Spiral Matrix Traversal in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Spiral Traversal on 3x3 Matrix

What is the output of the spiral traversal on the following 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([[1,2,3],[4,5,6],[7,8,9]]))

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

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

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

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

💡 Hint

Follow the edges of the matrix clockwise, moving inward layer by layer.

The spiral traversal starts at the top-left corner and moves right, then down, then left, then up, shrinking the boundaries after each direction until all elements are visited.

❓ Predict Output

intermediate

Spiral Traversal Output for Rectangular Matrix

What is the spiral traversal output of this 3x4 matrix?

Matrix:
1 2 3 4
5 6 7 8
9 10 11 12

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([[1,2,3,4],[5,6,7,8],[9,10,11,12]]))

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

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

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

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

💡 Hint

Traverse the outer layer first, then move inward.

The traversal goes right across the top row, down the last column, left across the bottom row, and up the first column, then repeats for the inner layers.

🔧 Debug

advanced

Identify the Error in Spiral Traversal Code

The following code is intended to perform spiral traversal on a matrix. What error will it raise when run on a 1x1 matrix?

Matrix:
[[1]]

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
        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
    return result

print(spiral_traversal([[1]]))

A [1, 2, 4, 3]

B [] (empty list)

C IndexError

D Infinite loop

💡 Hint

Check the loop condition for small matrices.

The loop condition uses < instead of <=, so for 1x1 matrix the while loop never runs and returns empty list.

🚀 Application

advanced

Number of Elements Visited in Spiral Traversal

Given a 4x5 matrix, how many elements will be visited during a complete spiral traversal?

A 9

B 18

C 20

D 10

💡 Hint

Count all elements in the matrix.

Spiral traversal visits every element exactly once, so total elements visited equals rows * columns = 4 * 5 = 20.

🧠 Conceptual

expert

Why Use Four Pointers in Spiral Traversal?

In spiral traversal of a matrix, why do we maintain four pointers (top, bottom, left, right)?

A To keep track of the current boundaries of the unvisited matrix area

B To store the values of the first row, last row, first column, and last column

C To count the total number of elements visited in each direction

D To mark the visited elements in the matrix

💡 Hint

Think about how the traversal shrinks the area it covers after each pass.

The four pointers define the current edges of the matrix section that still needs to be traversed in spiral order. They move inward after each side is processed.

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