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

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the middle element of a linked list

What is the output of the following TypeScript code that finds the middle element of a linked list?

DSA Typescript

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 -> 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);

console.log(findMiddle(head));

A 30

B 20

C 50

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

Find middle element in even-length linked list

What is the output of this code that finds the middle element in an even-length linked list?

DSA Typescript

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: 1 -> 2 -> 3 -> 4 -> null
const head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);

console.log(findMiddle(head));

A 2

B 3

C 1

D 4

💡 Hint

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

The slow pointer ends at the third node (value 3) when fast pointer reaches the end.

🔧 Debug

advanced

Identify the error in finding middle element

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

DSA Typescript

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.value;
}

const head = null;

console.log(findMiddle(head));

A Output: 15

B SyntaxError: Unexpected token '.'

C Output: 10

D TypeError: Cannot read property 'value' of null

💡 Hint

Check if slow can be null before accessing value.

If the list is empty or slow becomes null, accessing slow.value causes a TypeError.

🧠 Conceptual

advanced

Why use two pointers to find middle element?

Why is the two-pointer technique (slow and fast pointers) efficient for finding the middle element of a linked list?

A It uses recursion to find the middle element faster.

B It sorts the linked list first to find the middle element.

C It traverses the list only once, using O(n) time and O(1) space.

D It creates a new list with half the elements to find the middle.

💡 Hint

Think about how many times the list is traversed and memory used.

The two-pointer method moves through the list once, with one pointer moving twice as fast, so it finds the middle in one pass using constant extra space.

🚀 Application

expert

Find middle element after modifying linked list

Given this linked list and operations, what is the middle element after all operations?

DSA Typescript

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: 1 -> 3 -> 5 -> 7 -> 9 -> null
const head = new Node(1);
head.next = new Node(3);
head.next.next = new Node(5);
head.next.next.next = new Node(7);
head.next.next.next.next = new Node(9);

// Insert 6 after 5
let current = head;
while (current !== null && current.value !== 5) {
  current = current.next;
}
if (current !== null) {
  const newNode = new Node(6);
  newNode.next = current.next;
  current.next = newNode;
}

// Remove node with value 3
let prev = null;
current = head;
while (current !== null && current.value !== 3) {
  prev = current;
  current = current.next;
}
if (prev !== null && current !== null) {
  prev.next = current.next;
}

console.log(findMiddle(head));

A 6

B 5

C 9

D 7

💡 Hint

Update the list carefully after insert and delete, then find the middle.

After inserting 6 after 5 and removing 3, the list is 1 -> 5 -> 6 -> 7 -> 9. The middle is the third element, 6.

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