Spiral Matrix Traversal in DSA Java - Build from Scratch

Spiral Matrix Traversal

📖 Scenario: You have a 2D grid of numbers like a map or a table. You want to read all the numbers starting from the top-left corner and moving around the edges in a spiral, going inward until all numbers are read.

🎯 Goal: Build a Java program that takes a fixed 3x3 matrix and prints its elements in spiral order.

📋 What You'll Learn

Create a 3x3 integer matrix with exact values

Use variables to track the boundaries of the matrix

Use a loop to traverse the matrix in spiral order

Print the spiral order elements separated by spaces

💡 Why This Matters

🌍 Real World

Spiral traversal is useful in image processing, games, and reading data stored in layers.

💼 Career

Understanding matrix traversal helps in coding interviews and solving problems involving grids or 2D data.

1

Create the 3x3 matrix

Create a 2D integer array called matrix with these exact values: {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}

DSA Java

// Your code here

Hint

Use int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; inside the main method.

2

Set up boundary variables

Add four integer variables inside main: top set to 0, bottom set to 2, left set to 0, and right set to 2

DSA Java

public class SpiralMatrix {
    public static void main(String[] args) {
        int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        // Your code here
    }
}

Hint

These variables will help track the edges of the matrix as you move inward.

3

Traverse the matrix in spiral order

Add a while loop that runs while left <= right and top <= bottom. Inside the loop, use for loops to traverse from left to right on the top row, top to bottom on the right column, right to left on the bottom row, and bottom to top on the left column. Update top, bottom, left, and right accordingly. Collect the elements in a StringBuilder called result separated by spaces.

DSA Java

public class SpiralMatrix {
    public static void main(String[] args) {
        int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        int top = 0;
        int bottom = 2;
        int left = 0;
        int right = 2;
        // Add while loop and for loops to traverse spiral
    }
}

Hint

Use four loops inside the while loop to cover all four sides of the current matrix boundary.

4

Print the spiral order

Add a System.out.println statement to print the result string trimmed of trailing spaces.

DSA Java

public class SpiralMatrix {
    public static void main(String[] args) {
        int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
        int top = 0;
        int bottom = 2;
        int left = 0;
        int right = 2;
        StringBuilder result = new StringBuilder();
        while (left = left; i--) {
                    result.append(matrix[bottom][i]).append(" ");
                }
                bottom--;
            }
            if (left = top; i--) {
                    result.append(matrix[i][left]).append(" ");
                }
                left++;
            }
        }
        // Print the spiral order
    }
}

Hint

Use System.out.println(result.toString().trim()); to print the spiral order.

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