1368. Minimum Cost to Make at Least One Valid Path in a Grid

Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

• 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
• 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
• 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
• 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell (0, 0). A valid path in the grid is a path that starts from the upper left cell (0, 0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

• Time : O(nlogn)
• Space : O(n)

class Solution {
    vector<vector<int>> dp;
    int dx[5] = {0,1,-1,0,0}, dy[5] = {0,0,0,1,-1};
    int n, m;
    int solution(vector<vector<int>>& grid, int y, int x) {
        priority_queue<array<int,3>, vector<array<int,3>>, greater<array<int,3>>> q;
        q.push({0,y,x});
        while(!q.empty()) {
            auto [c, cy, cx] = q.top(); q.pop();
            if(cy == n-1 and cx == m-1) return c;
            for(int i = 1; i < 5; i++) {
                int ny = cy + dy[i], nx = cx + dx[i], nc = c + (i != grid[cy][cx]);
                if(0<=ny and ny < n and 0 <= nx and nx < m and dp[ny][nx] > nc) {
                    dp[ny][nx] = nc;
                    q.push({nc,ny,nx});
                }
            }
        }
        return -1;
    }
public:
    int minCost(vector<vector<int>>& grid) {
        n = grid.size(), m = grid[0].size();
        dp = vector<vector<int>>(n, vector<int>(m, INT_MAX));
        return solution(grid,0,0);
    }
};