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 opera

findMaxFish.js

/**
 * @param {number[][]} grid
 * @return {number}
 */
var findMaxFish = function(grid) {
     // Initialize the maximum fish count as 0
    let maxFish = 0;

    // Get the number of rows and columns in the grid
    const rows = grid.length; 
    const cols = grid[0].length;

    // Helper function to perform DFS and count the fish
    const dfs = (row, col) => {
        // Base case: return 0 if out of bounds or the cell is land
        if (row < 0 || col < 0 || row >= rows || col >= cols || grid[row][col] === 0) {
            return 0;
        }

        // Collect the fish in the current cell
        let fish = grid[row][col];
        // Mark the cell as visited by setting it to 0
        grid[row][col] = 0;

        // Recursively collect fish from adjacent cells
        let top = dfs(row + 1, col);
        let down = dfs(row - 1, col);
        let left = dfs(row, col - 1);
        let right = dfs(row, col + 1);

        // Return the total number of fish collected in this path
        return fish + top + down + left + right;
    }

    // Loop through all the cells in the grid
    for (let i = 0; i < rows; i++) {
        for (let j = 0; j < cols; j++) {
            // If it's a water cell with fish, perform DFS
            if (grid[i][j] !== 0) {
                maxFish = Math.max(dfs(i, j), maxFish); // Update the maximum fish count
            }
        }
    }

    // Return the maximum fish count
    return maxFish;
};