419. Battleships in a Board LeetCode Solution

In this guide, you will get 419. Battleships in a Board LeetCode Solution with the best time and space complexity. The solution to Battleships in a Board problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Battleships in a Board solution in C++
4. Battleships in a Board solution in Java
5. Battleships in a Board solution in Python
6. Additional Resources

Problem Statement of 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.length
n == board[i].length
1 <= m, n <= 200
board[i][j] is either ‘.’ or ‘X’.

Follow up: Could you do it in one-pass, using only O(1) extra memory and without modifying the values board?

Complexity Analysis

• Time Complexity: O(mn)
• Space Complexity: O(1)

419. Battleships in a Board LeetCode Solution in C++

class Solution {
 public:
  int countBattleships(vector<vector<char>>& board) {
    int ans = 0;

    for (int i = 0; i < board.size(); ++i)
      for (int j = 0; j < board[0].size(); ++j) {
        if (board[i][j] == '.')
          continue;
        if (i > 0 && board[i - 1][j] == 'X')
          continue;
        if (j > 0 && board[i][j - 1] == 'X')
          continue;
        ++ans;
      }

    return ans;
  }
};
/* code provided by PROGIEZ */

419. Battleships in a Board LeetCode Solution in Java

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

    for (int i = 0; i < board.length; ++i)
      for (int j = 0; j < board[0].length; ++j) {
        if (board[i][j] == '.')
          continue;
        if (i > 0 && board[i - 1][j] == 'X')
          continue;
        if (j > 0 && board[i][j - 1] == 'X')
          continue;
        ++ans;
      }

    return ans;
  }
}
// code provided by PROGIEZ

419. Battleships in a Board LeetCode Solution in Python

class Solution:
  def countBattleships(self, board: list[list[str]]) -> int:
    ans = 0

    for i, row in enumerate(board):
      for j, cell in enumerate(row):
        if cell == '.':
          continue
        if i > 0 and board[i - 1][j] == 'X':
          continue
        if j > 0 and board[i][j - 1] == 'X':
          continue
        ans += 1

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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