Introduction

In this blog post, we will explore how to solve the LeetCode problem "1030" 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

You are given four integers row, cols, rCenter, and cCenter. There is a rows x cols matrix and you are on the cell with the coordinates (rCenter, cCenter). Return the coordinates of all cells in the matrix, sorted by their distance from (rCenter, cCenter) from the smallest distance to the largest distance. You may return the answer in any order that satisfies this condition. The distance between two cells (r1, c1) and (r2, c2) is |r1 - r2| + |c1 - c2|. Example 1: Input: rows = 1, cols = 2, rCenter = 0, cCenter = 0 Output: [[0,0],[0,1]] Explanation: The distances from (0, 0) to other cells are: [0,1] Example 2: Input: rows = 2, cols = 2, rCenter = 0, cCenter = 1 Output: [[0,1],[0,0],[1,1],[1,0]] Explanation: The distances from (0, 1) to other cells are: [0,1,1,2] The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct. Example 3: Input: rows = 2, cols = 3, rCenter = 1, cCenter = 2 Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]] Explanation: The distances from (1, 2) to other cells are: [0,1,1,2,2,3] There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]]. Constraints: 1 <= rows, cols <= 100 0 <= rCenter < rows 0 <= cCenter < cols

Explanation

Here's a solution to the problem, along with explanations and complexity analysis:

• Key Idea: Since the constraints are relatively small (rows, cols <= 100), we can iterate through all cells, calculate their Manhattan distance to the center, and then sort the cells based on these distances.

• Optimized Sorting: Instead of using a general-purpose sorting algorithm (which would be O(n log n)), we can use a bucket sort (or counting sort) approach. Because the maximum possible distance is limited by the matrix dimensions, we can create buckets for each possible distance and place the corresponding cells into those buckets. This allows us to achieve linear time complexity.

• Complexity:

  • Runtime: O(rows * cols)
  • Storage: O(rows * cols)

Code

    def allCellsDistOrder(rows: int, cols: int, rCenter: int, cCenter: int) -> list[list[int]]:
    """
    Given dimensions of matrix, center and sort cells by distance.
    """

    max_distance = max(rCenter, rows - 1 - rCenter) + max(cCenter, cols - 1 - cCenter)
    buckets = [[] for _ in range(max_distance + 1)]

    for r in range(rows):
        for c in range(cols):
            dist = abs(r - rCenter) + abs(c - cCenter)
            buckets[dist].append([r, c])

    result = []
    for bucket in buckets:
        result.extend(bucket)

    return result