Spiral Matrix Traversal in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the starting row index for spiral traversal.

DSA C++

int top = [1];

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Dmatrix.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 as the starting index.

Using matrix.size() which is out of bounds.

The top row index starts at 0 because matrix indexing begins at 0.

2fill in blank

medium

Complete the code to move the left boundary right after traversing the top row.

DSA C++

left [1] 1;

Drag options to blanks, or click blank then click option'

A-=

B+=

C*=

D/=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-=' which moves boundary left instead of right.

Using '*=' or '/=' which are incorrect for index movement.

We increase the left boundary by 1 to move it right after traversing the top row.

3fill in blank

hard

Fix the error in the loop condition to traverse the bottom row from right to left.

DSA C++

for (int i = right; i [1] left; i--) {
    result.push_back(matrix[bottom][i]);
}

Drag options to blanks, or click blank then click option'

A==

B<=

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' which causes the loop to run incorrectly.

Using '==' which only runs one iteration.

The loop must continue while i is greater than or equal to left to include the left boundary element.

4fill in blank

hard

Fill both blanks to update the top and bottom boundaries after traversing rows.

DSA C++

top [1] 1;
bottom [2] 1;

Drag options to blanks, or click blank then click option'

A+=

B-=

C*=

D/=

Attempts:

3 left

💡 Hint

Common Mistakes

Decrementing top instead of incrementing.

Incrementing bottom instead of decrementing.

After traversing the top row, we move top down by adding 1. After traversing the bottom row, we move bottom up by subtracting 1.

5fill in blank

hard

Fill all three blanks to complete the while loop condition for spiral traversal.

DSA C++

while (top [1] bottom && left [2] right) {
    // traversal code
    left [3] 1;
}

Drag options to blanks, or click blank then click option'

A<=

B>=

C+=

D-=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' or '>=' which causes loop to end prematurely.

Using '<' which excludes boundary elements.

The loop continues while top is less than or equal to bottom and left is less than or equal to right. We increment left boundary inside the loop.

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