LeetCode-in-Java

2658. Maximum Number of Fish in a Grid

Medium

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

Solution

public class Solution {
    private int dfs(int[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i == grid.length || j == grid[i].length || grid[i][j] < 1) {
            return 0;
        }
        grid[i][j] -= 20;
        return 20
                + grid[i][j]
                + dfs(grid, i + 1, j)
                + dfs(grid, i - 1, j)
                + dfs(grid, i, j - 1)
                + dfs(grid, i, j + 1);
    }

    public int findMaxFish(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int max = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] > 0) {
                    max = Math.max(max, dfs(grid, i, j));
                }
            }
        }
        return max;
    }
}

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