Insert at End Tail Insert in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after inserting nodes at the tail?

Consider a singly linked list initially empty. We insert nodes with values 10, 20, and 30 at the tail in that order. What is the printed linked list after these insertions?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_at_tail(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        temp = self.head
        while temp.next:
            temp = temp.next
        temp.next = new_node

    def print_list(self):
        temp = self.head
        while temp:
            print(temp.data, end=' -> ')
            temp = temp.next
        print('null')

ll = LinkedList()
ll.insert_at_tail(10)
ll.insert_at_tail(20)
ll.insert_at_tail(30)
ll.print_list()

A 20 -> 30 -> 10 -> null

B 30 -> 20 -> 10 -> null

C 10 -> 20 -> null

D 10 -> 20 -> 30 -> null

💡 Hint

Remember that inserting at tail adds the new node at the end of the list.

Inserting at tail means the new node is added after the last node. So the order of nodes is the same as insertion order: 10, then 20, then 30.

❓ Predict Output

intermediate

What is the linked list after these tail insertions?

Given the following code inserting values 5, 15, and 25 at the tail of an empty linked list, what will be the printed output?

DSA Python

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

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_tail(self, val):
        node = Node(val)
        if self.head is None:
            self.head = node
            return
        current = self.head
        while current.next:
            current = current.next
        current.next = node

    def display(self):
        curr = self.head
        while curr:
            print(curr.val, end=' -> ')
            curr = curr.next
        print('null')

ll = LinkedList()
ll.insert_tail(5)
ll.insert_tail(15)
ll.insert_tail(25)
ll.display()

A 5 -> 15 -> 25 -> null

B 25 -> 15 -> 5 -> null

C 5 -> 25 -> 15 -> null

D 15 -> 5 -> 25 -> null

💡 Hint

Tail insertion keeps the order of insertion intact.

Each new node is added after the last node, so the list order is 5, then 15, then 25.

🔧 Debug

advanced

Why does this tail insertion code cause an infinite loop?

Examine the code below that tries to insert a node at the tail of a linked list. Why does it cause an infinite loop when inserting multiple nodes?

DSA Python

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_tail(self, data):
        new_node = Node(data)
        if self.head is None:
            self.head = new_node
        else:
            temp = self.head
            while temp.next != new_node:
                temp = temp.next
            temp.next = new_node

ll = LinkedList()
ll.insert_tail(1)
ll.insert_tail(2)
ll.insert_tail(3)

A The head is not updated after insertion, causing the loop.

B The new_node is not created properly; it should be created outside the method.

C The while loop condition is wrong; it should check temp.next != None, not temp.next != new_node.

D The temp pointer is not moved inside the loop, causing infinite loop.

💡 Hint

Check the condition inside the while loop carefully.

The loop condition compares temp.next to new_node, which is never true initially, so the loop never ends. It should check for temp.next != None to find the last node.

❓ Predict Output

advanced

What is the linked list after these tail insertions with duplicates?

A linked list is empty initially. We insert values 7, 7, 14, 21 at the tail. What is the printed linked list?

DSA Python

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

class LinkedList:
    def __init__(self):
        self.head = None

    def insert_tail(self, val):
        new_node = Node(val)
        if not self.head:
            self.head = new_node
            return
        curr = self.head
        while curr.next:
            curr = curr.next
        curr.next = new_node

    def print_list(self):
        curr = self.head
        while curr:
            print(curr.val, end=' -> ')
            curr = curr.next
        print('null')

ll = LinkedList()
ll.insert_tail(7)
ll.insert_tail(7)
ll.insert_tail(14)
ll.insert_tail(21)
ll.print_list()

A 21 -> 14 -> 7 -> 7 -> null

B 7 -> 7 -> 14 -> 21 -> null

C 7 -> 14 -> 21 -> null

D 7 -> 7 -> 21 -> 14 -> null

💡 Hint

Tail insertion preserves the order and duplicates are allowed.

Nodes are added at the end in the order inserted, so duplicates appear in the order inserted.

🧠 Conceptual

expert

What is the time complexity of inserting n nodes at the tail in a singly linked list without tail pointer?

If you insert n nodes one by one at the tail of a singly linked list that does NOT maintain a tail pointer, what is the total time complexity?

A O(n^2) because each insertion requires traversing the entire list to find the tail

B O(n) because each insertion is done in constant time

C O(log n) because traversal is optimized

D O(n log n) because of repeated traversals

💡 Hint

Think about how many nodes you traverse for each insertion.

Without a tail pointer, each insertion requires traversing the whole list to find the last node, which takes O(k) time for the k-th insertion. Summing over n insertions gives O(n^2).

Practice

1. What does the "Insert at End (Tail Insert)" operation do in a singly linked list?

easy

A. Adds a new node at the end of the list

B. Removes the last node from the list

C. Adds a new node at the beginning of the list

D. Deletes the first node of the list

Solution

1. Step 1: Understand the operation

   Insert at End means adding a new node after the last node in the list.

2. Step 2: Compare with other operations

   Other options describe removing or adding at the start, which is different.

3. Final Answer:

   Adds a new node at the end of the list -> Option A

4. Quick Check:

   Insert at End = Add node at end [OK]

Hint: Tail insert always adds node after the last one [OK]

Common Mistakes:

• Confusing insert at end with insert at beginning
• Thinking it removes nodes instead of adding
• Mixing up head and tail operations

2. Which of the following is the correct syntax to create a new node and insert it at the end of a singly linked list in Python?

easy

A. new_node = Node(data); new_node.next = head; head = new_node

B. new_node = Node(data); tail = new_node; new_node.next = tail

C. new_node = Node(data); head.next = new_node; head = new_node

D. new_node = Node(data); tail.next = new_node; tail = new_node

Solution

1. Step 1: Create new node and link it after tail

   We create new_node and set tail.next to it, then update tail to new_node.

2. Step 2: Check other options for correctness

   new_node = Node(data); new_node.next = head; head = new_node inserts at head, C wrongly updates head, D sets tail before linking.

3. Final Answer:

   new_node = Node(data); tail.next = new_node; tail = new_node -> Option D

4. Quick Check:

   Tail.next = new_node, tail = new_node [OK]

Hint: Tail insert updates tail.next then tail pointer [OK]

Common Mistakes:

• Assigning new_node.next before linking to tail
• Updating head instead of tail
• Not updating tail pointer after insertion

3. What will be the output linked list after executing this code?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(1)
tail = head

for value in [2, 3, 4]:
    new_node = Node(value)
    tail.next = new_node
    tail = new_node

# Print list
current = head
result = []
while current:
    result.append(current.data)
    current = current.next
print(result)

medium

A. [1, 2, 3, 4]

B. [4, 3, 2, 1]

C. [1, 2, 3]

D. [2, 3, 4]

Solution

1. Step 1: Build linked list by tail insertion

   Start with head=1, then add 2, 3, 4 at tail in order.

2. Step 2: Traverse and collect node data

   Traverse from head, collect [1, 2, 3, 4] in order.

3. Final Answer:

   [1, 2, 3, 4] -> Option A

4. Quick Check:

   Tail insert preserves order [OK]

Hint: Tail insert keeps order same as input list [OK]

Common Mistakes:

• Confusing order and reversing list
• Printing only last inserted nodes
• Not updating tail pointer correctly

4. Identify the error in this tail insert code snippet:

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

head = Node(10)
tail = head

new_node = Node(20)
new_node.next = tail
tail = new_node

medium

A. tail should be updated before assigning new_node.next

B. new_node.next should be None, not tail

C. head should be updated instead of tail

D. No error, code is correct

Solution

1. Step 1: Check linking of new_node

   new_node.next points to tail, which creates a cycle or wrong link.

2. Step 2: Correct linking for tail insert

   new_node.next should be None because it's new tail; tail.next should point to new_node.

3. Final Answer:

   new_node.next should be None, not tail -> Option B

4. Quick Check:

   New tail's next = None [OK]

Hint: New tail node's next must be None [OK]

Common Mistakes:

• Linking new_node.next to old tail
• Not updating tail.next to new_node
• Confusing head and tail pointers

5. Given a singly linked list with nodes containing values [5, 10, 15], write the sequence of steps to insert a new node with value 20 at the end, and then print the list. What will be the final printed output?

hard

A. [5, 10, 15]

B. [20, 5, 10, 15]

C. [5, 10, 15, 20]

D. [5, 10, 20, 15]

Solution

1. Step 1: Traverse to the last node (value 15)

   Start at head (5), move to 10, then 15 which is tail (next is None).

2. Step 2: Create new node with value 20 and link it

   Set last node's next to new node (20), new node's next is None.

3. Step 3: Traverse and print all nodes

   Traverse from head and collect values: 5, 10, 15, 20.

4. Final Answer:

   [5, 10, 15, 20] -> Option C

5. Quick Check:

   Tail insert adds node at end [OK]

Hint: Traverse to tail, link new node, print full list [OK]

Common Mistakes:

• Inserting new node in middle instead of end
• Not setting new node's next to None
• Forgetting to update last node's next pointer