Insert at Specific Position in Doubly Linked List in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after inserting at position 2 in a doubly linked list

What is the printed state of the doubly linked list after inserting the value 15 at position 2?

DSA Python

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

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

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node
        new_node.prev = last

    def insert_at_position(self, data, position):
        new_node = Node(data)
        if position == 1:
            new_node.next = self.head
            if self.head:
                self.head.prev = new_node
            self.head = new_node
            return
        temp = self.head
        for _ in range(position - 2):
            if temp is None:
                return
            temp = temp.next
        if temp is None:
            return
        new_node.next = temp.next
        if temp.next:
            temp.next.prev = new_node
        temp.next = new_node
        new_node.prev = temp

    def print_list(self):
        temp = self.head
        result = []
        while temp:
            result.append(temp.data)
            temp = temp.next
        print(' -> '.join(map(str, result)) + ' -> null')

dll = DoublyLinkedList()
dll.append(10)
dll.append(20)
dll.append(30)
dll.insert_at_position(15, 2)
dll.print_list()

A 10 -> 20 -> 15 -> 30 -> null

B 15 -> 10 -> 20 -> 30 -> null

C 10 -> 15 -> 20 -> 30 -> null

D 10 -> 20 -> 30 -> 15 -> null

💡 Hint

Remember that position 1 means inserting at the head, so position 2 means after the first node.

Inserting 15 at position 2 places it between 10 and 20. The list becomes 10 -> 15 -> 20 -> 30 -> null.

❓ Predict Output

intermediate

Output after inserting at head in doubly linked list

What is the printed state of the doubly linked list after inserting the value 5 at position 1?

DSA Python

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

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

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node
        new_node.prev = last

    def insert_at_position(self, data, position):
        new_node = Node(data)
        if position == 1:
            new_node.next = self.head
            if self.head:
                self.head.prev = new_node
            self.head = new_node
            return
        temp = self.head
        for _ in range(position - 2):
            if temp is None:
                return
            temp = temp.next
        if temp is None:
            return
        new_node.next = temp.next
        if temp.next:
            temp.next.prev = new_node
        temp.next = new_node
        new_node.prev = temp

    def print_list(self):
        temp = self.head
        result = []
        while temp:
            result.append(temp.data)
            temp = temp.next
        print(' -> '.join(map(str, result)) + ' -> null')

dll = DoublyLinkedList()
dll.append(10)
dll.append(20)
dll.append(30)
dll.insert_at_position(5, 1)
dll.print_list()

A 5 -> 10 -> 20 -> 30 -> null

B 10 -> 5 -> 20 -> 30 -> null

C 10 -> 20 -> 30 -> 5 -> null

D 10 -> 20 -> 5 -> 30 -> null

💡 Hint

Position 1 means inserting at the very start of the list.

Inserting 5 at position 1 makes it the new head, so the list starts with 5 followed by the original nodes.

🔧 Debug

advanced

Identify the error in insert_at_position method

What error will occur when trying to insert at position 5 in a list of length 3 using this code?

DSA Python

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

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

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node
        new_node.prev = last

    def insert_at_position(self, data, position):
        new_node = Node(data)
        if position == 1:
            new_node.next = self.head
            if self.head:
                self.head.prev = new_node
            self.head = new_node
            return
        temp = self.head
        for _ in range(position - 2):
            temp = temp.next
        new_node.next = temp.next
        if temp.next:
            temp.next.prev = new_node
        temp.next = new_node
        new_node.prev = temp

dll = DoublyLinkedList()
dll.append(10)
dll.append(20)
dll.append(30)
dll.insert_at_position(40, 5)

A TypeError: unsupported operand type(s)

B IndexError: list index out of range

C No error, inserts at the end

D AttributeError: 'NoneType' object has no attribute 'next'

💡 Hint

Check what happens when temp becomes None in the for loop.

When position is beyond the list length + 1, temp becomes None and accessing temp.next causes AttributeError.

🧠 Conceptual

advanced

Why update prev pointer when inserting in doubly linked list?

Why is it necessary to update the prev pointer of the node after the inserted node when inserting at a specific position in a doubly linked list?

A To free memory occupied by the previous node

B To maintain the backward link so traversal in reverse direction works correctly

C To avoid memory leaks in the list

D To speed up the insertion operation

💡 Hint

Think about how doubly linked lists allow moving backward.

Updating the prev pointer ensures the node after the inserted node points back correctly, preserving backward traversal.

❓ Predict Output

expert

Output after multiple insertions at various positions

What is the printed state of the doubly linked list after these operations? 1. Append 10 2. Append 20 3. Insert 15 at position 2 4. Insert 5 at position 1 5. Insert 25 at position 5

DSA Python

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

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

    def append(self, data):
        new_node = Node(data)
        if not self.head:
            self.head = new_node
            return
        last = self.head
        while last.next:
            last = last.next
        last.next = new_node
        new_node.prev = last

    def insert_at_position(self, data, position):
        new_node = Node(data)
        if position == 1:
            new_node.next = self.head
            if self.head:
                self.head.prev = new_node
            self.head = new_node
            return
        temp = self.head
        for _ in range(position - 2):
            if temp is None:
                return
            temp = temp.next
        if temp is None:
            return
        new_node.next = temp.next
        if temp.next:
            temp.next.prev = new_node
        temp.next = new_node
        new_node.prev = temp

    def print_list(self):
        temp = self.head
        result = []
        while temp:
            result.append(temp.data)
            temp = temp.next
        print(' -> '.join(map(str, result)) + ' -> null')

dll = DoublyLinkedList()
dll.append(10)
dll.append(20)
dll.insert_at_position(15, 2)
dll.insert_at_position(5, 1)
dll.insert_at_position(25, 5)
dll.print_list()

A 5 -> 10 -> 15 -> 20 -> 25 -> null

B 5 -> 10 -> 15 -> 25 -> 20 -> null

C 10 -> 5 -> 15 -> 20 -> 25 -> null

D 5 -> 15 -> 10 -> 20 -> 25 -> null

💡 Hint

Track each insertion carefully and remember positions start at 1.

The insertions place 5 at head, 15 between 10 and 20, and 25 at the end (position 5). The final list is 5 -> 10 -> 15 -> 20 -> 25 -> null.