<p>Given an <code>m x n</code> matrix <code>board</code> where each cell is a battleship <code>&#39;X&#39;</code> or empty <code>&#39;.&#39;</code>, return <em>the number of the <strong>battleships</strong> on</em> <code>board</code>.</p>

<p><strong>Battleships</strong> can only be placed horizontally or vertically on <code>board</code>. In other words, they can only be made of the shape <code>1 x k</code> (<code>1</code> row, <code>k</code> columns) or <code>k x 1</code> (<code>k</code> rows, <code>1</code> column), where <code>k</code> can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img height="333" src="https://assets.leetcode.com/uploads/2024/06/21/image.png" width="333" />
<pre>
<strong>Input:</strong> board = [[&quot;X&quot;,&quot;.&quot;,&quot;.&quot;,&quot;X&quot;],[&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;X&quot;],[&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;X&quot;]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> board = [[&quot;.&quot;]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 200</code></li>
	<li><code>board[i][j]</code> is either <code>&#39;.&#39;</code> or <code>&#39;X&#39;</code>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do it in one-pass, using only <code>O(1)</code> extra memory and without modifying the values <code>board</code>?</p>

class Solution:
    def countBattleships(self, board: List[List[str]]) -> int:
        if not board:
            return 0

        visited = set()
        ships = 0

        def bfs(x, y):
            queue = [[x,y]]
            visited.add((x,y))

            while queue:
                i, j = queue.pop(0)
                dirs = [[1,0], [-1, 0], [0, 1], [0, -1]]

                for dx, dy in dirs:
                    nx, ny = dx + i, dy + j
                    if 0 <= nx < len(board) and 0 <= ny < len(board[0]) and board[nx][ny] == "X" and (nx, ny) not in visited:
                        queue.append((nx, ny))
                        visited.add((nx, ny))

        for r in range(len(board)):
            for c in range(len(board[0])):  # Fixed to use len(board[0]) for column iteration
                if board[r][c] == "X" and (r, c) not in visited:
                    bfs(r, c)
                    ships += 1

        return ships

17800