Spiral Matrix

Given an m x n matrix of integers matrix, return a list of all elements within the matrix in spiral order.

Approach

Maintain four boundaries (top, bottom, left, right) and traverse the matrix in layers: top row, right column, bottom row, left column, shrinking boundaries after each pass. Stop when the boundaries cross.

Why It Works

Each boundary traversal visits a unique outer layer of the matrix. Shrinking the boundaries ensures each element is visited exactly once in spiral order.

Complexity

• Time: O(m * n)
• Space: O(1)

Reference Implementation (Python)

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        if not matrix or not matrix[0]:
            return []
        top, bottom = 0, len(matrix) - 1
        left, right = 0, len(matrix[0]) - 1
        result = []

        while top <= bottom and left <= right:
            for c in range(left, right + 1):
                result.append(matrix[top][c])
            top += 1

            for r in range(top, bottom + 1):
                result.append(matrix[r][right])
            right -= 1

            if top <= bottom:
                for c in range(right, left - 1, -1):
                    result.append(matrix[bottom][c])
                bottom -= 1

            if left <= right:
                for r in range(bottom, top - 1, -1):
                    result.append(matrix[r][left])
                left += 1

        return result