361. Bomb Enemy

361. Bomb Enemy (Medium)

Given a 2D grid, each cell is either a wall 'W', an enemy 'E' or empty '0' (the number zero), return the maximum enemies you can kill using one bomb.
The bomb kills all the enemies in the same row and column from the planted point until it hits the wall since the wall is too strong to be destroyed.
Note: You can only put the bomb at an empty cell.

Example:

Input: [["0","E","0","0"],["E","0","W","E"],["0","E","0","0"]]
Output: 3 
Explanation: For the given grid,
0 E 0 0
E 0 W E
0 E 0 0
Placing a bomb at (1,1) kills 3 enemies.

Companies:
Google, Uber

Related Topics:
Dynamic Programming

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/bomb-enemy/
// Author: github.com/lzl124631x
// Time: O(MN(M+N))
// Space: O(1)
class Solution {
private:
    int M, N;
    int dirs[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    int count(vector<vector<char>>& grid, int x, int y, int dir[2]) {
        int ans = 0;
        while (true) {
            x += dir[0];
            y += dir[1];
            if (x < 0 || x >= M || y < 0 || y >= N || grid[x][y] == 'W') break;
            if (grid[x][y] == 'E') ++ans;
        }
        return ans;
    }
    int count(vector<vector<char>>& grid, int x, int y) {
        int ans = 0;
        for (auto &dir : dirs) ans += count(grid, x, y, dir);
        return ans;
    }
public:
    int maxKilledEnemies(vector<vector<char>>& grid) {
        if (grid.empty() || grid[0].empty()) return 0;
        M = grid.size();
        N = grid[0].size();
        int ans = 0;
        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (grid[i][j] != '0') continue;
                ans = max(ans, count(grid, i, j));
            }
        }
        return ans;
    }
};