Question

1036. Escape a Large Maze

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Description

There is a 1 million by 1 million grid on an XY-plane, and the coordinates of each grid square are (x, y).

We start at the source = [sx, sy] square and want to reach the target = [tx, ty] square. There is also an array of blocked squares, where each blocked[i] = [xi, yi] represents a blocked square with coordinates (xi, yi).

Each move, we can walk one square north, east, south, or west if the square is not in the array of blocked squares. We are also not allowed to walk outside of the grid.

Return true_ if and only if it is possible to reach the _target_ square from the _source_ square through a sequence of valid moves_.

 

Example 1:

Input: blocked = [[0,1],[1,0]], source = [0,0], target = [0,2]
Output: false
Explanation: The target square is inaccessible starting from the source square because we cannot move.
We cannot move north or east because those squares are blocked.
We cannot move south or west because we cannot go outside of the grid.

Example 2:

Input: blocked = [], source = [0,0], target = [999999,999999]
Output: true
Explanation: Because there are no blocked cells, it is possible to reach the target square.

 

Constraints:

• 0 <= blocked.length <= 200
• blocked[i].length == 2
• 0 <= xi, yi < 106
• source.length == target.length == 2
• 0 <= sx, sy, tx, ty < 106
• source != target
• It is guaranteed that source and target are not blocked.

Solutions

Solution 1

class Solution:
  def isEscapePossible(
    self, blocked: List[List[int]], source: List[int], target: List[int]
  ) -> bool:
    def dfs(source, target, seen):
      x, y = source
      if (
        not (0 <= x < 10**6 and 0 <= y < 10**6)
        or (x, y) in blocked
        or (x, y) in seen
      ):
        return False
      seen.add((x, y))
      if len(seen) > 20000 or source == target:
        return True
      for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
        next = [x + a, y + b]
        if dfs(next, target, seen):
          return True
      return False

    blocked = set((x, y) for x, y in blocked)
    return dfs(source, target, set()) and dfs(target, source, set())

class Solution {
  private int[][] dirs = new int[][] {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
  private static final int N = (int) 1e6;
  private Set<Integer> blocked;

  public boolean isEscapePossible(int[][] blocked, int[] source, int[] target) {
    this.blocked = new HashSet<>();
    for (int[] b : blocked) {
      this.blocked.add(b[0] * N + b[1]);
    }
    return dfs(source, target, new HashSet<>()) && dfs(target, source, new HashSet<>());
  }

  private boolean dfs(int[] source, int[] target, Set<Integer> seen) {
    int sx = source[0], sy = source[1];
    int tx = target[0], ty = target[1];
    if (sx < 0 || sx >= N || sy < 0 || sy >= N || tx < 0 || tx >= N || ty < 0 || ty >= N
      || blocked.contains(sx * N + sy) || seen.contains(sx * N + sy)) {
      return false;
    }
    seen.add(sx * N + sy);
    if (seen.size() > 20000 || (sx == target[0] && sy == target[1])) {
      return true;
    }
    for (int[] dir : dirs) {
      if (dfs(new int[] {sx + dir[0], sy + dir[1]}, target, seen)) {
        return true;
      }
    }
    return false;
  }
}

typedef unsigned long long ULL;

class Solution {
public:
  vector<vector<int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
  unordered_set<ULL> blocked;
  int N = 1e6;

  bool isEscapePossible(vector<vector<int>>& blocked, vector<int>& source, vector<int>& target) {
    this->blocked.clear();
    for (auto& b : blocked) this->blocked.insert((ULL) b[0] * N + b[1]);
    unordered_set<ULL> s1;
    unordered_set<ULL> s2;
    return dfs(source, target, s1) && dfs(target, source, s2);
  }

  bool dfs(vector<int>& source, vector<int>& target, unordered_set<ULL>& seen) {
    int sx = source[0], sy = source[1];
    int tx = target[0], ty = target[1];
    if (sx < 0 || sx >= N || sy < 0 || sy >= N || tx < 0 || tx >= N || ty < 0 || ty >= N || blocked.count((ULL) sx * N + sy) || seen.count((ULL) sx * N + sy)) return 0;
    seen.insert((ULL) sx * N + sy);
    if (seen.size() > 20000 || (sx == target[0] && sy == target[1])) return 1;
    for (auto& dir : dirs) {
      vector<int> next = {sx + dir[0], sy + dir[1]};
      if (dfs(next, target, seen)) return 1;
    }
    return 0;
  }
};

func isEscapePossible(blocked [][]int, source []int, target []int) bool {
  const N = 1e6
  dirs := [4][2]int{{0, -1}, {0, 1}, {1, 0}, {-1, 0}}
  block := make(map[int]bool)
  for _, b := range blocked {
    block[b[0]*N+b[1]] = true
  }
  var dfs func(source, target []int, seen map[int]bool) bool
  dfs = func(source, target []int, seen map[int]bool) bool {
    sx, sy := source[0], source[1]
    tx, ty := target[0], target[1]
    if sx < 0 || sx >= N || sy < 0 || sy >= N || tx < 0 || tx >= N || ty < 0 || ty >= N || block[sx*N+sy] || seen[sx*N+sy] {
      return false
    }
    seen[sx*N+sy] = true
    if len(seen) > 20000 || (sx == target[0] && sy == target[1]) {
      return true
    }
    for _, dir := range dirs {
      next := []int{sx + dir[0], sy + dir[1]}
      if dfs(next, target, seen) {
        return true
      }
    }
    return false
  }
  s1, s2 := make(map[int]bool), make(map[int]bool)
  return dfs(source, target, s1) && dfs(target, source, s2)
}

use std::collections::{ HashSet, VecDeque };

const BOUNDARY: i32 = 1_000_000;
const MAX: usize = 20000;

impl Solution {
  pub fn is_escape_possible(blocked: Vec<Vec<i32>>, source: Vec<i32>, target: Vec<i32>) -> bool {
    let mut block = HashSet::with_capacity(blocked.len());
    for b in blocked.iter() {
      block.insert((b[0], b[1]));
    }
    bfs(&block, &source, &target) && bfs(&block, &target, &source)
  }
}

fn bfs(block: &HashSet<(i32, i32)>, source: &Vec<i32>, target: &Vec<i32>) -> bool {
  let dir = vec![(-1, 0), (1, 0), (0, -1), (0, 1)];

  let mut queue = VecDeque::new();
  let mut vis = HashSet::new();
  queue.push_back((source[0], source[1]));
  vis.insert((source[0], source[1]));

  while !queue.is_empty() && vis.len() < MAX {
    let (x, y) = queue.pop_front().unwrap();
    if x == target[0] && y == target[1] {
      return true;
    }
    for (dx, dy) in dir.iter() {
      let (nx, ny) = (x + dx, y + dy);
      if
        nx < 0 ||
        nx >= BOUNDARY ||
        ny < 0 ||
        ny >= BOUNDARY ||
        vis.contains(&(nx, ny)) ||
        block.contains(&(nx, ny))
      {
        continue;
      }
      queue.push_back((nx, ny));
      vis.insert((nx, ny));
    }
  }

  vis.len() >= MAX
}