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2814. Minimum Time Takes to Reach Destination Without Drowning
Description
You are given an n * m 0-indexed grid of string land. Right now, you are standing at the cell that contains "S", and you want to get to the cell containing "D". There are three other types of cells in this land:
".": These cells are empty."X": These cells are stone."*": These cells are flooded.
At each second, you can move to a cell that shares a side with your current cell (if it exists). Also, at each second, every empty cell that shares a side with a flooded cell becomes flooded as well.
There are two problems ahead of your journey:
- You can't step on stone cells.
- You can't step on flooded cells since you will drown (also, you can't step on a cell that will be flooded at the same time as you step on it).
Return the minimum time it takes you to reach the destination in seconds, or -1 if it is impossible.
Note that the destination will never be flooded.
Example 1:
Input: land = [["D",".","*"],[".",".","."],[".","S","."]] Output: 3 Explanation: The picture below shows the simulation of the land second by second. The blue cells are flooded, and the gray cells are stone. Picture (0) shows the initial state and picture (3) shows the final state when we reach destination. As you see, it takes us 3 second to reach destination and the answer would be 3. It can be shown that 3 is the minimum time needed to reach from S to D.
Example 2:
Input: land = [["D","X","*"],[".",".","."],[".",".","S"]] Output: -1 Explanation: The picture below shows the simulation of the land second by second. The blue cells are flooded, and the gray cells are stone. Picture (0) shows the initial state. As you see, no matter which paths we choose, we will drown at the 3rd second. Also the minimum path takes us 4 seconds to reach from S to D. So the answer would be -1.
Example 3:
Input: land = [["D",".",".",".","*","."],[".","X",".","X",".","."],[".",".",".",".","S","."]] Output: 6 Explanation: It can be shown that we can reach destination in 6 seconds. Also it can be shown that 6 is the minimum seconds one need to reach from S to D.
Constraints:
2 <= n, m <= 100landconsists only of"S","D",".","*"and"X".- Exactly one of the cells is equal to
"S". - Exactly one of the cells is equal to
"D".
Solutions
Solution 1: Two BFS Traversals
First, we run a BFS (Breadth-First Search) to calculate the shortest distance from each cell to the water, and record it in the array $g$. Then, we run another BFS starting from the cell $(s_i, s_j)$ to find the shortest distance to the target cell $(d_i, d_j)$. During this process, if the adjacent cell $(x, y)$ of the current cell $(i, j)$ satisfies $g[x][y] > t + 1$, then we can move from $(x, y)$ to $(i, j)$.
The time complexity is $O(m \times n)$ and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the number of rows and columns of the array $land$, respectively.
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class Solution { public int minimumSeconds(List<List<String>> land) { int m = land.size(), n = land.get(0).size(); boolean[][] vis = new boolean[m][n]; int[][] g = new int[m][n]; Deque<int[]> q = new ArrayDeque<>(); int si = 0, sj = 0; for (int i = 0; i < m; ++i) { Arrays.fill(g[i], 1 << 30); for (int j = 0; j < n; ++j) { String c = land.get(i).get(j); if ("*".equals(c)) { q.offer(new int[] {i, j}); } else if ("S".equals(c)) { si = i; sj = j; } } } int[] dirs = {-1, 0, 1, 0, -1}; for (int t = 0; !q.isEmpty(); ++t) { for (int k = q.size(); k > 0; --k) { int[] p = q.poll(); int i = p[0], j = p[1]; g[i][j] = t; for (int d = 0; d < 4; ++d) { int x = i + dirs[d], y = j + dirs[d + 1]; if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) { boolean empty = ".".equals(land.get(x).get(y)); boolean start = "S".equals(land.get(x).get(y)); if (empty || start) { vis[x][y] = true; q.offer(new int[] {x, y}); } } } } } q.offer(new int[] {si, sj}); vis = new boolean[m][n]; vis[si][sj] = true; for (int t = 0; !q.isEmpty(); ++t) { for (int k = q.size(); k > 0; --k) { int[] p = q.poll(); int i = p[0], j = p[1]; if ("D".equals(land.get(i).get(j))) { return t; } for (int d = 0; d < 4; ++d) { int x = i + dirs[d], y = j + dirs[d + 1]; if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && g[x][y] > t + 1) { boolean empty = ".".equals(land.get(x).get(y)); boolean dest = "D".equals(land.get(x).get(y)); if (empty || dest) { vis[x][y] = true; q.offer(new int[] {x, y}); } } } } } return -1; } } -
class Solution { public: int minimumSeconds(vector<vector<string>>& land) { int m = land.size(), n = land[0].size(); bool vis[m][n]; int g[m][n]; memset(vis, false, sizeof(vis)); memset(g, 0x3f, sizeof(g)); queue<pair<int, int>> q; int si = 0, sj = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { auto c = land[i][j]; if (c == "*") { q.emplace(i, j); } else if (c == "S") { si = i; sj = j; } } } int dirs[5] = {-1, 0, 1, 0, -1}; for (int t = 0; !q.empty(); ++t) { for (int k = q.size(); k; --k) { auto [i, j] = q.front(); q.pop(); g[i][j] = t; for (int d = 0; d < 4; ++d) { int x = i + dirs[d], y = j + dirs[d + 1]; if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y]) { bool empty = land[x][y] == "."; bool start = land[x][y] == "S"; if (empty || start) { vis[x][y] = true; q.emplace(x, y); } } } } } q.emplace(si, sj); memset(vis, false, sizeof(vis)); vis[si][sj] = true; for (int t = 0; !q.empty(); ++t) { for (int k = q.size(); k; --k) { auto [i, j] = q.front(); q.pop(); if (land[i][j] == "D") { return t; } for (int d = 0; d < 4; ++d) { int x = i + dirs[d], y = j + dirs[d + 1]; if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && g[x][y] > t + 1) { bool empty = land[x][y] == "."; bool dest = land[x][y] == "D"; if (empty || dest) { vis[x][y] = true; q.emplace(x, y); } } } } } return -1; } }; -
class Solution: def minimumSeconds(self, land: List[List[str]]) -> int: m, n = len(land), len(land[0]) vis = [[False] * n for _ in range(m)] g = [[inf] * n for _ in range(m)] q = deque() si = sj = 0 for i, row in enumerate(land): for j, c in enumerate(row): match c: case "*": q.append((i, j)) case "S": si, sj = i, j dirs = (-1, 0, 1, 0, -1) t = 0 while q: for _ in range(len(q)): i, j = q.popleft() g[i][j] = t for a, b in pairwise(dirs): x, y = i + a, j + b if ( 0 <= x < m and 0 <= y < n and not vis[x][y] and land[x][y] in ".S" ): vis[x][y] = True q.append((x, y)) t += 1 t = 0 q = deque([(si, sj)]) vis = [[False] * n for _ in range(m)] vis[si][sj] = True while q: for _ in range(len(q)): i, j = q.popleft() if land[i][j] == "D": return t for a, b in pairwise(dirs): x, y = i + a, j + b if ( 0 <= x < m and 0 <= y < n and g[x][y] > t + 1 and not vis[x][y] and land[x][y] in ".D" ): vis[x][y] = True q.append((x, y)) t += 1 return -1 -
func minimumSeconds(land [][]string) int { m, n := len(land), len(land[0]) vis := make([][]bool, m) g := make([][]int, m) q := [][2]int{} var si, sj int for i, row := range land { vis[i] = make([]bool, n) g[i] = make([]int, n) for j := range g[i] { g[i][j] = 1 << 30 } for j, c := range row { if c == "*" { q = append(q, [2]int{i, j}) } else if c == "S" { si, sj = i, j } } } dirs := [5]int{-1, 0, 1, 0, -1} for t := 0; len(q) > 0; t++ { for k := len(q); k > 0; k-- { p := q[0] q = q[1:] i, j := p[0], p[1] g[i][j] = t for d := 0; d < 4; d++ { x, y := i+dirs[d], j+dirs[d+1] if x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] { empty := land[x][y] == "." start := land[x][y] == "S" if empty || start { vis[x][y] = true q = append(q, [2]int{x, y}) } } } } } q = append(q, [2]int{si, sj}) vis = make([][]bool, m) for i := range vis { vis[i] = make([]bool, n) } vis[si][sj] = true for t := 0; len(q) > 0; t++ { for k := len(q); k > 0; k-- { p := q[0] q = q[1:] i, j := p[0], p[1] if land[i][j] == "D" { return t } for d := 0; d < 4; d++ { x, y := i+dirs[d], j+dirs[d+1] if x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && g[x][y] > t+1 { empty := land[x][y] == "." dest := land[x][y] == "D" if empty || dest { vis[x][y] = true q = append(q, [2]int{x, y}) } } } } } return -1 } -
function minimumSeconds(land: string[][]): number { const m = land.length; const n = land[0].length; const g: number[][] = Array(m) .fill(0) .map(() => Array(n).fill(1 << 30)); const vis: boolean[][] = Array(m) .fill(0) .map(() => Array(n).fill(false)); const q: number[][] = []; let [si, sj] = [0, 0]; for (let i = 0; i < m; ++i) { for (let j = 0; j < n; ++j) { const c = land[i][j]; if (c === '*') { q.push([i, j]); } else if (c === 'S') { [si, sj] = [i, j]; } } } const dirs = [-1, 0, 1, 0, -1]; for (let t = 0; q.length; ++t) { for (let k = q.length; k; --k) { const [i, j] = q.shift()!; g[i][j] = t; for (let d = 0; d < 4; ++d) { const [x, y] = [i + dirs[d], j + dirs[d + 1]]; if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && 'S.'.includes(land[x][y])) { vis[x][y] = true; q.push([x, y]); } } } } q.push([si, sj]); for (let i = 0; i < m; ++i) { vis[i].fill(false); } vis[si][sj] = true; for (let t = 0; q.length; ++t) { for (let k = q.length; k; --k) { const [i, j] = q.shift()!; if (land[i][j] === 'D') { return t; } for (let d = 0; d < 4; ++d) { const [x, y] = [i + dirs[d], j + dirs[d + 1]]; if ( x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && g[x][y] > t + 1 && 'D.'.includes(land[x][y]) && t + 1 < g[x][y] ) { vis[x][y] = true; q.push([x, y]); } } } } return -1; }