Spiral Matrix Traversal in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Spiral Traversal on 3x3 Matrix

What is the output of the following C++ code that prints the spiral traversal of a 3x3 matrix?

DSA C++

int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
int top = 0, bottom = 2, left = 0, right = 2;
std::vector<int> result;
while (top <= bottom && left <= right) {
  for (int i = left; i <= right; i++) result.push_back(matrix[top][i]);
  top++;
  for (int i = top; i <= bottom; i++) result.push_back(matrix[i][right]);
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--) result.push_back(matrix[bottom][i]);
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--) result.push_back(matrix[i][left]);
    left++;
  }
}
for (int x : result) std::cout << x << " ";

A [1, 2, 3, 6, 9, 8, 7, 4, 5]

B [1, 2, 3, 6, 9, 7, 8, 4, 5]

C [1, 2, 3, 6, 9, 8, 7, 5, 4]

D [1, 2, 3, 5, 6, 9, 8, 7, 4]

💡 Hint

Follow the spiral order: top row left to right, right column top to bottom, bottom row right to left, left column bottom to top.

The code traverses the matrix in a spiral order starting from the top-left corner. It moves right, down, left, and up, shrinking the boundaries after each direction until all elements are visited.

❓ Predict Output

intermediate

Spiral Traversal Output for 2x4 Matrix

What is the output of the spiral traversal for this 2x4 matrix?

DSA C++

int matrix[2][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}};
int top = 0, bottom = 1, left = 0, right = 3;
std::vector<int> result;
while (top <= bottom && left <= right) {
  for (int i = left; i <= right; i++) result.push_back(matrix[top][i]);
  top++;
  for (int i = top; i <= bottom; i++) result.push_back(matrix[i][right]);
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--) result.push_back(matrix[bottom][i]);
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--) result.push_back(matrix[i][left]);
    left++;
  }
}
for (int x : result) std::cout << x << " ";

A [1, 2, 3, 4, 8, 6, 7, 5]

B [1, 2, 3, 4, 8, 7, 6, 5]

C [1, 2, 3, 4, 7, 8, 6, 5]

D [1, 2, 3, 4, 5, 6, 7, 8]

💡 Hint

Remember to move right, down, left, then up, adjusting boundaries each time.

The spiral traversal visits top row left to right, then right column down, then bottom row right to left, and finally left column up, covering all elements.

🔧 Debug

advanced

Identify the Bug in Spiral Traversal Code

The following code is intended to print the spiral traversal of a matrix but produces incorrect output. What is the bug?

DSA C++

int matrix[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
int top = 0, bottom = 2, left = 0, right = 2;
std::vector<int> result;
while (top < bottom && left < right) {
  for (int i = left; i <= right; i++) result.push_back(matrix[top][i]);
  top++;
  for (int i = top; i <= bottom; i++) result.push_back(matrix[i][right]);
  right--;
  for (int i = right; i >= left; i--) result.push_back(matrix[bottom][i]);
  bottom--;
  for (int i = bottom; i >= top; i--) result.push_back(matrix[i][left]);
  left++;
}
for (int x : result) std::cout << x << " ";

A The matrix should be traversed column-wise first, not row-wise.

B The for loops should use '<' instead of '<=' in their conditions.

C The variables 'top' and 'left' should be initialized to 1 instead of 0.

D The while loop condition should be 'top <= bottom && left <= right' instead of 'top < bottom && left < right'.

💡 Hint

Check the loop condition carefully to ensure all elements are included.

Using '<' excludes the last row and column when top equals bottom or left equals right, missing elements. Using '<=' ensures all elements are covered.

🧠 Conceptual

advanced

Time Complexity of Spiral Matrix Traversal

What is the time complexity of traversing an MxN matrix in spiral order?

A O(M * N)

B O(M + N)

C O(log(M * N))

D O(M^2 + N^2)

💡 Hint

Consider how many elements are visited in total.

Each element of the matrix is visited exactly once during the spiral traversal, so the time complexity is proportional to the total number of elements, M times N.

🚀 Application

expert

Spiral Traversal Output for Non-Square Matrix 4x3

Given the 4x3 matrix below, what is the spiral traversal output?

DSA C++

int matrix[4][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}};
int top = 0, bottom = 3, left = 0, right = 2;
std::vector<int> result;
while (top <= bottom && left <= right) {
  for (int i = left; i <= right; i++) result.push_back(matrix[top][i]);
  top++;
  for (int i = top; i <= bottom; i++) result.push_back(matrix[i][right]);
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--) result.push_back(matrix[bottom][i]);
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--) result.push_back(matrix[i][left]);
    left++;
  }
}
for (int x : result) std::cout << x << " ";

A [1, 2, 3, 6, 9, 12, 10, 11, 7, 4, 5, 8]

B [1, 2, 3, 6, 9, 12, 11, 10, 7, 4, 8, 5]

C [1, 2, 3, 6, 9, 12, 11, 10, 7, 4, 5, 8]

D [1, 2, 3, 6, 9, 12, 11, 10, 7, 5, 4, 8]

💡 Hint

Follow the spiral carefully, adjusting boundaries after each direction.

The spiral traversal visits elements in order: top row left to right, right column top to bottom, bottom row right to left, left column bottom to top, repeating until all elements are visited.

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