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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]togrid[i][j + 1]) - 2 which means go to the cell to the left. (i.e go from
grid[i][j]togrid[i][j - 1]) - 3 which means go to the lower cell. (i.e go from
grid[i][j]togrid[i + 1][j]) - 4 which means go to the upper cell. (i.e go from
grid[i][j]togrid[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.lengthn == grid[i].length1 <= 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]; }