2257. Count Unguarded Cells in the Grid

Description

You are given two integers m and n representing a 0-indexed m x n grid. You are also given two 2D integer arrays guards and walls where guards[i] = [rowi, coli] and walls[j] = [rowj, colj] represent the positions of the ith guard and jth wall respectively.

A guard can see every cell in the four cardinal directions (north, east, south, or west) starting from their position unless obstructed by a wall or another guard. A cell is guarded if there is at least one guard that can see it.

Return the number of unoccupied cells that are not guarded.

 

Example 1:

Input: m = 4, n = 6, guards = [[0,0],[1,1],[2,3]], walls = [[0,1],[2,2],[1,4]]
Output: 7
Explanation: The guarded and unguarded cells are shown in red and green respectively in the above diagram.
There are a total of 7 unguarded cells, so we return 7.

Example 2:

Input: m = 3, n = 3, guards = [[1,1]], walls = [[0,1],[1,0],[2,1],[1,2]]
Output: 4
Explanation: The unguarded cells are shown in green in the above diagram.
There are a total of 4 unguarded cells, so we return 4.

 

Constraints:

• 1 <= m, n <= 105
• 2 <= m * n <= 105
• 1 <= guards.length, walls.length <= 5 * 104
• 2 <= guards.length + walls.length <= m * n
• guards[i].length == walls[j].length == 2
• 0 <= rowi, rowj < m
• 0 <= coli, colj < n
• All the positions in guards and walls are unique.

Solutions

Solution 1: Simulation

We create a two-dimensional array $g$ of size $m \times n$, where $g[i][j]$ represents the cell in row $i$ and column $j$. Initially, the value of $g[i][j]$ is $0$, indicating that the cell is not guarded.

Then, we traverse all guards and walls, and set the value of $g[i][j]$ to $2$, indicating that these positions cannot be accessed.

Next, we traverse all guard positions, simulate in four directions from that position until we encounter a wall or guard, or go out of bounds. During the simulation, we set the value of the encountered cell to $1$, indicating that the cell is guarded.

Finally, we traverse $g$ and count the number of cells with a value of $0$, which is the answer.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns in the grid, respectively.

Python3JavaC++GoTypeScriptJavaScript

class Solution:
    def countUnguarded(
        self, m: int, n: int, guards: List[List[int]], walls: List[List[int]]
    ) -> int:
        g = [[0] * n for _ in range(m)]
        for i, j in guards:
            g[i][j] = 2
        for i, j in walls:
            g[i][j] = 2
        dirs = (-1, 0, 1, 0, -1)
        for i, j in guards:
            for a, b in pairwise(dirs):
                x, y = i, j
                while 0 <= x + a < m and 0 <= y + b < n and g[x + a][y + b] < 2:
                    x, y = x + a, y + b
                    g[x][y] = 1
        return sum(v == 0 for row in g for v in row)

class Solution {
    public int countUnguarded(int m, int n, int[][] guards, int[][] walls) {
        int[][] g = new int[m][n];
        for (var e : guards) {
            g[e[0]][e[1]] = 2;
        }
        for (var e : walls) {
            g[e[0]][e[1]] = 2;
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        for (var e : guards) {
            for (int k = 0; k < 4; ++k) {
                int x = e[0], y = e[1];
                int a = dirs[k], b = dirs[k + 1];
                while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
                    x += a;
                    y += b;
                    g[x][y] = 1;
                }
            }
        }
        int ans = 0;
        for (var row : g) {
            for (int v : row) {
                if (v == 0) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    int countUnguarded(int m, int n, vector<vector<int>>& guards, vector<vector<int>>& walls) {
        int g[m][n];
        memset(g, 0, sizeof(g));
        for (auto& e : guards) {
            g[e[0]][e[1]] = 2;
        }
        for (auto& e : walls) {
            g[e[0]][e[1]] = 2;
        }
        int dirs[5] = {-1, 0, 1, 0, -1};
        for (auto& e : guards) {
            for (int k = 0; k < 4; ++k) {
                int x = e[0], y = e[1];
                int a = dirs[k], b = dirs[k + 1];
                while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
                    x += a;
                    y += b;
                    g[x][y] = 1;
                }
            }
        }
        int ans = 0;
        for (auto& row : g) {
            ans += count(row, row + n, 0);
        }
        return ans;
    }
};

func countUnguarded(m int, n int, guards [][]int, walls [][]int) (ans int) {
    g := make([][]int, m)
    for i := range g {
        g[i] = make([]int, n)
    }
    for _, e := range guards {
        g[e[0]][e[1]] = 2
    }
    for _, e := range walls {
        g[e[0]][e[1]] = 2
    }
    dirs := [5]int{-1, 0, 1, 0, -1}
    for _, e := range guards {
        for k := 0; k < 4; k++ {
            x, y := e[0], e[1]
            a, b := dirs[k], dirs[k+1]
            for x+a >= 0 && x+a < m && y+b >= 0 && y+b < n && g[x+a][y+b] < 2 {
                x, y = x+a, y+b
                g[x][y] = 1
            }
        }
    }
    for _, row := range g {
        for _, v := range row {
            if v == 0 {
                ans++
            }
        }
    }
    return
}

function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
    const g: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
    for (const [i, j] of guards) {
        g[i][j] = 2;
    }
    for (const [i, j] of walls) {
        g[i][j] = 2;
    }
    const dirs: number[] = [-1, 0, 1, 0, -1];
    for (const [i, j] of guards) {
        for (let k = 0; k < 4; ++k) {
            let [x, y] = [i, j];
            let [a, b] = [dirs[k], dirs[k + 1]];
            while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
                x += a;
                y += b;
                g[x][y] = 1;
            }
        }
    }
    let ans = 0;
    for (const row of g) {
        for (const v of row) {
            ans += v === 0 ? 1 : 0;
        }
    }
    return ans;
}

function countUnguarded(m, n, guards, walls) {
    const g = Array.from({ length: m }, () => Array.from({ length: n }, () => 0));
    for (const [i, j] of guards) {
        g[i][j] = 2;
    }
    for (const [i, j] of walls) {
        g[i][j] = 2;
    }
    const dirs = [-1, 0, 1, 0, -1];
    for (const [i, j] of guards) {
        for (let k = 0; k < 4; ++k) {
            let [x, y] = [i, j];
            let [a, b] = [dirs[k], dirs[k + 1]];
            while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && g[x + a][y + b] < 2) {
                x += a;
                y += b;
                g[x][y] = 1;
            }
        }
    }
    let ans = 0;
    for (const row of g) {
        for (const v of row) {
            ans += v === 0 ? 1 : 0;
        }
    }
    return ans;
}

Solution 2: DFS + Simulation

TypeScriptJavaScript

function countUnguarded(m: number, n: number, guards: number[][], walls: number[][]): number {
    let c = 0;
    const mtx = Array.from({ length: m }, () => Array(n).fill(0));
    for (const [i, j] of guards) mtx[i][j] = 2;
    for (const [i, j] of walls) mtx[i][j] = 2;

    const dfs = (i: number, j: number, dx: number, dy: number) => {
        [i, j] = [i + dx, j + dy];

        if (i < 0 || m <= i || j < 0 || n <= j || mtx[i][j] === 2) return;

        if (mtx[i][j] === 0) {
            mtx[i][j] = 1;
            c++;
        }

        dfs(i, j, dx, dy);
    };

    const DIRS = [-1, 0, 1, 0, -1];
    for (const [i, j] of guards) {
        for (let k = 0; k < 4; k++) {
            const [dx, dy] = [DIRS[k], DIRS[k + 1]];
            dfs(i, j, dx, dy);
        }
    }

    return m * n - guards.length - walls.length - c;
}

function countUnguarded(m, n, guards, walls) {
    let c = 0;
    const mtx = Array.from({ length: m }, () => Array(n).fill(0));
    for (const [i, j] of guards) mtx[i][j] = 2;
    for (const [i, j] of walls) mtx[i][j] = 2;

    const dfs = (i, j, dx, dy) => {
        [i, j] = [i + dx, j + dy];

        if (i < 0 || m <= i || j < 0 || n <= j || mtx[i][j] === 2) return;

        if (mtx[i][j] === 0) {
            mtx[i][j] = 1;
            c++;
        }

        dfs(i, j, dx, dy);
    };

    const DIRS = [-1, 0, 1, 0, -1];
    for (const [i, j] of guards) {
        for (let k = 0; k < 4; k++) {
            const [dx, dy] = [DIRS[k], DIRS[k + 1]];
            dfs(i, j, dx, dy);
        }
    }

    return m * n - guards.length - walls.length - c;
}

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