Insert at Middle Specific Position in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after inserting at position 3?

Consider a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> null. We insert a new node with value 25 at position 3 (1-based index). What is the resulting list?

DSA Python

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

def insert_at_position(head, data, pos):
    new_node = Node(data)
    if pos == 1:
        new_node.next = head
        return new_node
    current = head
    count = 1
    while current and count < pos - 1:
        current = current.next
        count += 1
    if not current:
        return head
    new_node.next = current.next
    current.next = new_node
    return head

def print_list(head):
    result = []
    current = head
    while current:
        result.append(current.data)
        current = current.next
    return result

head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)

head = insert_at_position(head, 25, 3)
print(print_list(head))

A [25, 10, 20, 30, 40]

B [10, 20, 25, 30, 40]

C [10, 25, 20, 30, 40]

D [10, 20, 30, 40, 25]

💡 Hint

Remember that position 3 means the new node should be the third node in the list.

The new node with value 25 is inserted after the second node (20), so it becomes the third node. The list updates to 10 -> 20 -> 25 -> 30 -> 40 -> null.

❓ Predict Output

intermediate

What is the output after inserting at position 1?

Given a linked list 5 -> 15 -> 25 -> null, insert a node with value 1 at position 1. What is the resulting list?

DSA Python

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

def insert_at_position(head, data, pos):
    new_node = Node(data)
    if pos == 1:
        new_node.next = head
        return new_node
    current = head
    count = 1
    while current and count < pos - 1:
        current = current.next
        count += 1
    if not current:
        return head
    new_node.next = current.next
    current.next = new_node
    return head

def print_list(head):
    result = []
    current = head
    while current:
        result.append(current.data)
        current = current.next
    return result

head = Node(5)
head.next = Node(15)
head.next.next = Node(25)

head = insert_at_position(head, 1, 1)
print(print_list(head))

A [1, 5, 15, 25]

B [5, 1, 15, 25]

C [5, 15, 25, 1]

D [1, 15, 25]

💡 Hint

Inserting at position 1 means the new node becomes the new head.

The new node with value 1 is inserted at the start, so it points to the old head (5). The list becomes 1 -> 5 -> 15 -> 25 -> null.

🔧 Debug

advanced

Why does this insertion code fail to insert at position 4?

The code below tries to insert a node with value 50 at position 4 in the list 10 -> 20 -> 30 -> null, but the list remains unchanged. What is the bug?

DSA Python

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

def insert_at_position(head, data, pos):
    new_node = Node(data)
    current = head
    count = 1
    while current and count < pos:
        current = current.next
        count += 1
    if not current:
        return head
    new_node.next = current.next
    current.next = new_node
    return head

head = Node(10)
head.next = Node(20)
head.next.next = Node(30)

head = insert_at_position(head, 50, 4)

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

A The loop condition should be count < pos - 1 to stop at the node before insertion point.

B The position variable is off by one; it should start counting from zero.

C The function should check if head is None before insertion.

D The new node's next should be set to None explicitly before insertion.

💡 Hint

Think about where the pointer 'current' stops before insertion.

The loop moves current to the node at position 4, but insertion requires stopping at position 3 to link the new node. Changing the loop to count < pos - 1 fixes this.

❓ Predict Output

advanced

What is the output after inserting at position beyond list length?

Given a linked list 1 -> 2 -> 3 -> null, what is the output after trying to insert 99 at position 5?

DSA Python

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

def insert_at_position(head, data, pos):
    new_node = Node(data)
    if pos == 1:
        new_node.next = head
        return new_node
    current = head
    count = 1
    while current and count < pos - 1:
        current = current.next
        count += 1
    if not current:
        return head
    new_node.next = current.next
    current.next = new_node
    return head

def print_list(head):
    result = []
    current = head
    while current:
        result.append(current.data)
        current = current.next
    return result

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)

head = insert_at_position(head, 99, 5)
print(print_list(head))

A [99, 1, 2, 3]

B [1, 2, 3, 99]

C [1, 2, 3]

D IndexError

💡 Hint

If the position is beyond the list length, the list remains unchanged.

The function returns the original list because position 5 is beyond the current list length of 3, so no insertion happens.

🧠 Conceptual

expert

What is the time complexity of inserting at a specific middle position in a singly linked list?

Assuming a singly linked list with n nodes, what is the time complexity to insert a new node at position k (1 <= k <= n+1)?

A O(log n) because linked lists use binary search to find positions.

B O(1) because insertion is always constant time in linked lists.

C O(n) because we always traverse the entire list regardless of position.

D O(k) because we must traverse to the (k-1)th node before insertion.

💡 Hint

Think about how many nodes you must visit before inserting.

To insert at position k, you must move through k-1 nodes to reach the insertion point, so the time is proportional to k, which is O(k).