Published: Oct 30, 2022

## Problem Description

You are given an

`m x n`

integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step.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.Constraints:

`m == grid.length`

`n == grid[i].length`

`1 <= m, n <= 40`

`1 <= k <= m * n`

`grid[i][j]`

is either 0 or 1.`grid[0][0] == grid[m - 1][n - 1] == 0`

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

## Examples

```
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.
```

## How to Solve

If this problem doesn’t have a condition, “at most k obstacles elimination,” it is a typical breadth-first search problem. Now, the condition of obstacles elimination has been added. Still, the way of solving the problem stays the same. Instead of adding (next_row, next_col) to the queue, it adds (next_row next_col, number_of_k_left). The number of k left will be decreased by one if the grid is the obstacle, otherwise keep the same value. While going over the cells, if the cell is the obstacle and k is 0, the cell won’t be added to the queue. This way, we can find the answer.

The problem has an extra condition, however, traversing cells stays the same. The algorithm is the breadth-first search since the problem asks the shortest path. The queue will have the combination of (distance, (next_row, next_col, number_of_k_left)). If the cell is the obstacle, the number of k left is decreased by one. Conveniently, the obstacle value is 1, so k - grid[row][col] does the decrement. When the popped out row and col is the one of right bottom, return the distance.

## Solution

```
class ShortestPathInAGridWithObstaclesElimination {
public:
int shortestPath(vector<vector<int>>& grid, int k) {
int rows = grid.size(), cols = grid[0].size();
queue<pair<int, tuple<int, int, int>>> q; // (dist, (row, col, k))
q.push({0, make_tuple(0, 0, k)});
vector<vector<int>> seen(rows, vector<int>(cols, -1));
seen[0][0] = k;
vector<pair<int, int>> dirs = { {-1, 0}, {0, -1}, {0, 1}, {1, 0} };
while (!q.empty()) {
auto [dist, tpl] = q.front();
auto [r, c, k] = tpl;
q.pop();
if (r == rows - 1 && c == cols - 1) { return dist; }
for (auto [dr, dc] : dirs) {
int next_r = r + dr, next_c = c + dc;
if (next_r < 0 || next_r >= rows || next_c < 0 || next_c >= cols ||
(grid[next_r][next_c] == 1 and k == 0) ||
seen[next_r][next_c] >= k) {
continue;
}
seen[next_r][next_c] = k;
q.push({dist + 1, make_tuple(next_r, next_c, k - grid[next_r][next_c])});
}
}
return -1;
}
};
```

```
```

```
```

```
class ShortestPathInAGridWithObstaclesElimination:
def shortestPath(self, grid: List[List[int]], k: int) -> int:
rows, cols = len(grid), len(grid[0])
queue = [(0, (0, 0, k))] # dist, (row, col, k)
seen = set((0, 0, k))
dirs = [[-1, 0], [0, -1], [0, 1], [1, 0]]
while queue:
dist, (r, c, k) = queue.pop(0)
if (r, c) == (rows - 1, cols - 1):
return dist
for dr, dc in dirs:
next_r, next_c = r + dr, c + dc
if next_r < 0 or next_r >= rows or\
next_c < 0 or next_c >= cols or\
(grid[next_r][next_c] == 1 and k == 0) or\
(next_r, next_c, k) in seen:
continue
seen.add((next_r, next_c, k))
queue.append((dist + 1, (next_r, next_c, k - grid[next_r][next_c])))
return -1
```

```
```

## Complexities

- Time:
`O(n * k)`

– n: number of cells in the grid, k: obstacle eliminate limit - Space:
`O(n * k)`