Array Traversal Patterns in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Nested Loop Traversal

What is the output of the following code that traverses a 2D array row-wise?

DSA Python

arr = [[1, 2], [3, 4]]
result = []
for row in arr:
    for val in row:
        result.append(val)
print(result)

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

B [1, 3, 2, 4]

C [1, 2, 3, 4]

D [4, 3, 2, 1]

💡 Hint

Think about how nested loops go through each element in order.

The outer loop goes row by row, and the inner loop goes through each element in the row, so the elements are added in order: 1, 2, 3, 4.

❓ Predict Output

intermediate

Output of Reverse Array Traversal

What is the output of this code that traverses an array backwards?

DSA Python

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

A [40, 30, 20, 10]

B [10, 20, 30, 40]

C [40, 20, 30, 10]

D [30, 40, 20, 10]

💡 Hint

The loop starts from the last index and goes to the first.

The loop goes from the last element to the first, so the result is the reversed array.

🔧 Debug

advanced

Identify the Error in Diagonal Traversal

What error does this code produce when trying to traverse the main diagonal of a square matrix?

DSA Python

matrix = [[1,2,3],[4,5,6],[7,8,9]]
for i in range(len(matrix)):
    print(matrix[i][i+1])

A No error, prints diagonal elements

B TypeError: unsupported operand type(s)

C SyntaxError: invalid syntax

D IndexError: list index out of range

💡 Hint

Check the index used inside the loop for accessing elements.

The code tries to access matrix[i][i+1], which goes out of range when i is at the last index.

🧠 Conceptual

advanced

Number of Elements Visited in Zigzag Traversal

Given a 3x3 matrix, how many elements are visited if we traverse it in a zigzag pattern row-wise (left to right on first row, right to left on second, and so on)?

A 9

B 6

C 3

D 12

💡 Hint

Zigzag means visiting every element but changing direction each row.

All elements are visited once, just the order changes, so total is 9.

🚀 Application

expert

Output After Spiral Traversal of 3x3 Matrix

What is the output list after performing a spiral traversal on this 3x3 matrix? matrix = [[1,2,3],[4,5,6],[7,8,9]]

DSA Python

def spiral_traversal(matrix):
    result = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1
    while left <= right and top <= bottom:
        for i in range(left, right + 1):
            result.append(matrix[top][i])
        top += 1
        for i in range(top, bottom + 1):
            result.append(matrix[i][right])
        right -= 1
        if top <= bottom:
            for i in range(right, left - 1, -1):
                result.append(matrix[bottom][i])
            bottom -= 1
        if left <= right:
            for i in range(bottom, top - 1, -1):
                result.append(matrix[i][left])
            left += 1
    return result

print(spiral_traversal(matrix))

A [9, 8, 7, 6, 5, 4, 3, 2, 1]

B [1, 2, 3, 6, 9, 8, 7, 4, 5]

C [1, 2, 3, 4, 5, 6, 7, 8, 9]

D [1, 4, 7, 8, 9, 6, 3, 2, 5]

💡 Hint

Spiral traversal goes around the edges inward in a clockwise direction.

The spiral order visits top row left to right, right column top to bottom, bottom row right to left, left column bottom to top, then repeats inward.

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