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

Find Middle Element of Linked List

📖 Scenario: Imagine you have a line of people waiting for a bus. You want to find the person standing exactly in the middle of the line.In programming, we can represent this line as a linked list, where each person is a node pointing to the next person.

🎯 Goal: You will create a linked list, then write code to find and print the middle element's value.

📋 What You'll Learn

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

Create a variable to help find the middle element

Write the logic to find the middle element using the two-pointer technique

Print the value of the middle node

💡 Why This Matters

🌍 Real World

Finding the middle element is useful in many real-world problems like splitting data, balancing tasks, or optimizing searches.

💼 Career

Understanding linked lists and pointer techniques is important for software engineering roles, especially in system design and algorithm optimization.

1

Create a Linked List with 5 Nodes

Create a class called ListNode with a constructor that takes a value of type number and a next pointer initialized to null. Then create 5 nodes with values 10, 20, 30, 40, and 50 linked together in a variable called head.

DSA Typescript

// Define ListNode class and create linked list nodes
// Your code here

Hint

Start by defining the ListNode class with value and next. Then link nodes one by one.

2

Create Two Pointer Variables

Create two variables called slow and fast and set both to head. These will help us find the middle element.

DSA Typescript

class ListNode {
    value: number;
    next: ListNode | null;
    constructor(value: number) {
        this.value = value;
        this.next = null;
    }
}

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

// Create slow and fast pointers
// Your code here

Hint

Both slow and fast start at the head of the list.

3

Find the Middle Element Using Two Pointers

Use a while loop with the condition fast !== null && fast.next !== null. Inside the loop, move slow one step forward (slow = slow.next) and fast two steps forward (fast = fast.next.next).

DSA Typescript

while (fast !== null && fast.next !== null) {
    slow = slow!.next;
    fast = fast.next.next;
// Print the middle element value

Hint

Move slow by one and fast by two steps until fast reaches the end.

4

Print the Middle Element's Value

Write console.log(slow!.value) to print the value of the middle node.

DSA Typescript

Hint

The middle node value is printed after the loop ends.

Practice

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

easy

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

B. Count all nodes first, then traverse again to the middle index.

C. Reverse the linked list and pick the middle node from the reversed list.

D. Use a stack to store all nodes and pop half of them to find the middle.

Solution

1. Step 1: Understand the two-pointer technique

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

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

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

3. Final Answer:

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

4. Quick Check:

   Two-pointer technique = slow at middle when fast ends [OK]

Hint: Move one pointer twice as fast to find middle easily [OK]

Common Mistakes:

• Trying to count nodes first wastes extra time
• Reversing list is unnecessary and inefficient
• Using extra space like stack is not optimal

2. Which of the following TypeScript code snippets correctly initializes a singly linked list node?

easy

A. class Node { value: number; next: any; constructor(value: number) { this.value = value; this.next = undefined; } }

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

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

D. class Node { value: number; next: Node | null; constructor(value: number) { this.value = value; this.next = null; } }

Solution

1. Step 1: Check the type of the next pointer

   It should be either another Node or null if it is the last node.

2. Step 2: Verify constructor initialization

   The constructor sets value and initializes next to null correctly.

3. Final Answer:

   class Node { value: number; next: Node | null; constructor(value: number) { this.value = value; this.next = null; } } -> Option D

4. Quick Check:

   Next pointer type includes null and initialized to null [OK]

Hint: Next pointer must allow null for list end [OK]

Common Mistakes:

• Not allowing next to be null causes type errors
• Initializing next to undefined instead of null
• Using any type loses type safety

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

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

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

// Create linked list: 10 -> 20 -> 30 -> 40 -> 50
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);

console.log(findMiddle(head));

medium

A. 20

B. 30

C. 40

D. 50

Solution

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

   Initially both at 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.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   Slow pointer stops at middle node value 30 [OK]

Hint: Fast pointer moves twice, slow pointer lands on middle [OK]

Common Mistakes:

• Returning fast pointer value instead of slow
• Not checking fast.next in while condition
• Off-by-one errors in pointer movement

4. Identify the error in the following TypeScript function that attempts to find the middle element of a linked list:

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

medium

A. The fast pointer should move one step at a time, not two.

B. The while loop condition should be while (fast !== null) only.

C. The function does not check if slow is null before accessing slow.value.

D. The function should return fast.value instead of slow.value.

Solution

1. Step 1: Analyze pointer null safety before accessing value

   Slow pointer can be null if the list is empty, so accessing slow.value without check can cause runtime error.

2. Step 2: Identify missing null check before return

   The function should check if slow is not null before returning slow.value to avoid errors.

3. Final Answer:

   The function does not check if slow is null before accessing slow.value. -> Option C

4. Quick Check:

   Always check pointer is not null before accessing properties [OK]

Hint: Check for null before accessing node properties [OK]

Common Mistakes:

• Ignoring null checks causes runtime errors
• Changing fast pointer movement breaks logic
• Returning fast.value misses middle element

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 to head.next instead of head and then proceed normally.

B. Move slow pointer two steps and fast pointer one step in each iteration.

C. Use a stack to store nodes and pop half to get the second middle.

D. Traverse the list twice: first to count nodes, second to middle+1 node.

Solution

1. Step 1: Understand difference between first and second middle in even-length list

   For even nodes, two middle nodes exist; the second middle is the one after the first middle.

2. Step 2: Modify fast pointer start to head.next

   Starting fast pointer at head.next shifts slow pointer to first middle when fast reaches end.

3. Final Answer:

   Initialize fast pointer to head.next instead of head and then proceed normally. -> Option A

4. Quick Check:

   Fast starts at second node to get first middle [OK]

Hint: Start fast pointer one step ahead for first middle [OK]

Common Mistakes:

• Swapping slow and fast pointer steps breaks logic
• Using stack wastes extra space
• Counting nodes twice is inefficient