leetcode Max Area of Island Solution (Max Area of Island) (leetcode june 2021 challenge)

Max Area of Island

You are given an m x n binary matrix grid. An island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

 

Example 1:

 

Example 1:

Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]

Output: 6

Explanation: The answer is not 11, because the island must be connected 4-directionally.

 

Example 2:

Input: grid = [[0,1,1,1,0,0,0,0]]

Output: 3

Constraints:

·         m == grid.length

·         n == grid[i].length

·         1 <= m, n <= 50

·         grid[i][j] is either 0 or 1.

 

Solution 1:

public class maxAreaOfIsland {

    public static int length(int[][] grid, int i, int j) {

        if (i >= grid.length || j >= grid[0].length || i < 0 || j < 0 || grid[i][j] != 1)

            return 0;

        int ans = 0;

        grid[i][j] = 2;

        if (j + 1 < grid[0].length && grid[i][j + 1] == 1)

            ans = ans + 1 + length(grid, i, j + 1);

        if (i + 1 < grid.length && grid[i + 1][j] == 1)

            ans = ans + 1 + length(grid, i + 1, j);

        if (j - 1 >= 0 && grid[i][j - 1] == 1)

            ans = ans + 1 + length(grid, i, j - 1);

        if (i - 1 >= 0 && grid[i - 1][j] == 1)

            ans = ans + 1 + length(grid, i - 1, j);

        return ans;

    }

    public static int Solution(int[][] grid) {

        int ans = 0;

        for (int i = 0; i < grid.length; i++) {

            for (int j = 0; j < grid[0].length; j++) {

                if (grid[i][j] == 1)

                    ans = Math.max(ans, length(grid, i, j) + 1);

            }

        }

        return ans;

    }