Spiral Matrix Traversal in DSA Python - Time & Space Complexity

Time Complexity: Spiral Matrix Traversal

O(n * m)

Understanding Time Complexity

We want to understand how the time needed to traverse a matrix in a spiral way grows as the matrix size increases.

Specifically, how does the number of steps change when the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code collects all elements of a 2D 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 visited once during the spiral traversal loops.

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 with the total number of elements, so if the matrix size doubles in each dimension, the operations increase by about four times.

Final Time Complexity

Time Complexity: O(n * m)

This means the time to complete the spiral traversal grows linearly with the total number of elements in the matrix.

Common Mistake

[X] Wrong: "The spiral traversal has nested loops that multiply to a higher time complexity like O(n² * m²)."

[OK] Correct: Although there are nested loops, each element is visited only once, so the total steps are just the number of elements, not multiplied repeatedly.

Interview Connect

Understanding how to analyze traversal of 2D structures like matrices is a key skill that shows you can reason about loops and data size clearly.

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 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