understand linked list structure

    A linked list is a chain of nodes where each node points to the next node.
  2. Step 2: Identify circular linked list property

    If the last node points back to any previous node, it creates a loop, making the list circular.
  3. Final Answer:

    The last node points back to one of the previous nodes, forming a loop -> Option A
  4. Quick Check:

    Circular linked list = last node points back [OK]
Hint: Circular means last node links back inside list [OK]
Common Mistakes:
  • Thinking circular means sorted
  • Confusing empty list with circular
  • Assuming single node list is always circular
2. Which of the following C code snippets correctly checks if a linked list is circular?
easy
A. for (int i = 0; i < 10; i++) { head = head->next; } return 0;
B. if (head == NULL) return 1; else return 0;
C. while (head->next != NULL) { head = head->next; } return 1;
D. while (head != NULL) { head = head->next; if (head == start) return 1; } return 0;

Solution

  1. Step 1: Analyze code logic for circular detection

    The code moves through nodes until it finds the start node again, indicating a loop.
  2. Step 2: Check other options for correctness

    if (head == NULL) return 1; else return 0; only checks if list is empty, not circular. while (head->next != NULL) { head = head->next; } return 1; stops at NULL, no loop check. for (int i = 0; i < 10; i++) { head = head->next; } return 0; blindly moves 10 steps without loop check.
  3. Final Answer:

    while (head != NULL) { head = head->next; if (head == start) return 1; } return 0; -> Option D
  4. Quick Check:

    Loop check by revisiting start node = while (head != NULL) { head = head->next; if (head == start) return 1; } return 0; [OK]
Hint: Check if traversal returns to start node [OK]
Common Mistakes:
  • Only checking for NULL without loop detection
  • Using fixed iteration count instead of loop condition
  • Returning true for empty list
3. What will be the output of the following C code snippet?
typedef struct Node {
  int data;
  struct Node* next;
} Node;

int isCircular(Node* head) {
  if (head == NULL) return 0;
  Node* temp = head->next;
  while (temp != NULL && temp != head) {
    temp = temp->next;
  }
  return (temp == head);
}

int main() {
  Node a, b, c;
  a.data = 1; b.data = 2; c.data = 3;
  a.next = &b; b.next = &c; c.next = &a;
  printf("%d", isCircular(&a));
  return 0;
}
medium
A. Runtime error
B. 0
C. 1
D. Compilation error

Solution

  1. Step 1: Understand linked list setup

    Nodes a, b, c are linked in a circle: a->b->c->a.
  2. Step 2: Trace isCircular function

    Function moves temp pointer until it finds head again or NULL. Here, temp will reach head again, so returns 1.
  3. Final Answer:

    1 -> Option C
  4. Quick Check:

    Loop detected, returns 1 [OK]
Hint: Returns 1 if traversal loops back to head [OK]
Common Mistakes:
  • Assuming NULL termination means circular
  • Confusing return values 0 and 1
  • Not initializing temp correctly
4. Identify the bug in the following code that tries to detect if a linked list is circular:
int isCircular(Node* head) {
  Node* slow = head;
  Node* fast = head->next;
  while (fast != NULL && fast->next != NULL) {
    if (slow == fast) return 1;
    slow = slow->next;
    fast = fast->next->next;
  }
  return 0;
}
medium
A. slow pointer should move two steps, fast one step
B. fast pointer initialization should be head, not head->next
C. The loop condition should check only fast != NULL
D. Return values are reversed

Solution

  1. Step 1: Understand Floyd's cycle detection setup

    Both slow and fast pointers should start at head to detect cycle properly.
  2. Step 2: Identify initialization error

    fast is initialized to head->next, which can miss cycle detection if list has only one node or cycle starts at head.
  3. Final Answer:

    fast pointer initialization should be head, not head->next -> Option B
  4. Quick Check:

    Initialize fast = head for correct cycle detection [OK]
Hint: Start both pointers at head for Floyd's cycle detection [OK]
Common Mistakes:
  • Starting fast pointer one step ahead
  • Swapping slow and fast pointer steps
  • Ignoring null checks in loop condition
5. You have a singly linked list where the last node's next pointer is unknown. How would you modify the list to make it circular and then detect the cycle using Floyd's algorithm? Choose the correct sequence of steps.
hard
A. 1. Traverse to last node 2. Set last node's next to head 3. Use Floyd's algorithm to detect cycle
B. 1. Set head's next to NULL 2. Use Floyd's algorithm 3. Traverse list
C. 1. Reverse the list 2. Set last node's next to NULL 3. Use Floyd's algorithm
D. 1. Traverse to middle node 2. Set middle node's next to head 3. Use Floyd's algorithm

Solution

  1. Step 1: Make the list circular by linking last node to head

    Traverse the list to find the last node, then set its next pointer to head to form a loop.
  2. Step 2: Apply Floyd's cycle detection algorithm

    Use slow and fast pointers starting at head to detect the cycle created.
  3. Final Answer:

    1. Traverse to last node 2. Set last node's next to head 3. Use Floyd's algorithm to detect cycle -> Option A
  4. Quick Check:

    Make circular then detect cycle with Floyd's algorithm [OK]
Hint: Link last node to head before running cycle detection [OK]
Common Mistakes:
  • Setting wrong node's next pointer
  • Not traversing to last node correctly
  • Using Floyd's algorithm before making list circular
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