Spiral Matrix Traversal in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Spiral Traversal on 3x3 Matrix

What is the output of the following C# code that prints elements of a 3x3 matrix in spiral order?

DSA C#

int[,] matrix = {
  {1, 2, 3},
  {4, 5, 6},
  {7, 8, 9}
};

List<int> result = new List<int>();
int top = 0, bottom = 2, left = 0, right = 2;

while (top <= bottom && left <= right) {
    for (int i = left; i <= right; i++) result.Add(matrix[top, i]);
    top++;
    for (int i = top; i <= bottom; i++) result.Add(matrix[i, right]);
    right--;
    if (top <= bottom) {
        for (int i = right; i >= left; i--) result.Add(matrix[bottom, i]);
        bottom--;
    }
    if (left <= right) {
        for (int i = bottom; i >= top; i--) result.Add(matrix[i, left]);
        left++;
    }
}

Console.WriteLine(string.Join(", ", result));

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

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

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

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

💡 Hint

Follow the boundaries top, bottom, left, right and move inward in spiral.

The code traverses the matrix in layers: top row left to right, right column top to bottom, bottom row right to left, left column bottom to top, then moves inward. This produces the spiral order: 1,2,3,6,9,8,7,4,5.

❓ Predict Output

intermediate

Spiral Traversal Output for 2x4 Matrix

What is the output of the spiral traversal code on this 2x4 matrix?

DSA C#

int[,] matrix = {
  {1, 2, 3, 4},
  {5, 6, 7, 8}
};

List<int> result = new List<int>();
int top = 0, bottom = 1, left = 0, right = 3;

while (top <= bottom && left <= right) {
    for (int i = left; i <= right; i++) result.Add(matrix[top, i]);
    top++;
    for (int i = top; i <= bottom; i++) result.Add(matrix[i, right]);
    right--;
    if (top <= bottom) {
        for (int i = right; i >= left; i--) result.Add(matrix[bottom, i]);
        bottom--;
    }
    if (left <= right) {
        for (int i = bottom; i >= top; i--) result.Add(matrix[i, left]);
        left++;
    }
}

Console.WriteLine(string.Join(", ", result));

A 1, 2, 3, 4, 8, 7, 6, 5

B 1, 2, 3, 4, 5, 6, 7, 8

C 1, 5, 6, 7, 8, 4, 3, 2

D 1, 4, 8, 7, 6, 5, 2, 3

💡 Hint

Traverse top row, then right column, then bottom row reversed, then left column reversed.

The spiral order for this 2x4 matrix is: top row (1,2,3,4), right column down (8), bottom row reversed (7,6,5), left column up (none left). So output is 1,2,3,4,8,7,6,5.

🔧 Debug

advanced

Identify the Bug in Spiral Traversal Code

The following code attempts spiral traversal but produces incorrect output. What is the bug?

DSA C#

int[,] matrix = {
  {1, 2, 3},
  {4, 5, 6},
  {7, 8, 9}
};

List<int> result = new List<int>();
int top = 0, bottom = 2, left = 0, right = 2;

while (top < bottom && left < right) {
    for (int i = left; i <= right; i++) result.Add(matrix[top, i]);
    top++;
    for (int i = top; i <= bottom; i++) result.Add(matrix[i, right]);
    right--;
    for (int i = right; i >= left; i--) result.Add(matrix[bottom, i]);
    bottom--;
    for (int i = bottom; i >= top; i--) result.Add(matrix[i, left]);
    left++;
}

Console.WriteLine(string.Join(", ", result));

A The result list is not cleared before traversal causing duplicate output

B The for loops should iterate in reverse order for top and right boundaries

C The variables top and bottom are swapped in the code causing wrong indexing

D The while loop condition should use <= instead of < to include last row and column

💡 Hint

Check the loop condition carefully to ensure all rows and columns are included.

Using 'top < bottom && left < right' excludes the last row and column when top == bottom or left == right. Changing to 'top <= bottom && left <= right' ensures all elements are covered.

🧠 Conceptual

advanced

Number of Layers in Spiral Traversal

For an MxN matrix, how many layers (or rings) does the spiral traversal algorithm process?

A Math.Min(M, N) / 2 rounded up

B Math.Max(M, N) / 2 rounded down

C Math.Min(M, N) / 2 rounded down

D Math.Max(M, N) / 2 rounded up

💡 Hint

Each layer removes one row from top and bottom and one column from left and right.

Each layer corresponds to one ring around the matrix. The number of layers is half the smaller dimension rounded down because each layer consumes two rows or columns.

🚀 Application

expert

Final Spiral Traversal Output for 4x4 Matrix

What is the output of spiral traversal on this 4x4 matrix?

DSA C#

int[,] matrix = {
  {1, 2, 3, 4},
  {5, 6, 7, 8},
  {9, 10, 11, 12},
  {13, 14, 15, 16}
};

List<int> result = new List<int>();
int top = 0, bottom = 3, left = 0, right = 3;

while (top <= bottom && left <= right) {
    for (int i = left; i <= right; i++) result.Add(matrix[top, i]);
    top++;
    for (int i = top; i <= bottom; i++) result.Add(matrix[i, right]);
    right--;
    if (top <= bottom) {
        for (int i = right; i >= left; i--) result.Add(matrix[bottom, i]);
        bottom--;
    }
    if (left <= right) {
        for (int i = bottom; i >= top; i--) result.Add(matrix[i, left]);
        left++;
    }
}

Console.WriteLine(string.Join(", ", result));

A 1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10

B 1, 5, 9, 13, 14, 15, 16, 12, 8, 4, 3, 2, 6, 10, 11, 7

C 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16

D 1, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10, 2, 3

💡 Hint

Follow the spiral from outer layer to inner layer carefully.

The spiral order is top row left to right (1,2,3,4), right column top to bottom (8,12,16), bottom row right to left (15,14,13), left column bottom to top (9,5), then inner layer top row (6,7), right column (11), bottom row reversed (10).

Practice

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

easy

A. To visit all elements of a matrix in a spiral order starting from the top-left corner

B. To sort the matrix elements in ascending order

C. To find the maximum element in each row of the matrix

D. To transpose the matrix by swapping rows and columns

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means visiting matrix elements starting from the outer layer and moving inward in a circular pattern.

2. Step 2: Identify the traversal goal

   The goal is to visit every element exactly once in a spiral order, usually starting from the top-left corner.

3. Final Answer:

   To visit all elements of a matrix in a spiral order starting from the top-left corner -> Option A

4. Quick Check:

   Spiral traversal = visiting elements in spiral order [OK]

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

Common Mistakes:

• Confusing spiral traversal with sorting
• Thinking it only visits diagonal elements
• Assuming it transposes the matrix

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

easy

A. int top = 0, bottom = matrix.Length - 1, left = 0, right = matrix.GetLength(0) - 1;

B. int top = 1, bottom = matrix.Length, left = 1, right = matrix.Length;

C. int top = matrix.GetLength(1), bottom = 0, left = matrix.GetLength(0), right = 0;

D. int top = 0, bottom = matrix.GetLength(0) - 1, left = 0, right = matrix.GetLength(1) - 1;

Solution

1. Step 1: Understand matrix dimensions in C#

   matrix.GetLength(0) gives number of rows, matrix.GetLength(1) gives number of columns.

2. Step 2: Set boundaries correctly

   Top and left start at 0; bottom and right are last indices (length - 1) for rows and columns respectively.

3. Final Answer:

   int top = 0, bottom = matrix.GetLength(0) - 1, left = 0, right = matrix.GetLength(1) - 1; -> Option D

4. Quick Check:

   Correct boundary initialization = int top = 0, bottom = matrix.GetLength(0) - 1, left = 0, right = matrix.GetLength(1) - 1; [OK]

Hint: Use GetLength(0) for rows, GetLength(1) for columns [OK]

Common Mistakes:

• Using matrix.Length which is total elements, not rows or columns
• Swapping rows and columns in GetLength calls
• Starting boundaries at 1 instead of 0

3. What is the output of the following C# code snippet for spiral traversal?

int[,] matrix = {
  {1, 2, 3},
  {4, 5, 6},
  {7, 8, 9}
};

List<int> result = new List<int>();
int top = 0, bottom = 2, left = 0, right = 2;

while (top <= bottom && left <= right) {
  for (int i = left; i <= right; i++) result.Add(matrix[top, i]);
  top++;
  for (int i = top; i <= bottom; i++) result.Add(matrix[i, right]);
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--) result.Add(matrix[bottom, i]);
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--) result.Add(matrix[i, left]);
    left++;
  }
}
Console.WriteLine(string.Join(", ", result));

medium

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

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

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

D. 9, 8, 7, 6, 5, 4, 3, 2, 1

Solution

1. Step 1: Traverse top row left to right

   Elements added: 1, 2, 3; then top moves from 0 to 1.

2. Step 2: Traverse right column top to bottom

   Elements added: 6, 9; right moves from 2 to 1.

3. Step 3: Traverse bottom row right to left

   Elements added: 8, 7; bottom moves from 2 to 1.

4. Step 4: Traverse left column bottom to top

   Elements added: 4; left moves from 0 to 1.

5. Step 5: Repeat for inner layer

   Now top=1, bottom=1, left=1, right=1; add element 5.

6. Final Answer:

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

7. Quick Check:

   Spiral order of 3x3 matrix = 1, 2, 3, 6, 9, 8, 7, 4, 5 [OK]

Hint: Follow boundaries and shrink after each side [OK]

Common Mistakes:

• Not updating boundaries correctly after each side
• Skipping the inner elements
• Reversing traversal directions incorrectly

4. Identify the bug in this spiral traversal code snippet and select the fix:

int top = 0, bottom = matrix.GetLength(0), left = 0, right = matrix.GetLength(1) - 1;
while (top <= bottom && left <= right) {
  for (int i = left; i <= right; i++) Console.Write(matrix[top, i] + " ");
  top++;
  for (int i = top; i <= bottom; i++) Console.Write(matrix[i, right] + " ");
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--) Console.Write(matrix[bottom, i] + " ");
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--) Console.Write(matrix[i, left] + " ");
    left++;
  }
}

medium

A. Change right to matrix.GetLength(1) to include last column

B. Change bottom to matrix.GetLength(0) - 1 to avoid index out of range

C. Remove the if conditions checking top <= bottom and left <= right

D. Initialize top and left to 1 instead of 0

Solution

1. Step 1: Check boundary initialization

   bottom is set to matrix.GetLength(0), which is out of index range since indexing is zero-based.

2. Step 2: Correct bottom boundary

   bottom should be matrix.GetLength(0) - 1 to point to the last valid row index.

3. Final Answer:

   Change bottom to matrix.GetLength(0) - 1 to avoid index out of range -> Option B

4. Quick Check:

   Zero-based indexing means last index = length - 1 [OK]

Hint: Remember zero-based indexing means last index = length - 1 [OK]

Common Mistakes:

• Using length instead of length - 1 for last index
• Removing boundary checks causing infinite loops
• Starting indices at 1 instead of 0

5. Given a rectangular matrix of size 3x4:

int[,] 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, 9, 10, 11, 5, 6, 7

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

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

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

Solution

1. Step 1: Traverse top row left to right

   Elements: 1, 2, 3, 4; top moves from 0 to 1.

2. Step 2: Traverse right column top to bottom

   Elements: 8, 12; right moves from 3 to 2.

3. Step 3: Traverse bottom row right to left

   Elements: 11, 10, 9; bottom moves from 2 to 1.

4. Step 4: Traverse left column bottom to top

   Elements: 5; left moves from 0 to 1.

5. Step 5: Traverse inner row left to right

   Elements: 6, 7; top moves from 1 to 2.

6. Final Answer:

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

7. Quick Check:

   Spiral order of 3x4 matrix = 1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7 [OK]

Hint: Shrink boundaries after each side, watch rectangular shape [OK]

Common Mistakes:

• Mixing order of inner elements
• Skipping last inner row or column
• Confusing left and right boundaries