Find Middle Element of Linked List in DSA Typescript - 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 Typescript

let slow = [1];

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

Aundefined

Bnull

Ctail

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing slow pointer to null or tail instead of head.

The slow pointer should start at the head of the linked list to begin traversal.

2fill in blank

medium

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

DSA Typescript

fast = fast[1]next[2]next;

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

A?.

B.

C->

D=>

Attempts:

3 left

💡 Hint

Common Mistakes

Using dot operator without checking for null causing runtime errors.

Using optional chaining (?.) safely moves the fast pointer two steps ahead without errors if next is null.

3fill in blank

hard

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

DSA Typescript

while (fast [1] null && fast.next [2] null) {

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

A!==

B==

C=

D!

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment (=) instead of comparison (!==).

Using equality (==) which is less strict.

The condition should check that fast and fast.next are not null to continue the loop.

4fill in blank

hard

Fill both blanks to move slow one step and fast two steps safely.

DSA Typescript

slow = slow[1]next;
fast = fast[2]next[3]next;

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

A?.

B.

C->

D=>

Attempts:

3 left

💡 Hint

Common Mistakes

Using optional chaining for slow pointer unnecessarily.

Using dot operator for fast pointer causing errors.

Slow moves one step with dot operator, fast moves two steps with optional chaining to avoid errors.

5fill in blank

hard

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

DSA Typescript

function findMiddle(head: ListNode | null): number | null {
  let slow = head;
  let fast = head;
  while (fast [1] null && fast.next [2] null) {
    slow = slow[3]next;
    fast = fast.next.next;
  }
  return slow ? slow.val : null;
}

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

A!==

B==

C===

D?.

Attempts:

3 left

💡 Hint

Common Mistakes

Using loose equality (==) instead of strict (!==).

Using optional chaining for slow pointer unnecessarily.

The while loop checks that fast and fast.next are not null using !==, and slow moves one step safely with dot operator.

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