Introduction

In this blog post, we will explore how to solve the LeetCode problem "54" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an m x n matrix, return all elements of the matrix in spiral order. Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5] Example 2: Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]] Output: [1,2,3,4,8,12,11,10,9,5,6,7] Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 10 -100 <= matrix[i][j] <= 100

Explanation

Here's a breakdown of the solution, followed by the Python code:

• Iterative Traversal: The solution iteratively traverses the matrix in a spiral pattern. It maintains boundaries (top, bottom, left, right) that shrink after each layer is processed.
• Direction Control: The algorithm moves in four directions (right, down, left, up) sequentially. After traversing one side, the corresponding boundary is updated.

• Boundary Checks: The algorithm ensures that the boundaries don't cross, preventing duplicate or out-of-bounds access.

• Runtime Complexity: O(m n) - visits each element once. *Storage Complexity: O(1) - excluding the output list.

Code

    def spiralOrder(matrix):
    """
    Given an m x n matrix, return all elements of the matrix in spiral order.

    Args:
        matrix (List[List[int]]): The input matrix.

    Returns:
        List[int]: A list containing the elements in spiral order.
    """
    result = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1
    direction = 0  # 0: right, 1: down, 2: left, 3: up

    while top <= bottom and left <= right:
        if direction == 0:  # right
            for i in range(left, right + 1):
                result.append(matrix[top][i])
            top += 1
        elif direction == 1:  # down
            for i in range(top, bottom + 1):
                result.append(matrix[i][right])
            right -= 1
        elif direction == 2:  # left
            for i in range(right, left - 1, -1):
                result.append(matrix[bottom][i])
            bottom -= 1
        elif direction == 3:  # up
            for i in range(bottom, top - 1, -1):
                result.append(matrix[i][left])
            left += 1
        direction = (direction + 1) % 4

    return result