Given a m * n grid, where each cell is either 0 (empty) or 1 (obstacle). In one step, you can move up, down, left or right from and to an empty cell.
Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m-1, n-1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.
Example 1:
Input:
grid =
[[0,0,0],
[1,1,0],
[0,0,0],
[0,1,1],
[0,0,0]],
k = 1
Output: 6
Explanation:
The shortest path without eliminating any obstacle is 10.
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).
Example 2:
Input:
grid =
[[0,1,1],
[1,1,1],
[1,0,0]],
k = 1
Output: -1
Explanation:
We need to eliminate at least two obstacles to find such a walk.
Constraints:
grid.length == m
grid[0].length == n
1 <= m, n <= 40
1 <= k <= m*n
grid[i][j] == 0 or 1
grid[0][0] == grid[m-1][n-1] == 0
ไปฃ็
Approach #1 BFS
Time: O(MNK) && Space: O(MNK)
classSolution {int[][] dirs =newint[][] {{0,1}, {0,-1}, {1,0}, {-1,0}};publicintshortestPath(int[][] grid,int k) {int n =grid.length;int m = grid[0].length;Queue<int[]> q =newLinkedList();boolean[][][] visited =newboolean[n][m][k+1]; visited[0][0][0] =true;q.offer(newint[]{0,0,0});int res =0;while (!q.isEmpty()) {int size =q.size();for (int i =0; i < size; i++) {int[] info =q.poll();int r = info[0];int c = info[1];int curK = info[2];if (r == n-1&& c == m-1) {return res; }for (int[] dir: dirs) {int nextR = dir[0] + r;int nextC = dir[1] + c;int nextK = curK;if (nextR >=0&& nextR < n && nextC >=0&& nextC < m) {if (grid[nextR][nextC] ==1) { nextK++; }if (nextK <= k &&!visited[nextR][nextC][nextK]) { visited[nextR][nextC][nextK] =true;q.offer(newint[]{nextR, nextC, nextK}); } } } } res++; }return-1; }}
Approach #2 DFS with Memorization
Starting from (0,0) have tried to go through all paths. Whenever we encounter a 1 and we have more than 0 moves available we choose to remove the obstacle and move ahead. Otherwise we skip the move and wait for a 0.