Find Middle Element of Linked List in DSA Javascript - Build from Scratch

Find Middle Element of Linked List

📖 Scenario: Imagine you have a line of people waiting to buy tickets. You want to find the person standing exactly in the middle of the line.

🎯 Goal: You will create a simple linked list and write code to find the middle element of this list.

📋 What You'll Learn

Create a linked list with exactly 5 nodes containing values 10, 20, 30, 40, 50

Create a variable called head pointing to the first node

Use two pointers called slow and fast to find the middle node

Print the value of the middle node

💡 Why This Matters

🌍 Real World

Finding the middle element is useful in many situations like splitting data, balancing tasks, or finding median values.

💼 Career

Understanding linked lists and pointer techniques is important for software developers working with data structures and algorithms.

1

Create the linked list nodes

Create 5 linked list nodes named node1, node2, node3, node4, and node5 with values 10, 20, 30, 40, and 50 respectively. Link them so that node1.next is node2, node2.next is node3, and so on. Finally, create a variable called head that points to node1.

DSA Javascript

# Your code here

Hint

Think of each node as a person holding a number and pointing to the next person in line.

2

Set up pointers to find the middle

Create two variables called slow and fast and set both to point to head.

DSA Javascript

const node1 = { value: 10, next: null };
const node2 = { value: 20, next: null };
const node3 = { value: 30, next: null };
const node4 = { value: 40, next: null };
const node5 = { value: 50, next: null };

node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;

const head = node1;

// Create slow and fast pointers pointing to head
// Your code here

Hint

Both pointers start at the beginning of the list.

3

Move pointers to find the middle node

Use a while loop that continues as long as fast and fast.next are not null. Inside the loop, move slow one step forward (slow = slow.next) and fast two steps forward (fast = fast.next.next).

DSA Javascript

const node1 = { value: 10, next: null };
const node2 = { value: 20, next: null };
const node3 = { value: 30, next: null };
const node4 = { value: 40, next: null };
const node5 = { value: 50, next: null };

node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;

const head = node1;

let slow = head;
let fast = head;

// Use a while loop to move slow and fast pointers
// Your code here

Hint

Fast pointer moves twice as fast as slow pointer. When fast reaches the end, slow is in the middle.

4

Print the middle element value

Write a console.log statement to print the value of the slow pointer, which is the middle node's value.

DSA Javascript

const node1 = { value: 10, next: null };
const node2 = { value: 20, next: null };
const node3 = { value: 30, next: null };
const node4 = { value: 40, next: null };
const node5 = { value: 50, next: null };

node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;

const head = node1;

let slow = head;
let fast = head;

while (fast !== null && fast.next !== null) {
    slow = slow.next;
    fast = fast.next.next;
}

// Print the value of the middle node
// Your code here

Hint

The middle node value is 30 because it is the third node in a list of five.

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