Array Traversal Patterns in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print all elements in the array using a for loop.

DSA Python

arr = [10, 20, 30, 40]
for [1] in arr:
    print([1])

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

Ax

Bi

Citem

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using an index variable instead of the element variable.

Using different variable names in the loop and print statement.

The variable x is used to represent each element in the array during traversal.

2fill in blank

medium

Complete the code to print elements at even indices of the array.

DSA Python

arr = [5, 10, 15, 20, 25]
for i in range(0, len(arr), [1]):
    print(arr[i])

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

A1

B2

C3

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using step size 1 which prints all elements.

Using step size 3 or 4 which skips too many elements.

Using a step of 2 in range allows traversal of even indices: 0, 2, 4, ...

3fill in blank

hard

Fix the error in the code to print elements in reverse order.

DSA Python

arr = [1, 2, 3, 4, 5]
for i in range(len(arr)-1, [1], -1):
    print(arr[i])

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

A1

B0

C-1

Dlen(arr)

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as stop which excludes index 0.

Using len(arr) which is out of range.

The stop value in range is exclusive, so to include index 0 when traversing backwards, use -1 as stop. This generates indices 4, 3, 2, 1, 0.

4fill in blank

hard

Fill both blanks to create a dictionary with elements as keys and their indices as values.

DSA Python

arr = ['a', 'b', 'c']
index_map = [1] for [2] in range(len(arr))}

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

A{arr[i]: i

Barr[i]

Ci

Drange(len(arr))

Attempts:

3 left

💡 Hint

Common Mistakes

Using the element as the loop variable instead of the index.

Swapping key and value in the dictionary.

The dictionary comprehension uses {arr[i]: i for i in range(len(arr))} to map elements to indices.

5fill in blank

hard

Fill all three blanks to create a list of squares of even numbers from the array.

DSA Python

arr = [1, 2, 3, 4, 5, 6]
squares = [[1] for [2] in arr if [3] % 2 == 0]

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

Ax**2

Bx

Dnum

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the loop and condition.

Not squaring the variable in the output expression.

The list comprehension squares each x in arr only if x is even.

Practice

1. What does the following code do?
for i in range(len(arr)):
  print(arr[i])
Assuming arr is a list of numbers.

easy

A. Prints only the first element of the array.

B. Prints all elements of the array from first to last.

C. Prints the indices of the array elements.

D. Prints the array in reverse order.

Solution

1. Step 1: Understand the loop range

   The loop runs from 0 to len(arr)-1, covering all valid indices of the array.

2. Step 2: Access and print each element

   Inside the loop, arr[i] accesses each element in order, printing them one by one.

3. Final Answer:

   Prints all elements of the array from first to last. -> Option B

4. Quick Check:

   Loop over indices prints elements in order [OK]

Hint: Loop over range(len(arr)) accesses all elements by index [OK]

Common Mistakes:

• Thinking it prints indices instead of elements
• Assuming it prints in reverse order
• Confusing range(len(arr)) with just range(arr)

2. Which of the following is the correct syntax to traverse all elements in a list arr using a for-each style loop in Python?

easy

A. for i to len(arr): print(arr[i])

B. for i in range(arr): print(i)

C. for element in arr: print(element)

D. for element in range(len(arr)): print(element)

Solution

1. Step 1: Identify correct for-each syntax

   Python uses for variable in iterable to loop over elements directly.

2. Step 2: Check each option

   for element in arr: print(element) uses for element in arr, which correctly loops over elements. Others have syntax errors or misuse range.

3. Final Answer:

   for element in arr: print(element) -> Option C

4. Quick Check:

   For-each loop uses 'for element in arr' [OK]

Hint: Use 'for element in arr' to loop elements directly [OK]

Common Mistakes:

• Using range(arr) instead of range(len(arr))
• Writing 'for i to len(arr)' which is invalid syntax
• Confusing indices with elements in for-each loops

3. What is the output of the following code?

arr = [10, 20, 30, 40]
for i in range(1, len(arr)):
   print(arr[i])

medium

A. 20 30 40

B. 40

C. 10 20 30

D. 10 20 30 40

Solution

1. Step 1: Analyze the loop range

   The loop runs from 1 to len(arr)-1, which is 1 to 3 for this array.

2. Step 2: Print elements at indices 1, 2, 3

   Elements are arr[1]=20, arr[2]=30, arr[3]=40, printed in order.

3. Final Answer:

   20 30 40 -> Option A

4. Quick Check:

   Range(1,4) prints elements from index 1 to 3 [OK]

Hint: range(1, len(arr)) skips first element [OK]

Common Mistakes:

• Including the first element (index 0) mistakenly
• Confusing range end as inclusive
• Printing indices instead of elements

4. Find the error in this code snippet that tries to print all elements of arr:

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

medium

A. No error, prints all elements correctly.

B. SyntaxError due to incorrect for loop syntax.

C. Prints elements in reverse order.

D. IndexError because i+1 goes out of range on last iteration.

Solution

1. Step 1: Check loop indices and access

   Loop runs i from 0 to 3 (len(arr)=4). Accessing arr[i+1] means indices 1 to 4.

2. Step 2: Identify out of range access

   When i=3, arr[4] is accessed, which is outside the array bounds causing IndexError.

3. Final Answer:

   IndexError because i+1 goes out of range on last iteration. -> Option D

4. Quick Check:

   Accessing arr[i+1] with i at last index causes error [OK]

Hint: Check index +1 can exceed array length [OK]

Common Mistakes:

• Ignoring index out of range error
• Thinking syntax is wrong instead
• Assuming it prints reversed elements

5. Given a list arr = [3, 5, 7, 9, 11], write code to print elements at even indices only using array traversal patterns. Which code snippet does this correctly?

hard

A. for i in range(0, len(arr), 2): print(arr[i])

B. for i in range(1, len(arr), 2): print(arr[i])

C. for element in arr: if element % 2 == 0: print(element)

D. for i in range(len(arr)): if i % 2 == 1: print(arr[i])

Solution

1. Step 1: Understand even index traversal

   Even indices are 0, 2, 4 for this list. Step size 2 starting at 0 covers these.

2. Step 2: Check each option

   for i in range(0, len(arr), 2): print(arr[i]) uses range(0, len(arr), 2) which correctly iterates over even indices. Others either use odd indices or check element values incorrectly.

3. Final Answer:

   for i in range(0, len(arr), 2): print(arr[i]) -> Option A

4. Quick Check:

   Step 2 in range picks even indices [OK]

Hint: Use range(start=0, end, step=2) for even indices [OK]

Common Mistakes:

• Using odd indices instead of even
• Checking element value parity instead of index
• Forgetting step argument in range