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Formatted question description: https://leetcode.ca/all/1091.html

1091. Shortest Path in Binary Matrix

Level

Medium

Description

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

  • Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
  • C_1 is at location (0, 0) (ie. has value grid[0][0])
  • C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
  • If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right. If such a path does not exist, return -1.

Example 1:

Input: [[0,1],[1,0]]

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Output: 2

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Example 2:

Input: [[0,0,0],[1,1,0],[1,1,0]]

Image text

Output: 4

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Note:

  1. 1 <= grid.length == grid[0].length <= 100
  2. grid[r][c] is 0 or 1

Solution

If grid has 0 rows or 0 columns, then such a path does not exist. Also if the top-left corner of grid is blocked or the bottom-right corner of grid is blocked, then such a path does not exist. In these cases, return -1.

Use breadth first search to find the length of the shortest path. The top-left corner has path length 1. Starting from the top-left corner, search all adjacent cells that are connected 8-directionally. If an adjacent cell is empty and not visited, update the distance of the adjacent cell and add it to the queue for further search.

If the bottom-right corner can be reached, return the length of the shortest path. Otherwise, return -1.

  • class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0)
            return -1;
        int side = grid.length;
        if (grid[0][0] == 1 || grid[side - 1][side - 1] == 1)
            return -1;
        final int BLOCK = -1;
        final int WHITE = 0;
        final int GRAY = 1;
        final int BLACK = 2;
        int[][] colors = new int[side][side];
        int[][] distances = new int[side][side];
        for (int i = 0; i < side; i++) {
            for (int j = 0; j < side; j++) {
                if (grid[i][j] == 0) {
                    colors[i][j] = WHITE;
                    distances[i][j] = Integer.MAX_VALUE;
                } else {
                    colors[i][j] = BLOCK;
                    distances[i][j] = -1;
                }
            }
        }
        colors[0][0] = GRAY;
        distances[0][0] = 1;
        int[][] directions = { {-1, -1}, {-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1} };
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(new int[]{0, 0});
        while (!queue.isEmpty()) {
            int[] cell = queue.poll();
            int row = cell[0], column = cell[1];
            int distance = distances[row][column];
            for (int[] direction : directions) {
                int newRow = row + direction[0], newColumn = column + direction[1];
                if (newRow >= 0 && newRow < side && newColumn >= 0 && newColumn < side && colors[newRow][newColumn] == WHITE) {
                    colors[newRow][newColumn] = GRAY;
                    distances[newRow][newColumn] = distance + 1;
                    queue.offer(new int[]{newRow, newColumn});
                }
            }
            colors[row][column] = BLACK;
        }
        return distances[side - 1][side - 1] == Integer.MAX_VALUE ? -1 : distances[side - 1][side - 1];
    }
}

  • // OJ: https://leetcode.com/problems/shortest-path-in-binary-matrix/
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& G) {
        if (G[0][0] == 1) return -1;
        int N = G.size();
        vector<vector<int>> dist(N, vector<int>(N, INT_MAX));
        queue<pair<int, int>> q;
        q.emplace(0, 0);
        dist[0][0] = 1;
        while (q.size()) {
            auto [x, y] = q.front();
            q.pop();
            for (int dx = -1; dx <= 1; ++dx) {
                for (int dy = -1; dy <= 1; ++dy) {
                    if (dx == 0 && dy == 0) continue;
                    int a = x + dx, b = y + dy;
                    if (a < 0 || a >= N || b < 0 || b >= N || G[a][b] == 1 || dist[a][b] != INT_MAX) continue;
                    dist[a][b] = dist[x][y] + 1;
                    q.emplace(a, b);
                }
            }
        }
        return dist[N - 1][N - 1] == INT_MAX ? -1 : dist[N - 1][N - 1];
    }
};

  • class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        if grid[0][0]:
            return -1
        n = len(grid)
        q = deque([(0, 0)])
        grid[0][0] = 1
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                i, j = q.popleft()
                if (i, j) == (n - 1, n - 1):
                    return ans
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < n and 0 <= y < n and grid[x][y] == 0:
                            q.append((x, y))
                            grid[x][y] = 1
        return -1

############

# 1091. Shortest Path in Binary Matrix
# https://leetcode.com/problems/shortest-path-in-binary-matrix/

class Solution:
    def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        if grid[0][0] == 1: return -1
        
        queue = collections.deque([(0, 0, 1)])
        visited = set()
        
        while queue:
            x, y, steps = queue.popleft()
            if x == rows - 1 and y == cols - 1: return steps
            
            for dx in range(x - 1, x + 2):
                for dy in range(y - 1, y + 2):
                    if 0 <= dx < rows and 0 <= dy < cols and grid[dx][dy] == 0 and (dx, dy) not in visited:
                        queue.append((dx, dy, steps + 1))
                        visited.add((dx, dy))
        
        return -1


  • func shortestPathBinaryMatrix(grid [][]int) int {
	if grid[0][0] == 1 {
		return -1
	}
	n := len(grid)
	q := [][]int{[]int{0, 0} }
	grid[0][0] = 1
	ans := 0
	for len(q) > 0 {
		ans++
		for m := len(q); m > 0; m-- {
			p := q[0]
			q = q[1:]
			i, j := p[0], p[1]
			if i == n-1 && j == n-1 {
				return ans
			}
			for x := i - 1; x <= i+1; x++ {
				for y := j - 1; y <= j+1; y++ {
					if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 0 {
						q = append(q, []int{x, y})
						grid[x][y] = 1
					}
				}
			}
		}
	}
	return -1
}

  • use std::collections::VecDeque;
impl Solution {
    pub fn shortest_path_binary_matrix(mut grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut queue = VecDeque::new();
        queue.push_back([0, 0]);
        let mut res = 0;
        while !queue.is_empty() {
            res += 1;
            for _ in 0..queue.len() {
                let [i, j] = queue.pop_front().unwrap();
                if grid[i][j] == 1 {
                    continue;
                }
                if i == n - 1 && j == n - 1 {
                    return res;
                }
                grid[i][j] = 1;
                for x in -1..=1 {
                    for y in -1..=1 {
                        let x = x + i as i32;
                        let y = y + j as i32;
                        if x < 0 || x == n as i32 || y < 0 || y == n as i32 {
                            continue;
                        }
                        queue.push_back([x as usize, y as usize]);
                    }
                }
            }
        }
        -1
    }
}


  • function shortestPathBinaryMatrix(grid: number[][]): number {
    if (grid[0][0]) {
        return -1;
    }
    const n = grid.length;
    grid[0][0] = 1;
    let q: number[][] = [[0, 0]];
    for (let ans = 1; q.length > 0; ++ans) {
        const nq: number[][] = [];
        for (const [i, j] of q) {
            if (i === n - 1 && j === n - 1) {
                return ans;
            }
            for (let x = i - 1; x <= i + 1; ++x) {
                for (let y = j - 1; y <= j + 1; ++y) {
                    if (x >= 0 && x < n && y >= 0 && y < n && !grid[x][y]) {
                        grid[x][y] = 1;
                        nq.push([x, y]);
                    }
                }
            }
        }
        q = nq;
    }
    return -1;
}

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