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.