# 419. Battleships in a Board (Medium)

https://leetcode.com/problems/battleships-in-a-board/

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.

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

## Solutions

class Solution {
public int countBattleships(char[][] board) {
int count = 0;

int row = board.length;
int col = board[0].length;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
// ignore slot with '.' since only 'X' comprises battleship
if (board[i][j] == '.') {
continue;
}

// The slot value of board[i][j] should be 'X' to reach here

if (i - 1 >= 0 && board[i - 1][j] == 'X') {
continue;
}

// upper slot with '.' indicates board[i][j] edge of the ship

if (j - 1 >= 0 && board[i][j - 1] == 'X') {
continue;
}

// left slot with '.' indicates board[i][j] edge of the ship

// To reach here, meets following two criterions

// 1. i == 0 || board[i-1][j] == '.'
// 2. j == 0 || board[i][j-1] == '.'

count++;
}
}

return count;
}
}