Spiral Matrix Traversal in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is Spiral Matrix Traversal?

It is a way to visit all elements of a matrix by moving in a spiral order, starting from the top-left corner and moving right, down, left, and up repeatedly until all elements are visited.

beginner

Which four directions are used in Spiral Matrix Traversal?

Right → Down → Left → Up, repeated in this order until all elements are covered.

intermediate

What variables are commonly used to track the boundaries in Spiral Matrix Traversal?

Top row index, bottom row index, left column index, and right column index are used to keep track of the current boundaries of the unvisited matrix area.

intermediate

Why do we update boundaries after each direction in Spiral Matrix Traversal?

Because after traversing one side (like top row), that row or column is fully visited and should not be visited again, so boundaries are moved inward to avoid repetition.

beginner

What is the time complexity of Spiral Matrix Traversal for an m x n matrix?

O(m * n), because every element in the matrix is visited exactly once.

In Spiral Matrix Traversal, which direction do we move first?

A Down

B Right

C Left

D Up

Traversal starts from the top-left corner and moves right first.

What happens to the top boundary after traversing the top row in Spiral Matrix Traversal?

A It stays the same

B It moves up by one row

C It moves down by one row

D It moves left by one column

After visiting the top row, the top boundary moves down to avoid revisiting.

Which condition stops the Spiral Matrix Traversal loop?

A When matrix is empty

B When top boundary < bottom boundary and left boundary < right boundary

C When all directions are visited once

D When top boundary > bottom boundary or left boundary > right boundary

Traversal ends when the boundaries cross, meaning all elements are visited.

What is the space complexity of Spiral Matrix Traversal if output is stored in a list?

A O(m * n)

B O(m + n)

C O(log(m * n))

D O(1)

The output list stores all elements, so space is proportional to total elements.

If the matrix has 3 rows and 4 columns, how many elements will Spiral Matrix Traversal visit?

A 12

B 10

C 7

D 9

3 rows * 4 columns = 12 elements total.

Explain how you would traverse a matrix in spiral order using boundary variables.

Think about moving right, down, left, and up while shrinking the boundaries.

Describe the stopping condition for a spiral matrix traversal loop and why it works.

When boundaries cross, no unvisited elements remain.

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 starting from the outer layer moving inward

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means moving around the edges of the matrix and then moving inward layer by layer.

2. Step 2: Compare options

   Only Traverse the matrix in a circular pattern starting from the outer layer moving inward 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 starting from the outer layer moving inward -> Option D

4. Quick Check:

   Spiral traversal = circular outer to inner [OK]

Hint: Spiral means circle around edges then go inside [OK]

Common Mistakes:

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

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

easy

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

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

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

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

Solution

1. Step 1: Understand boundary variables

   Top and bottom represent row boundaries, left and right represent column boundaries. For an m x n matrix, top starts at 0, bottom at m-1, left at 0, right at n-1.

2. Step 2: Check each option

   int top = 0, bottom = m - 1, left = 0, right = n - 1; correctly sets top=0, bottom=m-1, left=0, right=n-1. Others have reversed or incorrect assignments.

3. Final Answer:

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

4. Quick Check:

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

Hint: Top/left start at 0; bottom/right at size minus one [OK]

Common Mistakes:

• Swapping m and n in boundary initialization
• Starting boundaries at 1 instead of 0
• Confusing row and column boundaries

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

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++)
    std::cout << matrix[top][i] << " ";
  top++;
  for (int i = top; i <= bottom; i++)
    std::cout << matrix[i][right] << " ";
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--)
      std::cout << matrix[bottom][i] << " ";
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--)
      std::cout << matrix[i][left] << " ";
    left++;
  }
}

medium

A. 1 2 3 4 5 6 7 8 9

B. 1 4 7 8 9 6 3 2 5

C. 1 2 3 6 9 8 7 4 5

D. 9 8 7 6 5 4 3 2 1

Solution

1. Step 1: Trace the spiral traversal

   First top row: 1 2 3
   Right column down: 6 9
   Bottom row left: 8 7
   Left column up: 4
   Next inner layer top row: 5

2. Step 2: Combine all elements in order

   Combined output: 1 2 3 6 9 8 7 4 5

3. Final Answer:

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

4. Quick Check:

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

Hint: Follow edges clockwise, then move inward [OK]

Common Mistakes:

• Printing rows or columns in wrong order
• Skipping inner elements like 5
• Repeating elements or missing boundaries update

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

int top = 0, bottom = m - 1, left = 0, right = n - 1;
while (top < bottom && left < right) {
  for (int i = left; i <= right; i++) cout << matrix[top][i] << " ";
  top++;
  for (int i = top; i <= bottom; i++) cout << matrix[i][right] << " ";
  right--;
  for (int i = right; i >= left; i--) cout << matrix[bottom][i] << " ";
  bottom--;
  for (int i = bottom; i >= top; i--) cout << matrix[i][left] << " ";
  left++;
}

medium

A. The while loop condition should use <= instead of < to include all layers

B. The for loops should use < instead of <= to avoid out-of-bound errors

C. The variables top, bottom, left, right are not initialized correctly

D. The code misses printing the diagonal elements

Solution

1. Step 1: Analyze loop condition

   The while loop uses 'top < bottom && left < right', which skips the last layer when top == bottom or left == right.

2. Step 2: Correct condition for full traversal

   Changing to 'top <= bottom && left <= right' ensures the last single row or column is included in traversal.

3. Final Answer:

   The while loop condition should use <= instead of < to include all layers -> Option A

4. Quick Check:

   Use <= in loop condition to cover all layers [OK]

Hint: Use <= in while condition to cover last layer [OK]

Common Mistakes:

• Using < causes missing last row or column
• Changing for loops to < causes skipping elements
• Not updating boundaries after each traversal

5. Given a non-square matrix:

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

Which is the correct spiral traversal output?

hard

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

B. 1 2 3 4 8 12 11 10 9 5 6 7 0

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

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

Solution

1. Step 1: Trace spiral on 3x4 matrix

   Top row: 1 2 3 4
   Right column down: 8 12
   Bottom row left: 11 10 9
   Left column up: 5
   Inner top row: 6 7

2. Step 2: Combine elements in order

   Combined output: 1 2 3 4 8 12 11 10 9 5 6 7

3. Final Answer:

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

4. Quick Check:

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

Hint: Follow edges carefully, handle rectangular shape [OK]

Common Mistakes:

• Adding extra elements like 0
• Mixing inner elements order
• Swapping 6 and 7 positions