Given an n x n binary matrix grid, return the length of the shortest clear path in the matrix. If there is no clear path, return -1.
A clear path in a binary matrix is a path from the top-left cell (i.e., (0, 0)) to the bottom-right cell (i.e., (n - 1, n - 1)) such that:
- All the visited cells of the path are 0.
- All the adjacent cells of the path are 8-directionally connected (i.e., they are different, and they share an edge or a corner).
The length of a clear path is the number of visited cells of this path.
Test Cases
Example 1:
| 0 | 1 |
| 1 | 0 |
Input: grid = [[0,1],[1,0]]
Output: 2
Example 2:
| 0 | 0 | 0 |
| 1 | 1 | 0 |
| 1 | 1 | 0 |
Input: grid = [[0,0,0],[1,1,0],[1,1,0]]
Output: 4
Solution
class Solution {
public int shortestPathBinaryMatrix(int[][] grid) {
if (grid[0][0] == 1) return -1;
int n = grid.length, m = grid[0].length;
Queue<int[]> q = new LinkedList<>();
q.offer(new int[]{0, 0});
int steps = 1;
int[] dx = new int[]{-1,-1,-1,0,0,1,1,1};
int[] dy = new int[]{-1,0,1,-1,1,-1,0,1};
while(!q.isEmpty()) {
int qs = q.size();
for(int i=0; i<qs; i++) {
int[] first = q.poll();
if (grid[first[0]][first[1]] != 0) continue;
grid[first[0]][first[1]] = 1;
if (first[0] == n-1 && first[1] == m-1) {
return steps;
}
for(int j=0; j<8; j++) {
int nx = first[0] + dx[j], ny = first[1]+dy[j];
if (nx>=0 && ny >=0 && nx<n && ny<m && grid[nx][ny] == 0) {
q.offer(new int[]{nx, ny});
}
}
}
steps++;
}
return -1;
}
}