Formatted question description: https://leetcode.ca/all/1368.html

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

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

Related Topics: Breadth-first Search

Solution 1. BFS + DP

// OJ: https://leetcode.com/problems/minimum-cost-to-make-at-least-one-valid-path-in-a-grid/

// Time: O(MN)
// Space: O(MN)
class Solution {
public:
    int minCost(vector<vector<int>>& G) {
        int M = G.size(), N = G[0].size(), dp[100][100] = {}, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
        memset(dp, 0x3f, sizeof(dp));
        dp[0][0] = 0;
        queue<pair<int, int>> q;
        q.emplace(0, 0);
        while (q.size()) {
            auto [x, y] = q.front();
            q.pop();
            for (int i = 0; i < 4; ++i) {
                auto &[dx, dy] = dirs[i];
                int a = x + dx, b = y + dy;
                if (a < 0 || b < 0 || a >= M || b >= N) continue;
                if (dp[x][y] + (G[x][y] - 1 != i) < dp[a][b]) {
                    dp[a][b] = dp[x][y] + (G[x][y] - 1 != i);
                    q.emplace(a, b);
                }
            }
        }
        return dp[M - 1][N - 1];
    }
};

Java

class Solution {
    int[][] directions = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };

    public int minCost(int[][] grid) {
        int rows = grid.length, columns = grid[0].length;
        if (rows == 1 && columns == 1)
            return 0;
        int cost = 0;
        boolean[][] visited = new boolean[rows][columns];
        visited[0][0] = true;
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(new int[]{0, 0});
        List<int[]> levelList = new ArrayList<int[]>();
        levelList.add(new int[]{0, 0});
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0], column = cell[1];
            int directionIndex = grid[row][column] - 1;
            if (directionIndex >= 0) {
                int[] direction = directions[directionIndex];
                int newRow = row + direction[0], newColumn = column + direction[1];
                while (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && !visited[newRow][newColumn]) {
                    visited[newRow][newColumn] = true;
                    levelList.add(new int[]{newRow, newColumn});
                    int nextDirectionIndex = grid[newRow][newColumn] - 1;
                    if (nextDirectionIndex < 0)
                        break;
                    newRow += directions[nextDirectionIndex][0];
                    newColumn += directions[nextDirectionIndex][1];
                }
            } else {
                for (int i = 0; i < 4; i++) {
                    int newRow = row + directions[i][0], newColumn = column + directions[i][1];
                    while (newRow >= 0 && newRow < rows && newColumn >= 0 && newColumn < columns && !visited[newRow][newColumn]) {
                        visited[newRow][newColumn] = true;
                        levelList.add(new int[]{newRow, newColumn});
                        int nextDirectionIndex = grid[newRow][newColumn] - 1;
                        if (nextDirectionIndex < 0)
                            break;
                        newRow += directions[nextDirectionIndex][0];
                        newColumn += directions[nextDirectionIndex][1];
                    }
                }
            }
            if (queue.isEmpty() && !levelList.isEmpty()) {
                for (int[] levelCell : levelList) {
                    int curRow = levelCell[0], curColumn = levelCell[1];
                    if (curRow == rows - 1 && curColumn == columns - 1)
                        return cost;
                    queue.offer(levelCell);
                    grid[curRow][curColumn] = -1;
                }
                cost++;
                levelList.clear();
            }
        }
		return cost;
    }
}

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