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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

let slow = [1];

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

Ahead

Bnull

Ctail

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing slow to null or tail instead of head.

Using undefined which causes runtime errors.

The slow pointer should start at the head of the linked list to find the middle element.

2fill in blank

medium

Complete the code to move the fast pointer two steps ahead in the linked list.

DSA Javascript

fast = fast[1].next;

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

A.next

B.next.prev

C.prev

D.next.next

Attempts:

3 left

💡 Hint

Common Mistakes

Moving fast pointer only one step.

Using .prev which is for doubly linked lists.

Using .next.prev which is invalid.

The fast pointer moves two nodes ahead each time to reach the end faster.

3fill in blank

hard

Fix the error in the while loop condition to correctly check if fast and fast.next exist.

DSA Javascript

while (fast [1] fast.next) {

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

A||

B==

C&&

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using || which can cause runtime errors.

Using == or != which are not logical operators here.

Both fast and fast.next must exist to safely move the fast pointer two steps ahead.

4fill in blank

hard

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

DSA Javascript

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

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

A.next

B.next.next

C.prev

D.next.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Moving slow two steps or fast one step.

Using .prev which is invalid for singly linked lists.

Slow moves one step, fast moves two steps to find the middle node.

5fill in blank

hard

Fill all three blanks to return the value of the middle node after the loop.

DSA Javascript

function findMiddle(head) {
  let slow = head;
  let fast = head;
  while (fast [1] fast.next) {
    slow = slow[2];
    fast = fast[3];
  }
  return slow.value;
}

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

A&&

B.next

C.next.next

D||

Attempts:

3 left

💡 Hint

Common Mistakes

Using || in the while condition causing errors.

Moving slow or fast incorrectly.

Returning fast.value instead of slow.value.

The loop runs while fast and fast.next exist (using &&). Slow moves one step (.next), fast moves two steps (.next.next). Finally, slow.value is returned as the middle element.

Practice

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

easy

A. Reverse the list and pick the last element

B. Use one pointer moving twice as fast as the other; when fast reaches end, slow is at middle

C. Count all nodes first, then traverse again to the middle node

D. Use two pointers moving at the same speed and pick the first pointer's position

Solution

1. Step 1: Understand two-pointer technique

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

2. Step 2: Identify middle when fast pointer reaches end

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

3. Final Answer:

   Use one pointer moving twice as fast as the other; when fast reaches end, slow is at middle -> Option B

4. Quick Check:

   Two-pointer technique = fast & slow pointers [OK]

Hint: Fast pointer moves 2x speed, slow pointer at middle when fast ends [OK]

Common Mistakes:

• Trying to reverse the list unnecessarily
• Counting nodes twice instead of using two pointers
• Moving both pointers at the same speed

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

easy

A. let slow = head.next; let fast = head;

B. let slow = null; let fast = head.next;

C. let slow = head; let fast = head;

D. let slow = head; let fast = null;

Solution

1. Step 1: Initialize both pointers at head

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

2. Step 2: Confirm correct initialization

   Starting both at head allows fast to move ahead and slow to lag behind properly.

3. Final Answer:

   let slow = head; let fast = head; -> Option C

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both slow and fast pointers at head node [OK]

Common Mistakes:

• Starting slow or fast at null or head.next incorrectly
• Swapping slow and fast initialization
• Starting fast pointer at null causing errors

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

class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
  }
}

const head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head.next.next.next.next = new Node(50);

function findMiddle(head) {
  let slow = head;
  let fast = head;
  while (fast !== null && fast.next !== null) {
    slow = slow.next;
    fast = fast.next.next;
  }
  return slow.value;
}

console.log(findMiddle(head));

medium

A. 20

B. 50

C. 40

D. 30

Solution

1. Step 1: Trace slow and fast pointers through the list

   Initially, slow=10, fast=10. After first loop: slow=20, fast=30. Second loop: slow=30, fast=50. Third loop ends as fast.next is null.

2. Step 2: Identify slow pointer value at loop end

   Slow pointer is at node with value 30, which is the middle element.

3. Final Answer:

   30 -> Option D

4. Quick Check:

   Middle element = 30 [OK]

Hint: Fast moves 2 steps, slow moves 1; slow ends at middle [OK]

Common Mistakes:

• Returning fast pointer value instead of slow
• Off-by-one error in loop condition
• Confusing node values during traversal

4. The following code is intended to find the middle element of a linked list, but it has a bug. What is the bug?

function findMiddle(head) {
  let slow = head;
  let fast = head.next;
  while (fast !== null && fast.next !== null) {
    slow = slow.next;
    fast = fast.next.next;
  }
  return slow.value;
}

medium

A. Fast pointer should start at head, not head.next

B. Slow pointer should start at head.next, not head

C. Loop condition should be while (fast !== null)

D. Return statement should return fast.value

Solution

1. Step 1: Analyze pointer initialization

   Fast pointer starts at head.next, which causes slow pointer to point to the first middle in even-length lists incorrectly.

2. Step 2: Correct initialization for accurate middle

   Fast pointer should start at head to ensure slow pointer reaches the true middle node.

3. Final Answer:

   Fast pointer should start at head, not head.next -> Option A

4. Quick Check:

   Fast pointer must start at head [OK]

Hint: Start fast pointer at head to avoid off-by-one errors [OK]

Common Mistakes:

• Starting fast pointer at head.next causing wrong middle
• Changing loop condition incorrectly
• Returning fast pointer value instead of slow

5. Given a singly linked list with an even number of nodes, how can you modify the two-pointer approach to return the first middle element instead of the second middle element?

hard

A. Initialize fast pointer at head.next and slow pointer at head

B. Initialize slow pointer at head.next and fast pointer at head

C. Move slow pointer twice and fast pointer once in the loop

D. Return fast pointer's value instead of slow pointer's value

Solution

1. Step 1: Understand default behavior

   Normally, both pointers start at head, so slow points to second middle in even-length lists.

2. Step 2: Modify initialization to get first middle

   Starting fast pointer at head.next causes slow pointer to move fewer steps, landing on the first middle node.

3. Final Answer:

   Initialize fast pointer at head.next and slow pointer at head -> Option A

4. Quick Check:

   Fast starts at head.next for first middle [OK]

Hint: Start fast at head.next to get first middle in even lists [OK]

Common Mistakes:

• Swapping slow and fast initialization incorrectly
• Changing pointer movement speeds
• Returning fast pointer value without logic