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3286. Find a Safe Walk Through a Grid 👍

  • Time: $O(mn \cdot \texttt{health})$
  • Space: $O(mn \cdot \texttt{health})$
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class Solution {
 public:
  bool findSafeWalk(vector<vector<int>>& grid, int health) {
    constexpr int kDirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int m = grid.size();
    const int n = grid[0].size();
    const int initialHealth = health - grid[0][0];
    using T = tuple<int, int, int>;  // (i, j, h)
    queue<T> q{{{0, 0, initialHealth}}};
    vector<vector<vector<bool>>> seen(
        m, vector<vector<bool>>(n, vector<bool>(health + 1)));
    seen[0][0][initialHealth] = true;

    while (!q.empty())
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [i, j, h] = q.front();
        q.pop();
        if (i == m - 1 && j == n - 1 && h > 0)
          return true;
        for (const auto& [dx, dy] : kDirs) {
          const int x = i + dx;
          const int y = j + dy;
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          const int nextHealth = h - grid[x][y];
          if (nextHealth <= 0 || seen[x][y][nextHealth])
            continue;
          q.emplace(x, y, nextHealth);
          seen[x][y][nextHealth] = true;
        }
      }

    return false;
  }
};

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class Solution {
  public boolean findSafeWalk(List<List<Integer>> grid, int health) {
    record T(int i, int j, int h) {}
    final int[][] DIRS = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    final int m = grid.size();
    final int n = grid.get(0).size();
    final int initialHealth = health - grid.get(0).get(0);
    Queue<T> q = new ArrayDeque<>(List.of(new T(0, 0, initialHealth)));
    boolean[][][] seen = new boolean[m][n][health + 1];
    seen[0][0][initialHealth] = true;

    while (!q.isEmpty())
      for (int sz = q.size(); sz > 0; --sz) {
        final int i = q.peek().i;
        final int j = q.peek().j;
        final int h = q.poll().h;
        if (i == m - 1 && j == n - 1 && h > 0)
          return true;
        for (int k = 0; k < 4; ++k) {
          final int x = i + DIRS[k][0];
          final int y = j + DIRS[k][1];
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          final int nextHealth = h - grid.get(x).get(y);
          if (nextHealth <= 0 || seen[x][y][nextHealth])
            continue;
          q.offer(new T(x, y, nextHealth));
          seen[x][y][nextHealth] = true;
        }
      }

    return false;
  }
}

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class Solution:
  def findSafeWalk(self, grid: list[list[int]], health: int) -> bool:
    DIRS = ((0, 1), (1, 0), (0, -1), (-1, 0))
    m = len(grid)
    n = len(grid[0])
    initialHealth = health - grid[0][0]
    q = collections.deque([(0, 0, initialHealth)])
    seen = {(0, 0, initialHealth)}

    while q:
      for _ in range(len(q)):
        i, j, h = q.popleft()
        if i == m - 1 and j == n - 1 and h > 0:
          return True
        for dx, dy in DIRS:
          x = i + dx
          y = j + dy
          if x < 0 or x == m or y < 0 or y == n:
            continue
          nextHealth = h - grid[x][y]
          if nextHealth <= 0 or (x, y, nextHealth) in seen:
            continue
          q.append((x, y, nextHealth))
          seen.add((x, y, nextHealth))

    return False