Algorithm
Problem Name: 419. Battleships in a Board
Given an m x n matrix board where each cell is a battleship 'X' or empty '.', return the number of the battleships on board.
Battleships can only be placed horizontally or vertically on board. In other words, they can only be made of the shape 1 x k (1 row, k columns) or k x 1 (k rows, 1 column), where k can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).
Example 1:
Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]] Output: 2
Example 2:
Input: board = [["."]] Output: 0
Constraints:
m == board.lengthn == board[i].length1 <= m, n <= 200board[i][j]is either'.'or'X'.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int countBattleships(char[][] board) {
int count = 0;
int rows = board.length;
int cols = board[0].length;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (board[i][j] == '.' || (i > 0 && board[i - 1][j] == 'X') || (j > 0 && board[i][j - 1] == 'X')) {
continue;
}
count++;
}
}
return count;
}
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def countBattleships(self, board):
return sum(board[i][j] == "X" and (i == 0 or board[i - 1][j] == ".") and (j == 0 or board[i][j - 1] == ".") for i in range(len(board)) for j in range(len(board[0])))
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0419_BattleshipsInABoard
{
public int CountBattleships(char[][] board)
{
int rows = board.Length, cols = board[0].Length;
var count = 0;
for (int i = 0; i < rows; i++)
{
for (int j = 0; j < cols; j++)
{
if (board[i][j] == 'X')
{
if ((i == 0 || board[i - 1][j] != 'X') &&
(j == 0 || board[i][j - 1] != 'X'))
count++;
}
}
}
return count;
}
}
}
Input