# 1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/

## ้ข็ฎๆ่ฟฐ

``````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)

``````class Solution {
int[][] dirs = new int[][] {{0,1}, {0, -1}, {1,0}, {-1, 0}};
public int shortestPath(int[][] grid, int k) {
int n = grid.length;
int m = grid[0].length;
boolean[][][] visited = new boolean[n][m][k+1];
visited[0][0][0] = true;
q.offer(new int[]{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(new int[]{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.

``````class Solution {
public int shortestPath(int[][] grid, int k) {
if (grid.length == 0)            return 0;
Map<String, Integer> map = new HashMap<>();
boolean visited[][] = new boolean[grid.length][grid[0].length];
int min = dfs(grid, 0, 0, k, map, visited);
return min == Integer.MAX_VALUE ? -1 : min;
}

int[] row = {1, -1, 0, 0};
int[] col = {0, 0, 1, -1};
private int dfs(int[][] grid, int startI, int startJ, int k, Map<String, Integer> map, boolean visited[][]) {
if (k < 0)        return Integer.MAX_VALUE;
if (startI == grid.length-1 && startJ == grid[0].length-1) {
return 0;
}
String key = startI + "_" + startJ + "_" + k;
if (map.containsKey(key)) {
return map.get(key);
}

if (visited[startI][startJ])        return Integer.MAX_VALUE;
visited[startI][startJ] = true;

int min = Integer.MAX_VALUE;
for (int i = 0; i < 4; i++) {
int currRow = startI + row[i];
int currCol = startJ + col[i];
if (isSafe(grid, currRow, currCol)) {
int temp = Integer.MAX_VALUE;
if (grid[currRow][currCol] == 1) {
if (k > 0) {
temp = dfs(grid, currRow, currCol, k-1, map, visited);
} else {
continue;
}
} else {
temp = dfs(grid, currRow, currCol, k, map, visited);
}

if (temp != Integer.MAX_VALUE) {
min = Math.min(min, temp + 1);
}

}
}
visited[startI][startJ] = false;
map.put(key, min);
return min;
}

boolean isSafe(int grid[][], int i, int j) {
return i >= 0 && j >= 0
&& i < grid.length && j < grid[0].length;
}

}``````

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