Q5: How do you traverse an M x N 2D matrix in Spiral Order?

Maintain four boundary boundaries: top, bottom, left, and right. Traverse right along top, down along right, left along bottom (if top <= bottom), and up along left (if left <= right), shrinking boundaries inward after each direction traversal.

Spiral Matrix Traversal

import java.util.*;

public class SpiralMatrix {
    public static List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> result = new ArrayList<>();
        if (matrix == null || matrix.length == 0) return result;
        int top = 0, bottom = matrix.length - 1;
        int left = 0, right = matrix[0].length - 1;
        while (top <= bottom && left <= right) {
            for (int col = left; col <= right; col++) result.add(matrix[top][col]);
            top++;
            for (int row = top; row <= bottom; row++) result.add(matrix[row][right]);
            right--;
            if (top <= bottom) {
                for (int col = right; col >= left; col--) result.add(matrix[bottom][col]);
                bottom--;
            }
            if (left <= right) {
                for (int row = bottom; row >= top; row--) result.add(matrix[row][left]);
                left++;
            }
        }
        return result;
    }
}

💡 Scans grid perimeters clockwise while shrinking boundary limits.

Key Takeaways

Track four dynamic boundary boundaries.
Shrink boundaries inward and verify non-overlapping conditions before scanning reverse directions.

Knowledge Check

Why are the checks top <= bottom and left <= right required inside the spiral loop before processing bottom and left sweeps?

Q5: How do you traverse an M x N 2D matrix in Spiral Order?

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