Spiral Matrix Traversal in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C#

int top = [1];

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

A1

B-1

C0

Dmatrix.Length

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 for the starting row index.

Using matrix.Length which is the total number of rows, not the starting index.

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

2fill in blank

medium

Complete the code to initialize the ending column index for spiral traversal.

DSA C#

int right = [1] - 1;

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

Amatrix.Length - 1

Bmatrix[0].Length

Cmatrix.Length

Dmatrix[0].Length + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using matrix.Length which gives number of rows, not columns.

Adding 1 instead of subtracting 1.

The right column index is the number of columns minus one, so we use matrix[0].Length - 1.

3fill in blank

hard

Fix the error in the loop condition to traverse the top row from left to right.

DSA C#

for (int i = left; i [1]= right; i++)
{
    result.Add(matrix[top][i]);
}

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

A>

B>=

C<

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < which excludes the last column.

Using > or >= which causes the loop not to run.

The loop should run while i is less than or equal to right to include the last column.

4fill in blank

hard

Fill both blanks to traverse the right column from top to bottom.

DSA C#

for (int i = [1]; i [2]= bottom; i++)
{
    result.Add(matrix[i][right]);
}

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

Atop + 1

Bbottom - 1

C<=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from top instead of top + 1.

Using < which excludes the bottom row.

Start from the row below top (top + 1) and go up to bottom inclusive (i <= bottom).

5fill in blank

hard

Fill all three blanks to traverse the bottom row from right to left.

DSA C#

for (int i = [1]; i [2] [3]; i--)
{
    result.Add(matrix[bottom][i]);
}

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

Aright - 1

B>=

Cleft

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from right instead of right - 1.

Using <= or < which is wrong for decrementing loop.

Using left + 1 instead of left.

Start from right - 1, go while i is greater than or equal to left, decrement i each time.

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