Question

2257. Count Unguarded Cells in the Grid

中文文档

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:
  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;
}