Reorder Linked List in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

To rearrange a singly linked list so that nodes are ordered as: first node, last node, second node, second last node, and so on.

intermediate

Which three main steps are used to solve the Reorder Linked List problem efficiently?

1. Find the middle of the list.
2. Reverse the second half.
3. Merge the two halves alternating nodes.

beginner

How do you find the middle node of a singly linked list?

Use two pointers: a slow pointer moves one step at a time, and a fast pointer moves two steps. When fast reaches the end, slow is at the middle.

intermediate

What is the purpose of reversing the second half of the linked list in the reorder process?

Reversing the second half allows easy merging from the end towards the middle, enabling the alternating order required.

intermediate

What is the time complexity of the reorder linked list algorithm using the three-step approach?

O(n), where n is the number of nodes, because each step (finding middle, reversing, merging) takes linear time.

What does the fast pointer do when finding the middle of a linked list?

A Moves three steps at a time

B Moves one step at a time

C Moves two steps at a time

D Does not move

The fast pointer moves two steps at a time to reach the end faster than the slow pointer.

Why do we reverse the second half of the linked list in the reorder process?

A To sort the list

B To find the middle node

C To delete nodes

D To prepare for merging nodes from the end

Reversing the second half helps merge nodes from the end in alternating order.

After finding the middle, what is the next step in the reorder linked list algorithm?

A Reverse the second half

B Merge the two halves

C Reverse the first half

D Delete the middle node

The second half is reversed before merging.

What is the final step in the reorder linked list algorithm?

A Merge the two halves alternating nodes

B Reverse the entire list

C Find the middle node

D Delete the last node

Merging the two halves alternating nodes creates the reordered list.

What is the space complexity of the reorder linked list algorithm?

A O(log n)

B O(1)

C O(n)

D O(n^2)

The algorithm uses constant extra space by rearranging pointers in place.

Explain the three main steps to reorder a singly linked list.

Think about splitting, reversing, and merging.

Describe why reversing the second half of the list is necessary in the reorder linked list problem.

Consider how to access nodes from the end easily.

Practice

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

easy

A. To rearrange the list so nodes alternate from start and end

B. To sort the linked list in ascending order

C. To reverse the entire linked list

D. To remove duplicate nodes from the list

Solution

1. Step 1: Understand the reorder pattern

   The reorder linked list operation rearranges nodes so that the first node is followed by the last node, then the second node, then the second last node, and so on.

2. Step 2: Differentiate from other operations

   This is different from sorting, reversing, or removing duplicates because it mixes nodes from start and end alternatively.

3. Final Answer:

   To rearrange the list so nodes alternate from start and end -> Option A

4. Quick Check:

   Reorder = alternate start and end nodes [OK]

Hint: Reorder means alternate nodes from front and back [OK]

Common Mistakes:

• Confusing reorder with sorting
• Thinking reorder means reversing the list
• Assuming reorder removes duplicates

2. Which of the following C code snippets correctly finds the middle node of a singly linked list for reordering?

easy

A. while (fast && fast->next) { fast = fast->next; slow = slow->next; }

B. while (slow && slow->next) { fast = fast->next; slow = slow->next->next; }

C. while (fast && fast->next) { slow = slow->next; fast = fast->next->next; }

D. while (fast && slow) { slow = slow->next->next; fast = fast->next->next; }

Solution

1. Step 1: Identify the fast and slow pointer technique

   To find the middle node, fast pointer moves two steps, slow pointer moves one step until fast reaches end.

2. Step 2: Check each option for correct pointer movement

   while (fast && fast->next) { slow = slow->next; fast = fast->next->next; } correctly moves fast by two and slow by one. Others have incorrect pointer movements causing errors or wrong middle.

3. Final Answer:

   while (fast && fast->next) { slow = slow->next; fast = fast->next->next; } -> Option C

4. Quick Check:

   Fast moves 2 steps, slow moves 1 step [OK]

Hint: Fast pointer moves twice as fast as slow pointer [OK]

Common Mistakes:

• Swapping fast and slow pointer movements
• Not checking fast->next before fast->next->next
• Using slow pointer to move two steps

3. Given the linked list: 1 -> 2 -> 3 -> 4 -> 5 -> NULL, what is the reordered list after applying the reorder operation?

medium

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

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

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

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

Solution

1. Step 1: Identify the reorder pattern on the list

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

2. Step 2: Apply pattern to given list

   Start with 1, then 5, then 2, then 4, then 3. So reordered list is 1 -> 5 -> 2 -> 4 -> 3 -> NULL.

3. Final Answer:

   1 -> 5 -> 2 -> 4 -> 3 -> NULL -> Option D

4. Quick Check:

   Alternate front and back nodes [OK]

Hint: Alternate nodes from start and end [OK]

Common Mistakes:

• Reversing the entire list instead
• Keeping original order unchanged
• Skipping nodes or mixing order

4. Consider this buggy C code snippet for reordering a linked list. What is the main error?

void reorderList(Node* head) {
  Node *slow = head, *fast = head;
  while (fast && fast->next) {
    slow = slow->next;
    fast = fast->next;
  }
  // further code omitted
}

medium

A. Fast pointer moves only one step instead of two

B. Slow pointer moves two steps instead of one

C. Missing check for head being NULL

D. Using wrong variable names for pointers

Solution

1. Step 1: Analyze pointer movements in the loop

   Fast pointer should move two steps to find middle, but here it moves only one step.

2. Step 2: Identify impact of incorrect fast pointer movement

   This causes slow pointer to not reach the true middle, breaking reorder logic.

3. Final Answer:

   Fast pointer moves only one step instead of two -> Option A

4. Quick Check:

   Fast must move 2 steps, not 1 [OK]

Hint: Fast pointer moves twice as fast as slow [OK]

Common Mistakes:

• Not moving fast pointer two steps
• Ignoring NULL checks
• Confusing slow and fast pointer roles

5. You have a linked list: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> NULL. After reordering, what will be the list? Also, explain why the middle node is chosen as it is.

hard

A. 10 -> 60 -> 20 -> 50 -> 30 -> 40 -> NULL; middle is 40 because it's the node after center

B. 10 -> 60 -> 20 -> 50 -> 30 -> 40 -> NULL; middle is 30 because it's the exact center for even length

C. 10 -> 50 -> 20 -> 40 -> 30 -> 60 -> NULL; middle is 30 because it's the exact center

D. 10 -> 60 -> 20 -> 50 -> 30 -> 40 -> NULL; middle is 20 because it's the first half's end

Solution

1. Step 1: Find the middle node for even length list

   For 6 nodes, middle is node 3 (value 30) because slow pointer stops at exact center when fast moves two steps.

2. Step 2: Apply reorder pattern

   Reorder alternates nodes from start and end: 10, 60, 20, 50, 30, 40.

3. Final Answer:

   10 -> 60 -> 20 -> 50 -> 30 -> 40 -> NULL; middle is 30 because it's the exact center for even length -> Option B

4. Quick Check:

   Middle at center, reorder alternates front and back [OK]

Hint: Middle is center node; reorder alternates front and back [OK]

Common Mistakes:

• Choosing wrong middle for even length
• Mixing reorder order
• Confusing node positions