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

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Description

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.

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

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• 1 <= grid[i][j] <= 4

Solution

Dijkstra Algorithm

See reference (Chinese).

using PII = pair<int, int>;

class Solution {
private:
    static constexpr int dx[4] = {0, 0, 1, -1};
    static constexpr int dy[4] = {1, -1, 0, 0};

public:
    int minCost(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<int> dist(m * n, 0x3f3f3f3f);
        vector<bool> st(m * n, false);

        priority_queue<PII, vector<PII>, greater<PII>> heap;
        dist[0] = 0;
        heap.emplace(0, 0);

        while (!heap.empty()) {
            auto [distance, vertex] = heap.top();
            heap.pop();

            if (st[vertex]) continue;
            st[vertex] = true;

            int x = vertex / n, y = vertex % n;
            for (int i = 0; i < 4; i++) {
                int newX = x + dx[i];
                int newY = y + dy[i];
                int newVertex = newX * n + newY;
                int newDistance = distance + (grid[x][y] != i + 1);

                if (newX >= 0 && newX < m && newY >= 0 && newY < n && newDistance < dist[newVertex]) {
                    dist[newVertex] = newDistance;
                    heap.emplace(newDistance, newVertex);
                }
            }
        }

        return dist[m * n - 1];
    }
};

• Time complexity: $O(MN\log (MN))$;
• Space complexity: $O(MN)$.

0-1 BFS

using PII = pair<int, int>;

class Solution {
private:
    static constexpr int dx[4] = {0, 0, 1, -1};
    static constexpr int dy[4] = {1, -1, 0, 0};

public:
    int minCost(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<int> dist(m * n, 0x3f3f3f3f);
        vector<bool> st(m * n, false);

        deque<int> q;
        q.push_back(0);
        dist[0] = 0;

        while (!q.empty()) {
            auto vertex = q.front();
            q.pop_front();

            if (st[vertex]) continue;
            st[vertex] = true;

            int x = vertex / n, y = vertex % n;
            for (int i = 0; i < 4; i++) {
                int newX = x + dx[i], newY = y + dy[i];
                int newVertex = newX * n + newY;
                int newDistance = dist[vertex] + (grid[x][y] != i + 1);

                if (newX >= 0 && newX < m && newY >= 0 && newY < n && newDistance < dist[newVertex]) {
                    dist[newVertex] = newDistance;
                    if (grid[x][y] == i + 1)
                        q.push_front(newVertex);
                    else
                        q.push_back(newVertex);
                }
            }
        }

        return dist[m * n - 1];
    }
};

• Time complexity: $O(MN)$;
• Space complexity: $O(MN)$.