Find Middle Element of Linked List in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to traverse the linked list.

DSA C

struct Node* [1] = head;

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

Atemp

Bmiddle

Ccurrent

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using a pointer name that is not declared or inconsistent.

Not initializing the pointer to head.

We use a temporary pointer named temp to traverse the linked list from the head.

2fill in blank

medium

Complete the code to move the slow pointer one step forward in the linked list.

DSA C

slow = slow[1];

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A->next->next

B->next

C->prev

D->data

Attempts:

3 left

💡 Hint

Common Mistakes

Moving slow pointer two steps instead of one.

Using incorrect pointer member like prev or data.

To move the slow pointer one step forward, we use slow = slow->next;.

3fill in blank

hard

Fix the error in the while loop condition to correctly 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'

A->prev

B->next->next

C->next

D->data

Attempts:

3 left

💡 Hint

Common Mistakes

Checking fast->prev which may not exist in singly linked list.

Checking fast->data which is not a pointer.

The condition should check fast->next != NULL to avoid accessing NULL pointer.

4fill in blank

hard

Fill both blanks to move fast pointer two steps and slow pointer one step inside the loop.

DSA C

fast = fast[1];
slow = slow[2];

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

A->next->next

B->next

C->prev

D->data

Attempts:

3 left

💡 Hint

Common Mistakes

Moving fast pointer only one step.

Moving slow pointer two steps.

Using incorrect pointer members.

Inside the loop, fast moves two steps using fast->next->next, and slow moves one step using slow->next.

5fill in blank

hard

Fill all three blanks to print the middle element's data after finding it.

DSA C

printf("Middle element is %[1]\n", slow[2]); return [3];

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

Ad

B->data

C0

Ds

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong format specifier like %s.

Not accessing the data field correctly.

Returning wrong value.

Use %d to print an integer, access the data with slow->data, and return 0 to indicate success.

Practice

1. What is the main idea behind finding the middle element of a singly linked list using the two-pointer technique?

easy

A. Count all nodes first, then access the middle by index

B. Reverse the list and pick the first element

C. Use two pointers both moving one step at a time

D. Use one slow pointer moving one step and one fast pointer moving two steps until fast reaches the end

Solution

1. Step 1: Understand two-pointer technique

   One pointer moves one node at a time (slow), the other moves two nodes at a time (fast).

2. Step 2: Identify middle when fast reaches end

   When fast pointer reaches the end, slow pointer will be at the middle node.

3. Final Answer:

   Use one slow pointer moving one step and one fast pointer moving two steps until fast reaches the end -> Option D

4. Quick Check:

   Two-pointer technique = slow one step, fast two steps [OK]

Hint: Move fast pointer twice as fast as slow pointer [OK]

Common Mistakes:

• Moving both pointers at the same speed
• Counting nodes first instead of using pointers
• Reversing the list unnecessarily

2. Which of the following C code snippets correctly initializes pointers to find the middle of a linked list named head?

easy

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

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

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

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

Solution

1. Step 1: Initialize both pointers at head

   Both slow and fast pointers start at the head node to traverse the list properly.

2. Step 2: Check pointer initialization correctness

   Only Node *slow = head; Node *fast = head; initializes both pointers correctly at head; others start one pointer at NULL or next node incorrectly.

3. Final Answer:

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

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head node [OK]

Common Mistakes:

• Starting one pointer at NULL
• Starting pointers at next node instead of head
• Not initializing pointers before loop

3. What will be the output of the following C code snippet that finds the middle element of a linked list?

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

int main() {
  Node n3 = {30, NULL};
  Node n2 = {20, &n3};
  Node n1 = {10, &n2};
  Node *head = &n1;

  Node *slow = head, *fast = head;
  while (fast != NULL && fast->next != NULL) {
    slow = slow->next;
    fast = fast->next->next;
  }
  printf("%d", slow->data);
  return 0;
}

medium

A. 30

B. 10

C. 20

D. Runtime error

Solution

1. Step 1: Trace pointer movements

   Initially, slow and fast point to 10. After first loop, slow moves to 20, fast moves to 30. Next loop fast->next is NULL, loop ends.

2. Step 2: Identify slow pointer position

   Slow pointer is at node with data 20, which is the middle element.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Middle element = 20 [OK]

Hint: Slow pointer stops at middle when fast reaches end [OK]

Common Mistakes:

• Assuming middle is first node
• Confusing fast pointer position with slow
• Ignoring loop exit condition

4. Identify the error in the following code snippet intended to find the middle element of a linked list and select the correct fix.

Node *slow = head;
Node *fast = head;
while (fast != NULL) {
  slow = slow->next;
  fast = fast->next->next;
}
return slow->data;

medium

A. Change loop condition to while (fast != NULL && fast->next != NULL)

B. Initialize slow to head->next instead of head

C. Move slow pointer two steps instead of one

D. No error, code is correct

Solution

1. Step 1: Analyze loop condition

   The loop checks only fast != NULL, but fast->next could be NULL causing fast->next->next to crash.

2. Step 2: Correct loop condition

   Adding fast->next != NULL prevents dereferencing NULL pointer, avoiding runtime error.

3. Final Answer:

   Change loop condition to while (fast != NULL && fast->next != NULL) -> Option A

4. Quick Check:

   Check both fast and fast->next not NULL [OK]

Hint: Check both fast and fast->next in loop condition [OK]

Common Mistakes:

• Not checking fast->next before accessing
• Moving slow pointer incorrectly
• Assuming code runs without null checks

5. Given a linked list with an even number of nodes, which node will the two-pointer method return as the middle element? Consider the list: 5 -> 10 -> 15 -> 20 -> NULL

hard

A. 15 (the second middle node)

B. 10 (the first middle node)

C. 5 (the head node)

D. 20 (the last node)

Solution

1. Step 1: Trace pointers on even-length list

   Start slow and fast at 5. After first iteration, slow=10, fast=15. Next iteration, slow=15, fast=NULL (end).

2. Step 2: Identify middle node returned

   When fast reaches NULL, slow points to 15, which is the second middle node in even-length list.

3. Final Answer:

   15 (the second middle node) -> Option A

4. Quick Check:

   Two-pointer method returns second middle in even list [OK]

Hint: Two-pointer method returns second middle in even-length list [OK]

Common Mistakes:

• Assuming first middle is returned
• Confusing head or last node as middle
• Not tracing pointer steps carefully