Max Area of Island


#1

Given a non-empty 2D array grid of 0's and 1's, 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.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[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]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.


#2

The below code uses DFS. When we find an island, that is 1, we do a DFS on the grid starting with that indexes, count all neighbors and maintain a max.

var xa = []int{0, -1, 0, 1}
var ya =  []int{-1, 0, 1, 0}

func maxAreaOfIsland(grid [][]int) int {
    max := 0
    if len(grid)==0 {
        return max
    }
    for i:=0;i<len(grid);i++{
        for j:=0;j<len(grid[0]);j++{
            if grid[i][j]==1 {
                var count int
                dfs(grid, i, j, &count)
                if max<count{
                    max = count
                }
            }
        }
    }
    return max
}

func dfs(grid [][]int, i, j int, count *int) {
    if (i<0 || i>=len(grid) || j<0 || j>=len(grid[0]) || grid[i][j]==0) {
        return
    }
    grid[i][j] = 0
    *count++
    for m:=0;m<len(xa);m++{
        dfs(grid, i+xa[m], j+ya[m], count)
    }
    
  return
}

#3

Similar DFS approach in Java.

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        if(grid==null || grid.length==0) return 0;
        int max = 0;
        for(int i=0;i<grid.length;i++){
            for(int j=0;j<grid[0].length;j++){
                if(grid[i][j]==1){
                    max = Math.max(max, dfs(grid, i, j));
                }
            }
        }
        return max;
    }
    
    private int dfs(int[][] grid, int i, int j){
        if(i<0 || j<0 || i>=grid.length || j>=grid[0].length || grid[i][j]!=1) return 0;
        grid[i][j] = 0;
        return 1+dfs(grid, i, j-1)+dfs(grid, i, j+1)+dfs(grid, i-1, j)+dfs(grid, i+1, j);
    }
}