# Solving Leetcode Interviews in Seconds with AI: Spiral Matrix III


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "885" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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
	> You start at the cell (rStart, cStart) of an rows x cols grid facing east. The northwest corner is at the first row and column in the grid, and the southeast corner is at the last row and column. You will walk in a clockwise spiral shape to visit every position in this grid. Whenever you move outside the grid's boundary, we continue our walk outside the grid (but may return to the grid boundary later.). Eventually, we reach all rows * cols spaces of the grid. Return an array of coordinates representing the positions of the grid in the order you visited them.   Example 1:   Input: rows = 1, cols = 4, rStart = 0, cStart = 0 Output: [[0,0],[0,1],[0,2],[0,3]]  Example 2:   Input: rows = 5, cols = 6, rStart = 1, cStart = 4 Output: [[1,4],[1,5],[2,5],[2,4],[2,3],[1,3],[0,3],[0,4],[0,5],[3,5],[3,4],[3,3],[3,2],[2,2],[1,2],[0,2],[4,5],[4,4],[4,3],[4,2],[4,1],[3,1],[2,1],[1,1],[0,1],[4,0],[3,0],[2,0],[1,0],[0,0]]    Constraints:  1 <= rows, cols <= 100 0 <= rStart < rows 0 <= cStart < cols  

	# Explanation
	Here's the breakdown of the solution:

*   **Spiral Traversal:** The code simulates the clockwise spiral traversal. It uses a `direction` variable to keep track of the current direction (East, South, West, North) and updates coordinates accordingly.
*   **Grid Visit Tracking:** The `visited` counter ensures that the algorithm terminates only when all cells within the *actual* grid dimensions have been visited. This avoids infinite loops when walking outside of the grid boundaries.
*   **Coordinate Appending:** The current coordinates are appended to the `result` list at each step. The validity of the coordinates (whether they are within the grid) is checked before incrementing `visited`.

*   **Time & Space Complexity:** O(rows * cols) time complexity because each cell is visited. O(rows * cols) space complexity, because it stores the coordinates of each visited cell.

	
	# Code
	```python
	def spiralMatrixIII(rows: int, cols: int, rStart: int, cStart: int) -> list[list[int]]:
    result = []
    r, c = rStart, cStart
    direction = 0  # 0: East, 1: South, 2: West, 3: North
    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
    steps = 1
    visited = 0
    total_cells = rows * cols

    while visited < total_cells:
        for _ in range(2):
            for _ in range(steps):
                if 0 <= r < rows and 0 <= c < cols:
                    result.append([r, c])
                    visited += 1
                r += directions[direction][0]
                c += directions[direction][1]

            direction = (direction + 1) % 4
        steps += 1

    return result
	```