Reference: LeetCode
Difficulty: Medium

Problem

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.)

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

Example:

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[[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]] // 6

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

Analysis

DFS

For each position in the array, we do DFS and count how many 1 it can reach. Instead of using a mark array, we can simply set the visited position as 0.

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public int maxAreaOfIsland(int[][] grid) {
// assume grid is not null
if (grid.length == 0 || grid[0].length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;

int maxArea = 0;

// for each position in the array
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
int area = dfs(i, j, grid);
maxArea = Math.max(area, maxArea);
}
}

return maxArea;
}

int[][] direction = new int[][] { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };

// returns the area of island
private int dfs(int i, int j, int[][] grid) {
int m = grid.length;
int n = grid[0].length;
// base case (reject)
if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
return 0;
}
// visit
grid[i][j] = 0;
// for each direction
int area = 1;
for (int[] dir : direction) {
int x = i + dir[0];
int y = j + dir[1];
area += dfs(x, y, grid);
}
return area;
}

Time: $O(MN)$
Space: $O(MN)$


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