2257. Count Unguarded Cells in the Grid

Medium

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.

解題

func countUnguarded(m int, n int, guards [][]int, walls [][]int) int {
    land := make([][]byte, m)
    
    for i := range land {
        land[i] = make([]byte, n)
    }
    
    for _, wall := range walls {
        land[wall[0]][wall[1]] = 'W'
    }
    
    for _, guard := range guards {
        
        x := guard[0]
        y := guard[1]
        
        land[x][y] = 'G'
        
        for i:=x+1; i<m; i++ {
            if (land[i][y]=='W') || (land[i][y]=='G') { break }
            land[i][y] = 'A'
        }
        
        for i:=x-1; i>=0; i-- {
            if (land[i][y]=='W') || (land[i][y]=='G')  { break }
            land[i][y] = 'A'
        }
        
        for j:=y+1; j<n; j++ {
            if (land[x][j]=='W') || (land[x][j]=='G')  { break }
            land[x][j] = 'A'
        }
        
        for j:=y-1; j>=0; j-- {
            if (land[x][j]=='W') || (land[x][j]=='G')  { break }
            land[x][j] = 'A'
        }
    }
    
    ans := 0
    for i:=0; i<m; i++ {
        for j:=0; j<n; j++ {
            if land[i][j]!='W' && land[i][j]!='G' && land[i][j]!='A' {
                ans++
            } 
        }
    }
    
    return ans
}