1293. Shortest Path in a Grid with Obstacles Elimination

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1293. Shortest Path in a Grid with Obstacles Elimination

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.

Example 1:

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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:

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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:

  • 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

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    static constexpr int dirs[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};

    int shortestPath(vector<vector<int>>& grid, int k) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<vector<bool>>> visited(m, vector<vector<bool>>(n, vector<bool>(k + 1, false)));
        queue<vector<int>> q;
        q.push({0, 0, k});
        visited[0][0][k] = true;
        int step = 0;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                auto tri = q.front();
                q.pop();
                int x = tri[0], y = tri[1], remain = tri[2];
                if (x == m - 1 && y == n - 1) {
                    return step;
                }
                for (int k = 0; k < 4; k++) {
                    int dx = x + dirs[k][0], dy = y + dirs[k][1];
                    if (dx >= 0 && dx < m && dy >= 0 && dy < n) {
                        if (grid[dx][dy] == 0 && !visited[dx][dy][remain]) {
                            visited[dx][dy][remain] = true;
                            q.push({dx, dy, remain});
                        }
                        if (grid[dx][dy] == 1 && remain > 0 && !visited[dx][dy][remain - 1]) {
                            visited[dx][dy][remain - 1] = true;
                            q.push({dx, dy, remain - 1});
                        }
                    }
                }
            }
            step++;
        }
        return -1;
    }
};