Spiral Matrix Traversal in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the starting row index for spiral traversal.

DSA Python

top = [1]

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

A0

B-1

C1

Dlen(matrix)

Attempts:

3 left

💡 Hint

Common Mistakes

Starting top at 1 instead of 0 causes skipping the first row.

Using negative indices here confuses the traversal.

The top row index starts at 0 because matrix indexing begins from 0.

2fill in blank

medium

Complete the code to move the left boundary right after traversing the left column.

DSA Python

left [1]= 1

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Decreasing left boundary instead of increasing it.

Using multiplication or division operators here.

After traversing the left column, we move the left boundary right by increasing it by 1.

3fill in blank

hard

Fix the error in the loop condition to continue spiral traversal while boundaries are valid.

DSA Python

while top [1] bottom and left <= right:

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

A>

B<=

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < causes missing the last row.

Using > or >= reverses the logic and breaks the loop.

The loop should continue while the top boundary is less than or equal to the bottom boundary to cover all rows.

4fill in blank

hard

Fill both blanks to correctly traverse the top row from left to right.

DSA Python

for i in range([1], [2] + 1):
    result.append(matrix[top][i])

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

Aleft

Bright

Ctop

Dbottom

Attempts:

3 left

💡 Hint

Common Mistakes

Using top or bottom as loop range indices instead of left and right.

Not adding +1 to include the right boundary.

We traverse the top row from the left boundary to the right boundary inclusive.

5fill in blank

hard

Fill all three blanks to traverse the right column from top+1 to bottom and then update the right boundary.

DSA Python

for i in range([1] + 1, [2] + 1):
    result.append(matrix[i][[3]])
right -= 1

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

Atop

Bbottom

Cright

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from top instead of top + 1 causes duplicate elements.

Using left instead of right for the column index.

Not updating the right boundary after traversal.

Traverse from the row below top to bottom along the right column, then move the right boundary left.

Practice

1. What is the main idea behind Spiral Matrix Traversal?

easy

A. Traverse the matrix in a spiral order starting from the top-left corner

B. Traverse the matrix row by row from left to right only

C. Traverse the matrix column by column from top to bottom only

D. Traverse the matrix diagonally from top-left to bottom-right

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means moving around the matrix edges and gradually moving inward in a circular pattern.

2. Step 2: Identify starting point and direction

   It starts at the top-left corner and moves right, then down, then left, then up, repeating until all elements are visited.

3. Final Answer:

   Traverse the matrix in a spiral order starting from the top-left corner -> Option A

4. Quick Check:

   Spiral traversal = circular edge-wise movement [OK]

Hint: Think of peeling an onion layer by layer in a circle [OK]

Common Mistakes:

• Confusing spiral with row-wise or column-wise traversal
• Starting traversal from the wrong corner
• Ignoring direction changes after each edge

2. Which of the following Python code snippets correctly initializes the boundaries for spiral traversal of a matrix named matrix?

easy

A. top = 1; bottom = len(matrix); left = 1; right = len(matrix[0])

B. top = 0; bottom = len(matrix) - 1; left = 0; right = len(matrix[0]) - 1

C. top = 0; bottom = len(matrix); left = 0; right = len(matrix[0])

D. top = 1; bottom = len(matrix) - 1; left = 1; right = len(matrix[0]) - 1

Solution

1. Step 1: Understand matrix boundaries

   For a matrix, top and left start at 0, bottom is last row index (length - 1), right is last column index (length - 1).

2. Step 2: Check each option

   top = 0; bottom = len(matrix) - 1; left = 0; right = len(matrix[0]) - 1 correctly sets top=0, bottom=len(matrix)-1, left=0, right=len(matrix[0])-1 which matches the matrix indices.

3. Final Answer:

   top = 0; bottom = len(matrix) - 1; left = 0; right = len(matrix[0]) - 1 -> Option B

4. Quick Check:

   Boundary indices start at 0 and end at length-1 [OK]

Hint: Remember Python indexing starts at 0, so subtract 1 for last index [OK]

Common Mistakes:

• Using length instead of length - 1 for bottom and right
• Starting top or left at 1 instead of 0
• Mixing row and column lengths

3. What is the output of the following Python code?

matrix = [
  [1, 2, 3],
  [4, 5, 6],
  [7, 8, 9]
]
result = []
top, bottom = 0, 2
left, right = 0, 2
while top <= bottom and left <= right:
    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
    for i in range(right, left - 1, -1):
        result.append(matrix[bottom][i])
    bottom -= 1
    for i in range(bottom, top - 1, -1):
        result.append(matrix[i][left])
    left += 1
print(result)

medium

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

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

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

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

Solution

1. Step 1: Traverse top row left to right

   Elements: 1, 2, 3 added to result.

2. Step 2: Traverse right column top to bottom

   Elements: 6, 9 added to result.

3. Step 3: Traverse bottom row right to left

   Elements: 8, 7 added to result.

4. Step 4: Traverse left column bottom to top

   Element: 4 added to result.

5. Step 5: Inner layer traversal

   Only element 5 remains, added last.

6. Final Answer:

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

7. Quick Check:

   Spiral order = [1,2,3,6,9,8,7,4,5] [OK]

Hint: Follow edges clockwise, then move inward layer by layer [OK]

Common Mistakes:

• Skipping the last inner element
• Repeating elements or missing elements
• Wrong direction in bottom or left traversal

4. The following code attempts spiral traversal but produces an error. What is the main bug?

matrix = [[1,2,3],[4,5,6],[7,8,9]]
top, bottom = 0, 2
left, right = 0, 2
result = []
while top <= bottom and left <= right:
    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
    for i in range(right, left - 1, -1):
        result.append(matrix[bottom][i])
    bottom -= 1
    for i in range(bottom, top - 1, -1):
        result.append(matrix[i][left])
    left += 1
print(result)

medium

A. The code uses incorrect range in the first for loop

B. The code uses wrong indices for left and right boundaries

C. The code does not check if top <= bottom before traversing bottom row

D. The code misses incrementing the left boundary

Solution

1. Step 1: Analyze loop conditions

   The while loop checks top <= bottom and left <= right before each full cycle.

2. Step 2: Identify missing checks inside loop

   After updating top and right, the code traverses bottom row and left column without checking if top <= bottom or left <= right again, causing index errors when boundaries cross.

3. Final Answer:

   The code does not check if top <= bottom before traversing bottom row -> Option C

4. Quick Check:

   Boundary checks needed before each edge traversal [OK]

Hint: Check boundaries before each edge traversal inside the loop [OK]

Common Mistakes:

• Assuming one boundary check per loop is enough
• Not updating boundaries correctly
• Ignoring empty or single-row/column matrices

5. Given a rectangular matrix of size 3x4:

matrix = [
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9, 10, 11, 12]
]

What is the spiral traversal output?

hard

A. [1, 2, 3, 4, 8, 12, 11, 10, 9, 6, 5, 7]

B. [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 7, 6]

C. ]6 ,7 ,5 ,9 ,01 ,11 ,21 ,8 ,4 ,3 ,2 ,1[

D. [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

Solution

1. Step 1: Traverse top row left to right

   Elements: 1, 2, 3, 4 added.

2. Step 2: Traverse right column top to bottom

   Elements: 8, 12 added.

3. Step 3: Traverse bottom row right to left

   Elements: 11, 10, 9 added.

4. Step 4: Traverse left column bottom to top

   Element: 5 added.

5. Step 5: Traverse inner row left to right

   Elements: 6, 7 added.

6. Final Answer:

   [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7] -> Option D

7. Quick Check:

   Spiral order covers all elements layer by layer [OK]

Hint: Follow edges clockwise, then move inward carefully for rectangular matrices [OK]

Common Mistakes:

• Mixing order of inner elements
• Skipping or repeating elements in inner layers
• Confusing left column traversal direction