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

Given a 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 invalid signs on the cells of the grid which points outside the grid.

You will initially start at the upper left cell (0,0). A valid path in the grid is a path which 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 doesn't 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.

Example 1:

Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3
Explanation: You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
The total cost = 3.

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0
Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

Input: grid = [[1,2],[4,3]]
Output: 1

Example 4:

Input: grid = [[2,2,2],[2,2,2]]
Output: 3

Example 5:

Input: grid = [[4]]
Output: 0

Constraints:

• m == grid.length

• n == grid[i].length

• 1 <= m, n <= 100

Solution

The first idea is to use Dijkstra's Algo. Sorting the adjacent nodes by the path cost, and avoid cycle during the travel.

Time complexity: O(NM*logNM)

  int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    public int minCost(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        boolean[][] visited = new boolean[m][n];
        PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)->{
            if(a[2] != b[2]) return Integer.compare(a[2], b[2]); 
            return Integer.compare(b[0]+b[1], a[0]+a[1]);
        });
        pq.add(new int[]{0, 0, 0});
        
        while(!pq.isEmpty()) {
            int[] cur = pq.poll();
            int row = cur[0];
            int col = cur[1];
            if(row == m-1 && col == n-1) return cur[2];
            visited[row][col] = true;
            int index = grid[row][col] - 1;
            for(int i = 0; i < 4; i++){
                int newRow = row + dirs[i][0];
                int newCol = col + dirs[i][1];
                if(newRow < 0 || newRow >= m || newCol < 0 || newCol >= n || visited[newRow][newCol]) continue;
                pq.add(new int[]{newRow, newCol, cur[2] + ((i == index) ? 0 : 1)});
            }
        }
        return -1;
    }