Spiral Matrix Traversal in DSA C - Time & Space Complexity

Time Complexity: Spiral Matrix Traversal

O(rows * cols)

Understanding Time Complexity

We want to understand how the time needed to traverse a matrix in a spiral 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.

void spiralTraversal(int matrix[][N], int rows, int cols) {
    int top = 0, bottom = rows - 1;
    int left = 0, right = cols - 1;
    while (top <= bottom && left <= right) {
        for (int i = left; i <= right; i++) printf("%d ", matrix[top][i]);
        top++;
        for (int i = top; i <= bottom; i++) printf("%d ", matrix[i][right]);
        right--;
        if (top <= bottom) {
            for (int i = right; i >= left; i--) printf("%d ", matrix[bottom][i]);
            bottom--;
        }
        if (left <= right) {
            for (int i = bottom; i >= top; i--) printf("%d ", matrix[i][left]);
            left++;
        }
    }
}
    

This code prints all elements of a matrix in a spiral order starting from the top-left corner.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each element of the matrix exactly once.
• How many times: Each element is printed once during the spiral traversal loops.

How Execution Grows With Input

As the matrix size grows, the number of elements to visit grows too.

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

Pattern observation: The operations grow proportionally to the total number of elements, which is the square of the matrix dimension.

Final Time Complexity

Time Complexity: O(rows * cols)

This means the time grows directly with the total number of elements in the matrix.

Common Mistake

[X] Wrong: "Because there are nested loops, the time complexity must be O(n²) regardless of matrix shape."

[OK] Correct: The loops do not multiply the work repeatedly; each element is visited once, so the total work is proportional to the number of elements, not squared again.

Interview Connect

Understanding how to analyze traversal of 2D structures like matrices is a key skill. It shows you can reason about loops and data size clearly, which helps 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. Traverse only the diagonal elements of the matrix

B. Traverse the matrix row by row from top to bottom

C. Traverse the matrix column by column from left to right

D. Traverse the matrix in a circular pattern from outer to inner layers

Solution

1. Step 1: Understand traversal pattern

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

2. Step 2: Compare options

   Only Traverse the matrix in a circular pattern from outer to inner layers describes moving in a circular pattern from outer to inner layers, which matches spiral traversal.

3. Final Answer:

   Traverse the matrix in a circular pattern from outer to inner layers -> Option D

4. Quick Check:

   Spiral traversal = circular outer to inner [OK]

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

Common Mistakes:

• Confusing spiral with row-wise traversal
• Thinking spiral means diagonal traversal
• Ignoring the inward movement after outer layer

2. Which of the following C code snippets correctly initializes variables for spiral traversal of an m x n matrix?

easy

A. int top = 0, bottom = m-1, left = 0, right = n-1;

B. int top = m-1, bottom = 0, left = n-1, right = 0;

C. int top = 0, bottom = n-1, left = 0, right = m-1;

D. int top = n-1, bottom = m-1, left = 0, right = 0;

Solution

1. Step 1: Identify correct boundary initialization

   For spiral traversal, top starts at 0, bottom at last row (m-1), left at 0, right at last column (n-1).

2. Step 2: Match with options

   int top = 0, bottom = m-1, left = 0, right = n-1; matches this initialization exactly; others swap or mix row and column indices incorrectly.

3. Final Answer:

   int top = 0, bottom = m-1, left = 0, right = n-1; -> Option A

4. Quick Check:

   Boundaries = top=0, bottom=m-1, left=0, right=n-1 [OK]

Hint: Top and left start at 0; bottom and right at max index [OK]

Common Mistakes:

• Swapping m and n in bottom or right
• Starting top or left at max index
• Confusing row and column boundaries

3. What is the output of the following C code snippet for spiral traversal of a 3x3 matrix?

int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
int top=0, bottom=2, left=0, right=2;
while(top <= bottom && left <= right) {
  for(int i=left; i<=right; i++) printf("%d ", matrix[top][i]);
  top++;
  for(int i=top; i<=bottom; i++) printf("%d ", matrix[i][right]);
  right--;
  if(top <= bottom) {
    for(int i=right; i>=left; i--) printf("%d ", matrix[bottom][i]);
    bottom--;
  }
  if(left <= right) {
    for(int i=bottom; i>=top; i--) printf("%d ", matrix[i][left]);
    left++;
  }
}

medium

A. 1 4 7 8 9 6 3 2 5

B. 1 2 3 6 9 8 7 4 5

C. 1 2 3 4 5 6 7 8 9

D. 1 3 5 7 9 2 4 6 8

Solution

1. Step 1: Trace first outer layer traversal

   Top row left to right: 1 2 3; then right column top to bottom: 6 9; bottom row right to left: 8 7; left column bottom to top: 4.

2. Step 2: Move inward and print center

   Now top=1, bottom=1, left=1, right=1; print center element matrix[1][1] = 5.

3. Final Answer:

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

4. Quick Check:

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

Hint: Follow boundaries and print edges clockwise [OK]

Common Mistakes:

• Skipping center element
• Mixing row and column indices
• Not updating boundaries correctly

4. Find the error in this spiral traversal code snippet for a 3x3 matrix:

int top=0, bottom=2, left=0, right=2;
while(top < bottom && left < right) {
  for(int i=left; i<=right; i++) printf("%d ", matrix[top][i]);
  top++;
  for(int i=top; i<=bottom; i++) printf("%d ", matrix[i][right]);
  right--;
  for(int i=right; i>=left; i--) printf("%d ", matrix[bottom][i]);
  bottom--;
  for(int i=bottom; i>=top; i--) printf("%d ", matrix[i][left]);
  left++;
}

medium

A. The variables top and bottom are swapped

B. The for loops should use < instead of <=

C. The while loop condition should use <= instead of <

D. The matrix indices are accessed incorrectly

Solution

1. Step 1: Analyze loop condition

   The condition top < bottom && left < right stops before the center element is processed in odd-sized matrices.

2. Step 2: Correct condition for full traversal

   Using <= allows the loop to run when top equals bottom or left equals right, ensuring center elements are included.

3. Final Answer:

   The while loop condition should use <= instead of < -> Option C

4. Quick Check:

   Use <= in while condition for complete spiral [OK]

Hint: Use <= in while to include center elements [OK]

Common Mistakes:

• Stopping traversal too early
• Changing for loop bounds incorrectly
• Swapping top and bottom variables

5. Given a 4x4 matrix:

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

Write the spiral traversal output if you start from the bottom-left corner and move clockwise inward.

hard

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

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

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

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

Solution

1. Step 1: Understand starting point and direction

   Start at bottom-left (matrix[3][0] = 13) and move clockwise: up, right, down, left, inward.

2. Step 2: Trace spiral clockwise from bottom-left

   Up column: 13 9 5 1; right row: 2 3 4; down column: 8 12 16; left row: 15 14 10 6; then inner spiral: 7 11.

3. Final Answer:

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

4. Quick Check:

   Start bottom-left clockwise spiral = 13 9 5 1 2 3 4 8 12 16 15 14 10 6 7 11 [OK]

Hint: Change start and direction, then trace edges inward [OK]

Common Mistakes:

• Starting from top-left instead of bottom-left
• Moving counter-clockwise instead of clockwise
• Mixing row and column indices