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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.

  • 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;
        }
    };
    
  • 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)
    
    
  • 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;
    }
    
    

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