Array Insertion at Beginning in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after inserting elements at the beginning?

Consider the following Python code that inserts elements at the beginning of a list. What will be printed?

DSA Python

arr = [3, 4, 5]
arr.insert(0, 2)
arr.insert(0, 1)
print(arr)

A [1, 2, 3, 4, 5]

B [3, 4, 5, 2, 1]

C [2, 1, 3, 4, 5]

D [1, 3, 4, 5, 2]

💡 Hint

Remember that insert(0, x) adds x at the start of the list.

Each insert at index 0 places the new element at the front, pushing existing elements right.

❓ Predict Output

intermediate

What is the final list after multiple insertions at the beginning?

What will be the output of this code snippet?

DSA Python

lst = []
lst.insert(0, 'c')
lst.insert(0, 'b')
lst.insert(0, 'a')
print(lst)

A ['c', 'b', 'a']

B ['c', 'a', 'b']

C ['b', 'a', 'c']

D ['a', 'b', 'c']

💡 Hint

Each insert at index 0 adds the element to the front.

Inserting 'c' first makes ['c'], then 'b' at front makes ['b','c'], then 'a' at front makes ['a','b','c'].

🔧 Debug

advanced

Why does this code not insert at the beginning as expected?

The code below is intended to insert 0 at the beginning of the list but does not work as expected. What is the problem?

DSA Python

numbers = [1, 2, 3]
numbers.append(0)
print(numbers)

A It inserts 0 at the beginning but prints the list reversed.

B It raises a TypeError because append needs two arguments.

C It adds 0 at the end, not the beginning, because append adds to the end.

D It removes the first element instead of inserting.

💡 Hint

Check what append does compared to insert.

append adds an element at the end of the list, not at the beginning. To add at the start, use insert(0, value).

❓ Predict Output

advanced

What is the output after inserting elements at the beginning in a loop?

What will be printed after running this code?

DSA Python

arr = []
for i in range(3):
    arr.insert(0, i)
print(arr)

A [2, 1, 0]

B [0, 1, 2]

C [0, 2, 1]

D [1, 2, 0]

💡 Hint

Each new element is added at the front, pushing others right.

Loop inserts 0 at front → [0], then 1 at front → [1,0], then 2 at front → [2,1,0].

🧠 Conceptual

expert

What is the time complexity of inserting an element at the beginning of a Python list?

Consider a Python list with n elements. What is the time complexity of inserting an element at the beginning using insert(0, value)?

A O(n^2) because insertion requires nested loops.

B O(n) because all elements must be shifted right to make space.

C O(log n) because Python uses binary search internally.

D O(1) because insert at index 0 is constant time.

💡 Hint

Think about what happens to existing elements when inserting at the start.

Inserting at the start requires shifting all existing elements one position to the right, which takes linear time.

Practice

1. What does inserting an element at the beginning of an array do to the existing elements?

easy

A. Shifts all existing elements one position to the right

B. Deletes the last element of the array

C. Replaces the first element with the new element

D. Leaves the array unchanged

Solution

1. Step 1: Understand array insertion at beginning

   When we insert at the start, the new element takes index 0.

2. Step 2: Effect on existing elements

   All existing elements move one position right to make space.

3. Final Answer:

   Shifts all existing elements one position to the right -> Option A

4. Quick Check:

   Insert at start = shift right [OK]

Hint: New start means shift all right by one [OK]

Common Mistakes:

• Thinking insertion replaces first element
• Assuming array size shrinks
• Believing array stays unchanged

2. Which of the following Python statements correctly inserts the value 10 at the beginning of list arr?

easy

A. arr.add(10)

B. arr.append(10)

C. arr[0] = 10

D. arr.insert(0, 10)

Solution

1. Step 1: Identify correct method for insertion at index

   Python list method insert(index, value) inserts at given index.

2. Step 2: Check each option

   arr.insert(0, 10) uses insert(0, 10) which inserts 10 at start. Others either append at end, replace element, or invalid method.

3. Final Answer:

   arr.insert(0, 10) -> Option D

4. Quick Check:

   Use insert(0, value) to add at start [OK]

Hint: Use insert(0, value) to add at beginning [OK]

Common Mistakes:

• Using append() adds at end, not start
• Assigning arr[0] replaces element, no shift
• Using non-existent method add()

3. What is the output of the following code?

arr = [2, 3, 4]
arr.insert(0, 1)
print(arr)

medium

A. [1, 3, 4]

B. [2, 3, 4, 1]

C. [1, 2, 3, 4]

D. [2, 3, 4]

Solution

1. Step 1: Initial array and insertion

   Original array is [2, 3, 4]. Using insert(0, 1) adds 1 at index 0.

2. Step 2: Resulting array after insertion

   All elements shift right. New array is [1, 2, 3, 4].

3. Final Answer:

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

4. Quick Check:

   Insert at start shifts elements right [OK]

Hint: Insert at 0 puts element first, shifts others [OK]

Common Mistakes:

• Thinking insert adds at end
• Replacing first element instead of shifting
• Ignoring shift effect on elements

4. Find the error in the code below that tries to insert 5 at the beginning of the list nums:

nums = [10, 20, 30]
nums.add(5)
print(nums)

medium

A. Method add() does not exist for lists

B. The list nums is empty

C. The print statement is incorrect

D. The value 5 cannot be inserted into a list

Solution

1. Step 1: Check method used for insertion

   Python lists do not have an add() method; this causes an error.

2. Step 2: Correct method to insert at beginning

   Use insert(0, 5) to add 5 at start of list.

3. Final Answer:

   Method add() does not exist for lists -> Option A

4. Quick Check:

   Use insert() not add() for list insertion [OK]

Hint: Use insert() for lists, add() is invalid [OK]

Common Mistakes:

• Assuming add() works like append()
• Thinking print statement causes error
• Believing list cannot hold new values

5. Given the list data = [4, 5, 6], which code snippet correctly inserts the elements of another list new_items = [1, 2, 3] at the beginning so that the final list is [1, 2, 3, 4, 5, 6]?

hard

A. for item in new_items:
data.insert(0, item)

B. for item in reversed(new_items):
data.insert(0, item)

C. data.extend(new_items)

D. data.append(new_items)

Solution

1. Step 1: Understand insertion order impact

   Inserting each item at index 0 shifts existing elements right. To keep order, insert reversed list.

2. Step 2: Analyze each option

   for item in new_items: data.insert(0, item) inserts 1, then 2 at front, reversing order. for item in reversed(new_items): data.insert(0, item) inserts reversed list, preserving order. Options C and D add at end or nested list.

3. Final Answer:

   for item in reversed(new_items): data.insert(0, item) -> Option B

4. Quick Check:

   Insert reversed to keep order at start [OK]

Hint: Insert reversed list at 0 to keep order [OK]

Common Mistakes:

• Inserting forward order reverses new items
• Using extend() adds at end, not start
• Appending list nests it inside