package array;

/**

* Created by gouthamvidyapradhan on 12/08/2017.

* Given an 2D board, count how many battleships are in it. The battleships are represented with 'X's, empty slots are represented with '.'s. You may assume the following rules:

You receive a valid board, made of only battleships or empty slots.

Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), where N can be of any size.

At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.

Example:

X..X

...X

...X

In the above board there are 2 battleships.

Invalid Example:

...X

XXXX

...X

This is an invalid board that you will not receive - as battleships will always have a cell separating between them.

Follow up:

Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

Solution:

The below solution works in one pass using only O(1) memory.

Iterate through each cell and add one to count if and only if the current cell equals 'X' and its adjacent upper and

left cell does not contain 'X'

*/

public class BattleshipsInABoard {

/**

* Main method

* @param args

* @throws Exception

*/

public static void main(String[] args) throws Exception{

char[][] board = {{'X', '.', '.', 'X'}, {'.', '.', '.', 'X'}, {'.', '.', '.', 'X'}};

System.out.println(new BattleshipsInABoard().countBattleships(board));

}

public int countBattleships(char[][] board) {

int count = 0;

for(int i = 0; i < board.length; i ++){

for(int j = 0; j < board[0].length; j ++){

if(board[i][j] == 'X'){

if(i - 1 >= 0){ //check for the boundary condition

if(board[i - 1][j] == 'X')

continue;

}

if(j - 1 >= 0){

if(board[i][j - 1] == 'X'){

continue;

}

}

count++;

}

}

}

return count;

}

}