Reorder Linked List in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find the middle node of the linked list using slow and fast pointers.

DSA Javascript

let slow = head;
let fast = head;
while (fast && fast.next) {
  slow = slow.[1];
  fast = fast.next.next;
}
return slow;

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

Anext

Bprev

Cval

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' causes the pointer to move backward.

Using 'val' or 'data' accesses the node value instead of the next pointer.

To move the slow pointer one step forward, move to slow.next.

2fill in blank

medium

Complete the code to reverse the second half of the linked list starting from the middle node.

DSA Javascript

let prev = null;
let curr = middle;
while (curr) {
  let nextTemp = curr.next;
  curr.next = [1];
  prev = curr;
  curr = nextTemp;
}
return prev;

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

AnextTemp

Bcurr

Cprev

Dmiddle

Attempts:

3 left

💡 Hint

Common Mistakes

Setting curr.next to curr or nextTemp breaks the reversal.

Using middle instead of prev causes incorrect links.

To reverse the list, each node's next pointer should point to the previous node.

3fill in blank

hard

Fix the error in the code that merges two halves of the linked list alternately.

DSA Javascript

let first = head;
let second = reversed;
while (second) {
  let temp1 = first.next;
  let temp2 = second.next;
  first.next = second;
  second.next = [1];
  first = temp1;
  second = temp2;
}

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

Asecond

Btemp2

Cfirst

Dtemp1

Attempts:

3 left

💡 Hint

Common Mistakes

Using temp2 or second causes cycles or wrong links.

Using first instead of temp1 breaks the chain.

After linking second node, its next should point to the next node of first half (temp1).

4fill in blank

hard

Fill both blanks to split the linked list into two halves by breaking the link at the middle.

DSA Javascript

let slow = head;
let fast = head;
while (fast && fast.next) {
  slow = slow.next;
  fast = fast.next.next;
}
let secondHalf = slow.next;
slow.[1] = null;
return [2];

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

Anext

Bprev

CsecondHalf

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Setting slow.prev instead of slow.next causes errors.

Returning head instead of secondHalf returns the first half.

Setting slow.next to null breaks the list into two halves. We return the second half starting at secondHalf.

5fill in blank

hard

Fill all three blanks to reorder the linked list by merging first and reversed second halves alternately.

DSA Javascript

let first = head;
let second = reversed;
while (second) {
  let temp1 = first.next;
  let temp2 = second.next;
  first.next = [1];
  second.next = [2];
  first = [3];
  second = temp2;
}

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

Asecond

Btemp1

Dfirst

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping temp1 and second causes wrong links.

Moving first to first instead of temp1 causes infinite loop.

Link first.next to second, second.next to temp1, then move first to temp1 to continue merging.

Practice

1. What is the main goal of the Reorder Linked List problem?

easy

A. Rearrange the list so nodes alternate from start and end towards the center

B. Sort the linked list in ascending order

C. Reverse the entire linked list

D. Remove duplicate nodes from the linked list

Solution

1. Step 1: Understand the problem goal

   The problem asks to reorder nodes so the first node is followed by the last node, then second node, then second last, and so on.

2. Step 2: Differentiate from other operations

   This is not sorting or reversing the whole list, but alternating nodes from front and back.

3. Final Answer:

   Rearrange the list so nodes alternate from start and end towards the center -> Option A

4. Quick Check:

   Reorder Linked List = Alternate front and back nodes [OK]

Hint: Think of weaving nodes from front and back alternately [OK]

Common Mistakes:

• Confusing reorder with sorting
• Trying to reverse the entire list only
• Ignoring the alternating pattern

2. Which of the following code snippets correctly finds the middle node of a singly linked list in JavaScript?

easy

A. let slow = head; let fast = head.next; while(fast) { slow = slow.next; fast = fast.next; }

B. let slow = head; while(slow.next) { slow = slow.next.next; }

C. let fast = head; while(fast.next) { fast = fast.next.next; }

D. let slow = head, fast = head; while(fast && fast.next) { slow = slow.next; fast = fast.next.next; }

Solution

1. Step 1: Use two pointers to find middle

   Initialize two pointers: slow moves one step, fast moves two steps until fast reaches end.

2. Step 2: Check each option

   let slow = head, fast = head; while(fast && fast.next) { slow = slow.next; fast = fast.next.next; } correctly moves slow by one and fast by two steps in the loop condition.

3. Final Answer:

   let slow = head, fast = head; while(fast && fast.next) { slow = slow.next; fast = fast.next.next; } -> Option D

4. Quick Check:

   Fast moves twice, slow moves once to find middle [OK]

Hint: Use slow and fast pointers; fast moves twice as fast [OK]

Common Mistakes:

• Moving slow pointer incorrectly
• Not checking fast and fast.next in loop condition
• Starting fast pointer incorrectly

3. Given the linked list 1 -> 2 -> 3 -> 4 -> null, what will be the reordered list after applying the reorder algorithm?

medium

A. 1 -> 2 -> 3 -> 4 -> null

B. 4 -> 3 -> 2 -> 1 -> null

C. 1 -> 4 -> 2 -> 3 -> null

D. 1 -> 3 -> 2 -> 4 -> null

Solution

1. Step 1: Identify the reorder pattern

   The reorder pattern is first node, last node, second node, second last node, and so on.

2. Step 2: Apply pattern to list 1->2->3->4

   Start with 1, then 4, then 2, then 3, resulting in 1 -> 4 -> 2 -> 3 -> null.

3. Final Answer:

   1 -> 4 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Alternate front and back nodes = 1->4->2->3 [OK]

Hint: Alternate nodes from front and back to reorder [OK]

Common Mistakes:

• Reversing the entire list instead
• Not alternating nodes correctly
• Leaving list unchanged

4. Consider this buggy code snippet for reordering a linked list:

function reorderList(head) {
  if (!head) return;
  let slow = head, fast = head;
  while (fast.next && fast.next.next) {
    slow = slow.next;
    fast = fast.next.next;
  }
  let second = slow.next;
  slow.next = null;
  second = reverseList(second);
  let first = head;
  while (second) {
    let temp1 = first.next;
    let temp2 = second.next;
    first.next = second;
    second.next = temp1;
    first = temp1;
    second = temp2;
  }
}

function reverseList(head) {
  let prev = null;
  while (head) {
    let next = head.next;
    head.next = prev;
    prev = head;
    head = next;
  }
  return prev;
}

What is the bug in this code?

medium

A. The merging loop does not handle the case when first becomes null

B. The loop condition for finding middle is incorrect and may skip nodes

C. The reverseList function does not return the reversed list

D. The slow pointer is not moved correctly to split the list

Solution

1. Step 1: Analyze the merging loop

   The merging loop assumes first is never null, but if first becomes null before second, it causes errors.

2. Step 2: Identify missing null check

   There is no check if first is null before accessing first.next, which can cause runtime errors.

3. Final Answer:

   The merging loop does not handle the case when first becomes null -> Option A

4. Quick Check:

   Merging loop must check first != null to avoid errors [OK]

Hint: Check null before accessing next pointers in merge loop [OK]

Common Mistakes:

• Ignoring null checks in merging loop
• Thinking reverseList is faulty
• Misreading middle finding loop condition

5. You have a linked list with nodes: 1 -> 2 -> 3 -> 4 -> 5 -> null. After reordering, what will be the final linked list?

hard

A. 5 -> 4 -> 3 -> 2 -> 1 -> null

B. 1 -> 5 -> 2 -> 4 -> 3 -> null

C. 1 -> 3 -> 5 -> 2 -> 4 -> null

D. 1 -> 2 -> 3 -> 4 -> 5 -> null

Solution

1. Step 1: Understand reorder pattern for odd length list

   For odd length, reorder as first node, last node, second node, second last node, then middle node.

2. Step 2: Apply to list 1->2->3->4->5

   Sequence: 1 (first), 5 (last), 2 (second), 4 (second last), 3 (middle).

3. Final Answer:

   1 -> 5 -> 2 -> 4 -> 3 -> null -> Option B

4. Quick Check:

   Alternate front and back nodes with middle last [OK]

Hint: Alternate front and back nodes, middle stays last if odd length [OK]

Common Mistakes:

• Reversing entire list
• Not handling odd length middle node correctly
• Leaving list unchanged