Spiral Matrix Traversal in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Java

int top = [1];

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

A-1

B1

C0

Dmatrix.length

Attempts:

3 left

💡 Hint

Common Mistakes

Starting top at 1 causes skipping the first row.

Using negative index causes errors.

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

2fill in blank

medium

Complete the code to update the left column index after traversing the top row.

DSA Java

left = [1] + 1;

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

Aleft

Bright

Ctop

Dbottom

Attempts:

3 left

💡 Hint

Common Mistakes

Using right instead of left causes wrong boundary update.

Not updating left causes infinite loops.

After traversing the top row, we move the left boundary right by increasing left by 1.

3fill in blank

hard

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

DSA Java

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

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

Aright

Bleft

Cbottom

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using top or bottom as loop boundary causes index errors.

Using right as lower bound causes infinite loop.

The bottom row is traversed from right to left, so the loop runs while i >= left.

4fill in blank

hard

Fill both blanks to correctly update the boundaries after traversing the right column.

DSA Java

for (int i = top; i <= bottom; i++) {
    result.add(matrix[i][right]);
}
right = [1] - 1;
top = [2] + 1;

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

Aright

Bbottom

Ctop

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping top and bottom updates causes wrong boundaries.

Not updating boundaries causes infinite loops.

After traversing the right column, we move the right boundary left by 1 and move the top boundary down by 1.

5fill in blank

hard

Fill all three blanks to complete the while loop condition and the final boundary update after traversing the left column.

DSA Java

while (top <= [1] && left <= [2]) {
    // traversal code
    for (int i = bottom; i >= top; i--) {
        result.add(matrix[i][left]);
    }
    left = [3] + 1;
}

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

Abottom

Bright

Cleft

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variables in loop condition causes incorrect traversal.

Not updating left boundary causes infinite loops.

The loop continues while top is less or equal to bottom and left is less or equal to right. After traversing the left column, left boundary moves right by 1.

Practice

1. What is the main idea behind Spiral Matrix Traversal?

easy

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

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

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

D. Traverse only the diagonal elements of the matrix

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 A

4. Quick Check:

   Spiral traversal = circular outer to inner [OK]

Hint: Spiral means circle inward from edges [OK]

Common Mistakes:

• Confusing spiral with row-wise traversal
• Thinking spiral means diagonal only
• Mixing column-wise with spiral

2. Which of the following Java code snippets correctly initializes the boundaries for spiral traversal of a matrix int[][] matrix?

easy

A. int top = 1, bottom = matrix.length, left = 1, right = matrix[0].length;

B. int top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1;

C. int top = 0, bottom = matrix[0].length - 1, left = 0, right = matrix.length - 1;

D. int top = matrix.length, bottom = 0, left = matrix[0].length, right = 0;

Solution

1. Step 1: Identify matrix dimensions

   For a 2D array, rows = matrix.length, columns = matrix[0].length.

2. Step 2: Set boundaries correctly

   Top and left start at 0, bottom is last row index (length-1), right is last column index (length-1). int top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1; matches this.

3. Final Answer:

   int top = 0, bottom = matrix.length - 1, left = 0, right = matrix[0].length - 1; -> Option B

4. Quick Check:

   Boundaries start at 0 and end at length-1 [OK]

Hint: Boundaries start at 0, end at length-1 [OK]

Common Mistakes:

• Using 1-based indices instead of 0-based
• Swapping rows and columns in boundaries
• Setting bottom or right greater than max index

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

int[][] matrix = {
  {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++) System.out.print(matrix[top][i] + " ");
  top++;
  for (int i = top; i <= bottom; i++) System.out.print(matrix[i][right] + " ");
  right--;
  if (top <= bottom) {
    for (int i = right; i >= left; i--) System.out.print(matrix[bottom][i] + " ");
    bottom--;
  }
  if (left <= right) {
    for (int i = bottom; i >= top; i--) System.out.print(matrix[i][left] + " ");
    left++;
  }
}

medium

A. 1 4 7 8 9 6 3 2 5

B. 1 2 3 4 5 6 7 8 9

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 first outer layer

   Top row: 1 2 3, right column down: 6 9, bottom row left: 8 7, left column up: 4

2. Step 2: Trace inner layer

   Now top=1, bottom=1, left=1, right=1, print center element: 5

3. Final Answer:

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

4. Quick Check:

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

Hint: Follow boundaries and print edges inward [OK]

Common Mistakes:

• Printing rows or columns in wrong order
• Missing center element in odd-sized matrix
• Not updating boundaries correctly

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

int top = 0, bottom = matrix.length - 1;
int left = 0, right = matrix[0].length - 1;
while (top < bottom && left < right) {
  for (int i = left; i <= right; i++) System.out.print(matrix[top][i] + " ");
  top++;
  for (int i = top; i <= bottom; i++) System.out.print(matrix[i][right] + " ");
  right--;
  for (int i = right; i >= left; i--) System.out.print(matrix[bottom][i] + " ");
  bottom--;
  for (int i = bottom; i >= top; i--) System.out.print(matrix[i][left] + " ");
  left++;
}

medium

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

B. The for loops should use < instead of <= to avoid index errors

C. The boundaries are initialized incorrectly

D. The print statements should be inside if conditions to avoid duplicates

Solution

1. Step 1: Analyze while loop condition

   Using < excludes the case when top == bottom or left == right, missing last layer in odd-sized matrices.

2. Step 2: Correct condition to include all layers

   Changing to <= ensures the center or last layer is processed.

3. Final Answer:

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

4. Quick Check:

   Use <= in while to cover all layers [OK]

Hint: Use <= in while to cover all matrix layers [OK]

Common Mistakes:

• Using < excludes center element in odd matrices
• Not updating boundaries after each loop
• Missing if checks for inner loops

5. Given a rectangular matrix of size 3x4:

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

What is the spiral traversal output?

hard

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

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

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

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

Solution

1. Step 1: Traverse top row left to right

   Print 1 2 3 4

2. Step 2: Traverse right column top to bottom

   Print 8 12

3. Step 3: Traverse bottom row right to left

   Print 11 10 9

4. Step 4: Traverse left column bottom to top

   Print 5

5. Step 5: Traverse inner row left to right

   Print 6 7

6. Final Answer:

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

7. Quick Check:

   Follow edges inward for rectangular matrix [OK]

Hint: Print edges clockwise, move inward layer by layer [OK]

Common Mistakes:

• Mixing order of inner elements
• Skipping last inner row
• Confusing left column traversal direction