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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];
    }
};
  • 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;
        }
    }
    
    ############
    
    class Solution {
        public int minCost(int[][] grid) {
            int m = grid.length, n = grid[0].length;
            boolean[][] vis = new boolean[m][n];
            Deque<int[]> q = new ArrayDeque<>();
            q.offer(new int[] {0, 0, 0});
            int[][] dirs = { {0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            while (!q.isEmpty()) {
                int[] p = q.poll();
                int i = p[0], j = p[1], d = p[2];
                if (i == m - 1 && j == n - 1) {
                    return d;
                }
                if (vis[i][j]) {
                    continue;
                }
                vis[i][j] = true;
                for (int k = 1; k <= 4; ++k) {
                    int x = i + dirs[k][0], y = j + dirs[k][1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        if (grid[i][j] == k) {
                            q.offerFirst(new int[] {x, y, d});
                        } else {
                            q.offer(new int[] {x, y, d + 1});
                        }
                    }
                }
            }
            return -1;
        }
    }
    
  • // 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{ { {0, 0} } };
            while (q.size()) {
                auto [x, y] = q.front();
                q.pop();
                for (int i = 0; i < 4; ++i) {
                    int a = x + dirs[i][0], b = y + dirs[i][1];
                    if (a < 0 || b < 0 || a >= M || b >= N) continue;
                    int cost = dp[x][y] + (G[x][y] - 1 != i);
                    if (dp[a][b] > cost) {
                        dp[a][b] = cost;
                        q.emplace(a, b);
                    }
                }
            }
            return dp[M - 1][N - 1];
        }
    };
    
  • class Solution:
        def minCost(self, grid: List[List[int]]) -> int:
            m, n = len(grid), len(grid[0])
            dirs = [[0, 0], [0, 1], [0, -1], [1, 0], [-1, 0]]
            q = deque([(0, 0, 0)])
            vis = set()
            while q:
                i, j, d = q.popleft()
                if (i, j) in vis:
                    continue
                vis.add((i, j))
                if i == m - 1 and j == n - 1:
                    return d
                for k in range(1, 5):
                    x, y = i + dirs[k][0], j + dirs[k][1]
                    if 0 <= x < m and 0 <= y < n:
                        if grid[i][j] == k:
                            q.appendleft((x, y, d))
                        else:
                            q.append((x, y, d + 1))
            return -1
    
    
    
  • func minCost(grid [][]int) int {
    	m, n := len(grid), len(grid[0])
    	q := doublylinkedlist.New()
    	q.Add([]int{0, 0, 0})
    	dirs := [][]int{ {0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0} }
    	vis := make([][]bool, m)
    	for i := range vis {
    		vis[i] = make([]bool, n)
    	}
    	for !q.Empty() {
    		v, _ := q.Get(0)
    		p := v.([]int)
    		q.Remove(0)
    		i, j, d := p[0], p[1], p[2]
    		if i == m-1 && j == n-1 {
    			return d
    		}
    		if vis[i][j] {
    			continue
    		}
    		vis[i][j] = true
    		for k := 1; k <= 4; k++ {
    			x, y := i+dirs[k][0], j+dirs[k][1]
    			if x >= 0 && x < m && y >= 0 && y < n {
    				if grid[i][j] == k {
    					q.Insert(0, []int{x, y, d})
    				} else {
    					q.Add([]int{x, y, d + 1})
    				}
    			}
    		}
    	}
    	return -1
    }
    
  • function minCost(grid: number[][]): number {
        const m = grid.length,
            n = grid[0].length;
        let ans = Array.from({ length: m }, v => new Array(n).fill(Infinity));
        ans[0][0] = 0;
        let queue = [[0, 0]];
        const dirs = [
            [0, 1],
            [0, -1],
            [1, 0],
            [-1, 0],
        ];
        while (queue.length) {
            let [x, y] = queue.shift();
            for (let step = 1; step < 5; step++) {
                let [dx, dy] = dirs[step - 1];
                let [i, j] = [x + dx, y + dy];
                if (i < 0 || i >= m || j < 0 || j >= n) continue;
                let cost = ~~(grid[x][y] != step) + ans[x][y];
                if (cost >= ans[i][j]) continue;
                ans[i][j] = cost;
                if (grid[x][y] == step) {
                    queue.unshift([i, j]);
                } else {
                    queue.push([i, j]);
                }
            }
        }
        return ans[m - 1][n - 1];
    }
    
    

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