Array Access and Update at Index in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after updating array element at index

What is the output of the following Python code after updating the element at index 2?

DSA Python

arr = [5, 10, 15, 20, 25]
arr[2] = 100
print(arr)

A [5, 10, 15, 20, 25]

B [5, 100, 15, 20, 25]

C [100, 10, 15, 20, 25]

D [5, 10, 100, 20, 25]

💡 Hint

Remember that list indices start at 0 in Python.

The element at index 2 (third element) is updated from 15 to 100. The rest remain unchanged.

❓ Predict Output

intermediate

Output after multiple updates on array

What will be printed after these updates to the array?

DSA Python

arr = [1, 2, 3, 4, 5]
arr[0] = 10
arr[4] = 50
print(arr)

A [10, 2, 3, 4, 50]

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

C [10, 2, 3, 4, 5]

D [1, 2, 3, 4, 50]

💡 Hint

Check which indices are updated and what values they get.

The first element (index 0) is changed to 10 and the last element (index 4) to 50.

🔧 Debug

advanced

Identify the error when updating an array index

What error will this code produce when run?

DSA Python

arr = [7, 8, 9]
arr[3] = 10
print(arr)

A No error, output: [7, 8, 9, 10]

B TypeError: 'int' object is not subscriptable

C IndexError: list assignment index out of range

D SyntaxError: invalid syntax

💡 Hint

Check the valid indices for the array of length 3.

Index 3 does not exist in the array of length 3 (indices 0,1,2). Trying to assign to it causes IndexError.

🧠 Conceptual

advanced

Effect of negative index update on array

What is the output after updating the array element at index -1?

DSA Python

arr = [3, 6, 9, 12]
arr[-1] = 100
print(arr)

A [100, 6, 9, 12]

B [3, 6, 9, 100]

C [3, 6, 100, 12]

D IndexError: list assignment index out of range

💡 Hint

Negative indices count from the end of the list in Python.

Index -1 refers to the last element, so it updates 12 to 100.

🚀 Application

expert

Final array after sequence of updates and accesses

Given the code below, what is the final printed array?

DSA Python

arr = [0, 1, 2, 3, 4, 5]
for i in range(2, 5):
    arr[i] = arr[i] * 2
arr[0] = arr[5] + arr[1]
print(arr)

A [6, 1, 4, 6, 8, 5]

B [5, 1, 4, 6, 8, 5]

C [6, 1, 2, 3, 4, 5]

D [0, 1, 4, 6, 8, 5]

💡 Hint

Trace each update step carefully, especially the loop and the final assignment.

Elements at indices 2,3,4 are doubled: 2->4, 3->6, 4->8. Then arr[0] = arr[5] + arr[1] = 5 + 1 = 6.

Practice

1. What does the expression arr[2] do in an array arr?

easy

A. Returns the length of the array

B. Adds a new element at index 2

C. Deletes the element at index 2

D. Accesses the element at index 2 in the array

Solution

1. Step 1: Understand array indexing

   Array elements are accessed using their index inside square brackets, starting from 0.

2. Step 2: Access element at index 2

   The expression arr[2] retrieves the third element in the array (index 2).

3. Final Answer:

   Accesses the element at index 2 in the array -> Option D

4. Quick Check:

   arr[2] means element at position 2 [OK]

Hint: Indexing with brackets gets element at that position [OK]

Common Mistakes:

• Thinking index starts at 1 instead of 0
• Confusing access with insertion or deletion
• Assuming arr[2] returns length

2. Which of the following is the correct syntax to update the element at index 1 of array arr to 10?

easy

A. arr[1] = 10

B. arr{1} = 10

C. arr(1) = 10

D. arr.1 = 10

Solution

1. Step 1: Identify correct syntax for array update

   In Python, arrays or lists use square brackets to access or update elements by index.

2. Step 2: Use square brackets for assignment

   To update element at index 1, write arr[1] = 10.

3. Final Answer:

   arr[1] = 10 -> Option A

4. Quick Check:

   Use square brackets for indexing and assignment [OK]

Hint: Use square brackets [] to access or update array elements [OK]

Common Mistakes:

• Using parentheses () instead of brackets []
• Using curly braces {} which are for sets or dicts
• Trying dot notation for indexing

3. What will be the output of the following code?

arr = [5, 10, 15, 20]
arr[2] = 25
print(arr)

medium

A. [5, 10, 25, 20]

B. [5, 10, 15, 20]

C. [25, 10, 15, 20]

D. Error

Solution

1. Step 1: Understand initial array and update

   Initial array is [5, 10, 15, 20]. The statement arr[2] = 25 updates the element at index 2 (third element) to 25.

2. Step 2: Print updated array

   After update, array becomes [5, 10, 25, 20]. Printing it shows this updated list.

3. Final Answer:

   [5, 10, 25, 20] -> Option A

4. Quick Check:

   arr[2] changed to 25, rest unchanged [OK]

Hint: Update changes element at index, print shows new array [OK]

Common Mistakes:

• Thinking arr[2] updates first element
• Expecting original array without change
• Assuming syntax error

4. Identify the error in the following code and choose the correct fix:

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

medium

A. TypeError because arr is immutable; fix by converting to list

B. SyntaxError due to wrong assignment; fix by using arr(3) = 4

C. IndexError because index 3 is out of range; fix by using arr.append(4)

D. No error; output is [1, 2, 3, 4]

Solution

1. Step 1: Analyze index used for update

   Array has indices 0,1,2. Trying to assign at index 3 causes IndexError (out of range).

2. Step 2: Correct way to add new element

   To add element at end, use arr.append(4) instead of assigning at invalid index.

3. Final Answer:

   IndexError because index 3 is out of range; fix by using arr.append(4) -> Option C

4. Quick Check:

   Index 3 invalid, append adds element safely [OK]

Hint: Use append() to add beyond last index, not assignment [OK]

Common Mistakes:

• Assigning to index beyond array length
• Using parentheses instead of brackets
• Assuming no error occurs

5. Given arr = [2, 4, 6, 8, 10], which code snippet correctly doubles the value at index 3 and then swaps it with the value at index 1?

hard

A. arr[3] *= 2
arr[1] = arr[3]
arr[3] = arr[1]

B. arr[3] = arr[3] * 2
arr[1], arr[3] = arr[3], arr[1]

C. arr[3] = arr[1]
arr[1] = arr[3] * 2

D. arr[1], arr[3] = arr[3] * 2, arr[1]

Solution

1. Step 1: Double value at index 3

   Original arr[3] is 8. Doubling: arr[3] = 8 * 2 = 16.

2. Step 2: Swap values at index 1 and 3

   Swap arr[1] (4) and arr[3] (16) using tuple unpacking: arr[1], arr[3] = arr[3], arr[1].

3. Final Answer:

   arr[3] = arr[3] * 2 arr[1], arr[3] = arr[3], arr[1] -> Option B

4. Quick Check:

   Double then swap using tuple unpacking [OK]

Hint: Double first, then swap with tuple unpacking [OK]

Common Mistakes:

• Overwriting values before swap
• Incorrect swap order or syntax
• Not doubling before swapping