[LeetCode] 2257. Count Unguarded Cells in the Grid

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.

統計網格圖中沒有被保衛的格子數。

給你兩個整數 m 和 n 表示一個下標從 0 開始的 m x n 網格圖。同時給你兩個二維整數陣列 guards 和 walls ，其中 guards[i] = [rowi, coli] 且 walls[j] = [rowj, colj] ，分別表示第 i 個警衛和第 j 座牆所在的位置。

一個警衛能看到 4 個座標軸方向（即東、南、西、北）的 所有 格子，除非他們被一座牆或者另外一個警衛 擋住 了視線。如果一個格子能被 至少 一個警衛看到，那麼我們說這個格子被 保衛 了。

請你返回空格子中，有多少個格子是 沒被保衛 的。

思路

需要建立一個額外的 m x n 二維陣列，用來記錄每個格子上是否有守衛，是否有牆和是否有被警衛看到，如果都不是，則說明這個格子是沒被保衛的。

複雜度

時間O(mn)
空間O(mn)

程式碼

Java實現

class Solution {
    int[][] dirs = new int[][] {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    public int countUnguarded(int m, int n, int[][] guards, int[][] walls) {
        char[][] grid = new char[m][n];
		// 標註guard和wall的位置
        for (int[] g : guards) {
            int x = g[0];
            int y = g[1];
            grid[x][y] = 'G';
        }
        for (int[] w : walls) {
            int x = w[0];
            int y = w[1];
            grid[x][y] = 'W';
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
				// 往guard的四個方向看
                if (grid[i][j] == 'G') {
                    for (int[] dir : dirs) {
                        int x = i + dir[0];
                        int y = j + dir[1];
                        while (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] != 'G' && grid[x][y] != 'W') {
                            grid[x][y] = 'I';
                            x += dir[0];
                            y += dir[1];
                        }
                    }
                }
            }
        }

        int res = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] != 'I' && grid[i][j] != 'G' && grid[i][j] != 'W') {
                    res++;
                }
            }
        }
        return res;
    }
}