[LeetCode] 361. Bomb Enemy

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 that you can only put the bomb at an empty cell.
Example:
For the given grid

0 E 0 0
E 0 W E
0 E 0 0

return 3. (Placing a bomb at (1,1) kills 3 enemies)

Thought process:
Iterate through the matrix. Use global variables to keep track of row hits and column hits. These will propagate until we reach a wall.

Solution:

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class Solution {
    public int maxKilledEnemies(char[][] grid) {
        int height = grid.length;
        if (height == 0) {
            return 0;
        }
        int width = grid[0].length;
        
        int max = 0;
        int rowHits = 0;
        int[] colHits = new int[width];
        
        for (int i = 0; i < height; i++) {
            for (int j = 0; j < width; j++) {
                if (j == 0 || grid[i][j - 1] == 'W') {
                    rowHits = 0;
                    for (int k = j; k < width && grid[i][k] != 'W'; k++) {
                        rowHits += grid[i][k] == 'E' ? 1 : 0;
                    }
                }
                if (i == 0 || grid[i - 1][j] == 'W') {
                    colHits[j] = 0;
                    for (int k = i; k < height && grid[k][j] != 'W'; k++) {
                        colHits[j] += grid[k][j] == 'E' ? 1 : 0;
                    }
                }
                if (grid[i][j] == '0') {
                    max = Math.max(max, rowHits + colHits[j]);
                }
            }
        }
        return max;
    }
}

Time complexity:
O(height * width). The inner-most for-loops only execute when we're at the first row / column or when we reach a wall.
Location: Philadelphia, PA, USA

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