Find Middle Element of Linked List in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the middle element output

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

DSA Javascript

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

class LinkedList {
  constructor() {
    this.head = null;
  }

  append(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  findMiddle() {
    let slow = this.head;
    let fast = this.head;
    while (fast && fast.next) {
      slow = slow.next;
      fast = fast.next.next;
    }
    return slow ? slow.value : null;
  }
}

const list = new LinkedList();
[10, 20, 30, 40, 50].forEach(num => list.append(num));
console.log(list.findMiddle());

A 50

B 20

C 30

D 40

💡 Hint

Use two pointers: one moves twice as fast as the other.

The slow pointer moves one step at a time, the fast pointer moves two steps. When fast reaches the end, slow is at the middle.

❓ Predict Output

intermediate

Middle element with even number of nodes

What will be the output of the code when the linked list has an even number of nodes?

DSA Javascript

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

class LinkedList {
  constructor() {
    this.head = null;
  }

  append(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  findMiddle() {
    let slow = this.head;
    let fast = this.head;
    while (fast && fast.next) {
      slow = slow.next;
      fast = fast.next.next;
    }
    return slow ? slow.value : null;
  }
}

const list = new LinkedList();
[5, 10, 15, 20, 25, 30].forEach(num => list.append(num));
console.log(list.findMiddle());

A 20

B 15

C 25

D 10

💡 Hint

For even length, the middle is the second of the two middle nodes.

When the list has even nodes, the slow pointer ends at the second middle node.

🔧 Debug

advanced

Identify the error in middle element code

What error will this code produce when trying to find the middle element of a linked list?

DSA Javascript

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

class LinkedList {
  constructor() {
    this.head = null;
  }

  append(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

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

const list = new LinkedList();
[1].forEach(num => list.append(num));
console.log(list.findMiddle());

A TypeError: Cannot read property 'next' of null

B 1

C undefined

D ReferenceError: slow is not defined

💡 Hint

Check the while loop condition when list has only one node.

The condition fast.next.next fails when fast.next is null, causing a TypeError.

🧠 Conceptual

advanced

Why use two pointers to find middle?

Why does using two pointers (slow and fast) help find the middle element of a linked list efficiently?

A Because fast pointer counts nodes and slow pointer stores values

B Because slow pointer skips nodes to reach middle faster

C Because slow pointer moves randomly and fast pointer moves sequentially

D Because fast pointer moves twice as fast, slow pointer reaches middle when fast reaches end

💡 Hint

Think about relative speeds of pointers.

The fast pointer moves two steps at a time, so when it reaches the end, the slow pointer is at the middle.

❓ Predict Output

expert

Output after multiple middle element calls

What will be the output after these operations on the linked list?

DSA Javascript

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

class LinkedList {
  constructor() {
    this.head = null;
  }

  append(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = newNode;
      return;
    }
    let current = this.head;
    while (current.next) {
      current = current.next;
    }
    current.next = newNode;
  }

  findMiddle() {
    let slow = this.head;
    let fast = this.head;
    while (fast && fast.next) {
      slow = slow.next;
      fast = fast.next.next;
    }
    return slow ? slow.value : null;
  }
}

const list = new LinkedList();
[1, 2, 3, 4, 5].forEach(num => list.append(num));
const firstMiddle = list.findMiddle();
list.append(6);
const secondMiddle = list.findMiddle();
console.log([firstMiddle, secondMiddle]);

A [3, 3]

B [3, 4]

C [2, 3]

D [4, 5]

💡 Hint

Check middle for odd and then even length list.

Initially, list has 5 elements, middle is 3. After adding 6th element, middle is 4.

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