Given an N x N grid containing only values 0 and 1, where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized and return the distance.
The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|.
If no land or water exists in the grid, return -1.
Example 1:
Input: [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation:
The cell (1, 1) is as far as possible from all the land with distance 2.
Example 2:
Input: [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation:
The cell (2, 2) is as far as possible from all the land with distance 4.
Note:
1 <= grid.length == grid[0].length <= 100
grid[i][j] is 0 or 1
class Solution {
int[][] directions = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };
public int maxDistance(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0)
return -1;
boolean[][] visited = new boolean[grid.length][grid[0].length];
Queue<int[]> queue = new ArrayDeque<>();
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
queue.add(new int[] { i, j });
visited[i][j] = true;
}
}
}
int level = -1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] start = queue.poll();
int x = start[0];
int y = start[1];
for (int[] dir : directions) {
int newX = x + dir[0];
int newY = y + dir[1];
if (newX >= 0 && newX < grid.length && newY >= 0 && newY < grid[0].length && !visited[newX][newY]) {
visited[newX][newY] = true;
queue.add(new int[] { newX, newY });
}
}
}
level++;
}
// If no land or water exists in the grid, return -1
return level <= 0 ? -1 : level;
}
}
