Remove Nth Node from End of List in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after removing 2nd node from end

What is the output linked list after removing the 2nd node from the end in the given list?

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeNthFromEnd(head, n):
    dummy = ListNode(0, head)
    first = dummy
    second = dummy
    for _ in range(n + 1):
        first = first.next
    while first:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

# Create linked list 1 -> 2 -> 3 -> 4 -> 5 -> null
head = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))

new_head = removeNthFromEnd(head, 2)

# Print list
result = []
while new_head:
    result.append(new_head.val)
    new_head = new_head.next
print(result)

A [1, 2, 4, 5]

B [1, 3, 4, 5]

C [1, 2, 3, 5]

D [2, 3, 4, 5]

💡 Hint

Remember the 2nd node from the end is the 4th node from the start in a 5-node list.

Removing the 2nd node from the end removes the node with value 4. The resulting list is 1 -> 2 -> 3 -> 5 -> null.

❓ Predict Output

intermediate

Output after removing last node

What is the output linked list after removing the 1st node from the end (last node) in the given list?

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeNthFromEnd(head, n):
    dummy = ListNode(0, head)
    first = dummy
    second = dummy
    for _ in range(n + 1):
        first = first.next
    while first:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

# Create linked list 10 -> 20 -> 30 -> null
head = ListNode(10, ListNode(20, ListNode(30)))

new_head = removeNthFromEnd(head, 1)

# Print list
result = []
while new_head:
    result.append(new_head.val)
    new_head = new_head.next
print(result)

A [10, 20, 30]

B [10, 30]

C [20, 30]

D [10, 20]

💡 Hint

Removing the last node means removing the node with value 30.

The last node (value 30) is removed, so the list becomes 10 -> 20 -> null.

🔧 Debug

advanced

Identify the error in removing nth node from end

What error will this code raise when trying to remove the 3rd node from the end of a 2-node list?

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeNthFromEnd(head, n):
    dummy = ListNode(0, head)
    first = dummy
    second = dummy
    for _ in range(n):
        first = first.next
    while first.next:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

# Create linked list 1 -> 2 -> null
head = ListNode(1, ListNode(2))

new_head = removeNthFromEnd(head, 3)

A AttributeError

B NoneType AttributeError

C IndexError

D No error, returns [2]

💡 Hint

Check the for loop that moves the first pointer n times.

The for loop tries to move first pointer 3 times, but the list has only 2 nodes plus dummy, so first becomes None and accessing first.next causes AttributeError on NoneType.

🧠 Conceptual

advanced

Why use a dummy node in removing nth node from end?

What is the main reason to use a dummy node before the head when removing the nth node from the end of a linked list?

A To simplify edge cases like removing the head node

B To reduce memory usage

C To speed up the traversal

D To avoid using recursion

💡 Hint

Think about what happens if the node to remove is the first node.

Using a dummy node allows uniform handling of all nodes including the head, so removing the head node doesn't require special code.

❓ Predict Output

expert

Output after removing 4th node from end in a 5-node list

What is the output linked list after removing the 4th node from the end in the list 5 -> 4 -> 3 -> 2 -> 1 -> null?

DSA Python

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def removeNthFromEnd(head, n):
    dummy = ListNode(0, head)
    first = dummy
    second = dummy
    for _ in range(n + 1):
        first = first.next
    while first:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

# Create linked list 5 -> 4 -> 3 -> 2 -> 1 -> null
head = ListNode(5, ListNode(4, ListNode(3, ListNode(2, ListNode(1)))))

new_head = removeNthFromEnd(head, 4)

# Print list
result = []
while new_head:
    result.append(new_head.val)
    new_head = new_head.next
print(result)

A [5, 3, 2, 1]

B [4, 3, 2, 1]

C [5, 4, 2, 1]

D [5, 4, 3, 1]

💡 Hint

The 4th node from the end is the 2nd node from the start.

Removing the 4th node from the end removes the node with value 4. The resulting list is 5 -> 3 -> 2 -> 1 -> null.

Practice

1. What does the problem Remove Nth Node from End of List ask you to do?

easy

A. Add a new node at the Nth position from the start

B. Find the value of the Nth node from the start

C. Reverse the linked list up to the Nth node

D. Delete the node that is the Nth from the end in a linked list

Solution

1. Step 1: Understand the problem statement

   The problem requires removing a node counting from the end of the linked list, not from the start.

2. Step 2: Identify the correct action

   Deleting the Nth node from the end means locating it and removing it from the list.

3. Final Answer:

   Delete the node that is the Nth from the end in a linked list -> Option D

4. Quick Check:

   Remove Nth node from end = Delete node Nth from end [OK]

Hint: Focus on 'from end' means counting backwards [OK]

Common Mistakes:

• Confusing Nth from start with Nth from end
• Thinking the problem is about reversing the list
• Trying to add nodes instead of removing

2. Which of the following is the correct Python syntax to create a dummy node pointing to the head of a linked list named head?

easy

A. dummy = head.next

B. dummy = ListNode(head)

C. dummy = ListNode(0); dummy.next = head

D. dummy = ListNode(); dummy = head

Solution

1. Step 1: Understand dummy node creation

   A dummy node is usually created with a value (often 0) and its next pointer set to the head.

2. Step 2: Check each option

   dummy = ListNode(0); dummy.next = head correctly creates a dummy node with value 0 and points its next to head.

3. Final Answer:

   dummy = ListNode(0); dummy.next = head -> Option C

4. Quick Check:

   Dummy node points to head = dummy.next = head [OK]

Hint: Dummy node value 0 and next points to head [OK]

Common Mistakes:

• Passing head as value instead of next pointer
• Overwriting dummy after creation
• Not linking dummy to head

3. What will be the output linked list after removing the 2nd node from the end in the list 1 -> 2 -> 3 -> 4 -> 5 -> null?

medium

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

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

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

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

Solution

1. Step 1: Identify the 2nd node from the end

   Counting from the end: 5 (1st), 4 (2nd). So node with value 4 is to be removed.

2. Step 2: Remove node 4 and link previous node to next

   Node 3's next pointer should now point to node 5, skipping node 4.

3. Final Answer:

   1 -> 2 -> 3 -> 5 -> null -> Option A

4. Quick Check:

   Remove 2nd from end (4) = 1->2->3->5 [OK]

Hint: Count nodes from end, remove that node [OK]

Common Mistakes:

• Removing the wrong node counting from start
• Not updating the previous node's next pointer
• Confusing node values

4. Given this code snippet to remove the Nth node from end, what is the error?

def removeNthFromEnd(head, n):
    dummy = ListNode(0)
    dummy.next = head
    first = dummy
    second = dummy
    for i in range(n):
        first = first.next
    while first.next:
        first = first.next
        second = second.next
    second.next = second.next.next
    return dummy.next

medium

A. The for loop should run n+1 times to move first pointer

B. The while loop condition should be while first is not None

C. second.next = second.next.next causes error if second.next is None

D. dummy node is not needed for this problem

Solution

1. Step 1: Analyze the for loop moving first pointer

   To position first pointer n nodes ahead, it must move n+1 times to maintain gap with second pointer.

2. Step 2: Understand the impact of moving first pointer only n times

   If first moves only n times, the gap is off by one, causing wrong node removal.

3. Final Answer:

   The for loop should run n+1 times to move first pointer -> Option A

4. Quick Check:

   Move first pointer n+1 times for correct gap [OK]

Hint: Move first pointer n+1 times to keep gap [OK]

Common Mistakes:

• Moving first pointer only n times
• Incorrect while loop condition
• Ignoring dummy node usage

5. You have a linked list 1 -> 2 -> 3 -> 4 -> 5 -> null. If you remove the 5th node from the end, what will be the resulting list?

hard

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

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

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

D. 5 -> null

Solution

1. Step 1: Identify the 5th node from the end

   The list length is 5, so the 5th node from the end is the first node with value 1.

2. Step 2: Remove the first node and link dummy to second node

   Removing node 1 means the new head is node 2, so list becomes 2 -> 3 -> 4 -> 5 -> null.

3. Final Answer:

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

4. Quick Check:

   Remove 5th from end (head) = 2->3->4->5 [OK]

Hint: Removing Nth from end where N equals list length removes head [OK]

Common Mistakes:

• Removing last node instead of first
• Not updating head pointer after removal
• Confusing node positions