# Bomb Enemy

Bomb Enemy dynamic programming

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)
``````

Bomb Enemy Solution

`````` public class BombEnemy {
public int maxKilledEnemies(char[][] grid) {
if(grid == null || grid.length == 0 ||  grid[0].length == 0) {
return 0;
}

int max = 0;
int row = 0;
int[] col = new int[grid[0].length];

for(int i = 0; i<grid.length; i++) {
for(int j = 0; j<grid[0].length;j++) {
if(grid[i][j] == 'W') {
continue;
}

if(j == 0 || grid[i][j-1] == 'W') {
row = killedEnemiesRow(grid, i, j);
}

if(i == 0 || grid[i-1][j] == 'W') {
col[j] = killedEnemiesCol(grid,i,j);
}

if(grid[i][j] == '0') {
max = (row + col[j] > max) ? row + col[j] : max;
}
}
}

return max;
}

calculate killed enemies for row i from column j
private int killedEnemiesRow(char[][] grid, int i, int j) {
int num = 0;

while(j <= grid[0].length-1 && grid[i][j] != 'W') {
if(grid[i][j] == 'E') {
num++;
}

j++;
}

return num;
}

calculate killed enemies for  column j from row i
private int killedEnemiesCol(char[][] grid, int i, int j) {
int num = 0;

while(i <= grid.length -1 && grid[i][j] != 'W'){
if(grid[i][j] == 'E') {
num++;
}

i++;
}

return num;
}
}``````