Introduction

In this blog post, we will explore how to solve the LeetCode problem "2658" 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 a 0-indexed 2D matrix grid of size m x n, where (r, c) represents: A land cell if grid[r][c] = 0, or A water cell containing grid[r][c] fish, if grid[r][c] > 0. A fisher can start at any water cell (r, c) and can do the following operations any number of times: Catch all the fish at cell (r, c), or Move to any adjacent water cell. Return the maximum number of fish the fisher can catch if he chooses his starting cell optimally, or 0 if no water cell exists. An adjacent cell of the cell (r, c), is one of the cells (r, c + 1), (r, c - 1), (r + 1, c) or (r - 1, c) if it exists. Example 1: Input: grid = [[0,2,1,0],[4,0,0,3],[1,0,0,4],[0,3,2,0]] Output: 7 Explanation: The fisher can start at cell (1,3) and collect 3 fish, then move to cell (2,3) and collect 4 fish. Example 2: Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]] Output: 1 Explanation: The fisher can start at cells (0,0) or (3,3) and collect a single fish. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 10 0 <= grid[i][j] <= 10

Explanation

Here's the breakdown of the solution:

• Connected Components: Identify connected components of water cells. A connected component is a group of water cells that can be reached from each other by moving to adjacent water cells.
• Calculate Component Sums: Calculate the sum of fish in each connected component. This sum represents the maximum fish a fisher can catch if they start in any cell within that component.

• Find Maximum: Return the maximum sum among all connected components.

• Runtime Complexity: O(m * n), where m is the number of rows and n is the number of columns.

• Space Complexity: O(m * n)

Code

    def findMaxFish(grid):
    """
    Finds the maximum number of fish a fisher can catch.

    Args:
        grid: A 2D list of integers representing the grid.

    Returns:
        The maximum number of fish the fisher can catch.
    """

    m = len(grid)
    n = len(grid[0])
    visited = [[False] * n for _ in range(m)]
    max_fish = 0

    def dfs(row, col):
        """
        Depth-first search to explore connected water cells.

        Args:
            row: The row index of the current cell.
            col: The column index of the current cell.

        Returns:
            The total number of fish in the connected component.
        """

        if row < 0 or row >= m or col < 0 or col >= n or visited[row][col] or grid[row][col] == 0:
            return 0

        visited[row][col] = True
        fish = grid[row][col]

        fish += dfs(row + 1, col)
        fish += dfs(row - 1, col)
        fish += dfs(row, col + 1)
        fish += dfs(row, col - 1)

        return fish

    for i in range(m):
        for j in range(n):
            if grid[i][j] > 0 and not visited[i][j]:
                max_fish = max(max_fish, dfs(i, j))

    return max_fish