1036. Escape a Large Maze

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 = [s_x, s_y] square and want to reach the target = [t_x, t_y] square. There is also an array of blocked squares, where each blocked[i] = [x_i, y_i] represents a blocked square with coordinates (x_i, y_i).

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 <= x_i, y_i < 10⁶
source.length == target.length == 2
0 <= s_x, s_y, t_x, t_y < 10⁶
source != target
It is guaranteed that source and target are not blocked.

Solutions

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;
    }
};

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())

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
}

1036 - Escape a Large Maze

1036. Escape a Large Maze

Description

Solutions

All Problems

All Solutions

Welcome to Subscribe On Youtube

1036. Escape a Large Maze

Description

Solutions

All Problems

All Solutions