Spiral Matrix Traversal in DSA C# - Time & Space Complexity

Time Complexity: Spiral Matrix Traversal

O(n * m)

Understanding Time Complexity

We want to understand how the time needed to read all elements in a matrix in a spiral order grows as the matrix size increases.

How does the number of steps change when the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public List<int> SpiralOrder(int[][] matrix) {
    List<int> result = new List<int>();
    int top = 0, bottom = matrix.Length - 1;
    int left = 0, right = matrix[0].Length - 1;

    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++;
        }
    }
    return result;
}
    

This code reads all elements of a 2D matrix in a spiral order, starting from the top-left corner and moving inward.

Identify Repeating Operations

• Primary operation: Visiting each element of the matrix exactly once.
• How many times: Each element is accessed once in the nested loops inside the while loop.

How Execution Grows With Input

As the matrix size grows, the number of elements to visit grows too, so the steps increase proportionally.

Input Size (n x n)	Approx. Operations
10 x 10	100
100 x 100	10,000
1000 x 1000	1,000,000

Pattern observation: The operations grow roughly by the square of the matrix dimension because every element is visited once.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows directly with the total number of elements in the matrix, visiting each one once.

Common Mistake

[X] Wrong: "The spiral traversal takes more time than just reading the matrix row by row because of the multiple loops inside the while loop."

[OK] Correct: Even though there are multiple loops, each element is visited only once, so the total steps are still proportional to the number of elements.

Interview Connect

Understanding how to analyze traversal of 2D structures like matrices is a key skill. It shows you can reason about nested loops and how they relate to input size, which is common in many coding problems.

Self-Check

"What if the matrix was not rectangular but jagged (rows of different lengths)? How would the time complexity change?"

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