Array Access and Update at Index in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to access the element at index 2 in the list.

DSA Python

my_list = [10, 20, 30, 40, 50]
print(my_list[[1]])

Drag options to blanks, or click blank then click option'

A2

B3

C1

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 3 instead of 2, which accesses the fourth element.

Using index 1, which accesses the second element.

Index 2 accesses the third element in the list, which is 30.

2fill in blank

medium

Complete the code to update the element at index 1 to 99.

DSA Python

numbers = [5, 10, 15, 20]
numbers[[1]] = 99
print(numbers)

Drag options to blanks, or click blank then click option'

A0

B1

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Updating index 0 instead of 1.

Using an index out of range.

Index 1 refers to the second element, which is updated to 99.

3fill in blank

hard

Fix the error in the code to correctly update the last element to 100.

DSA Python

data = [1, 2, 3, 4, 5]
data[[1]] = 100
print(data)

Drag options to blanks, or click blank then click option'

A4

B-1

C5

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 5 which is out of range.

Using index -1 when positive index is expected.

The last element is at index 4 (length - 1), so data[4] updates it correctly.

4fill in blank

hard

Fill both blanks to access and update the middle element of the list.

DSA Python

arr = [10, 20, 30, 40, 50]
mid_index = len(arr) [1] 2
arr[mid_index] = [2]
print(arr)

Drag options to blanks, or click blank then click option'

A//

B/

C25

D30

Attempts:

3 left

💡 Hint

Common Mistakes

Using '/' which results in float index causing error.

Updating the middle element to wrong value.

Using integer division '//' by 2 finds the middle index. Updating that index to 25 changes the middle element.

5fill in blank

hard

Fill all three blanks to create a dictionary with keys as uppercase letters and values as elements greater than 10.

DSA Python

values = {'a': 5, 'b': 15, 'c': 25, 'd': 8}
result = { [1]: [2] for k, v in values.items() if v [3] 10}
print(result)

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'k' instead of 'k.upper()' for keys.

Using '<' instead of '>' in the condition.

Using keys as values or vice versa.

Keys are converted to uppercase with k.upper(), values are v, and only values greater than 10 are included.

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