<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:
        # dfs solution
        # seen = set()

        # def dfs(row, col):
        #     i = row
        #     j = col
        #     count = 0
        #     while j < len(board[0]) and board[row][j] == "X":
        #         count += 1
        #         seen.add((row, j))
        #         j += 1
        #     if count == 1:
        #         while i < len(board) and board[i][col] == "X":
        #             seen.add((i, col))
        #             i += 1
        #     return 1
        
        # if not board or len(board) == 0 or len(board[0]) == 0:
        #     return 0
        # total = 0
        # for row in range(len(board)):
        #     for col in range(len(board[0])):
        #         if (row, col) not in seen and board[row][col] == "X":
        #             total += dfs(row, col)
        # return total
        # Time (O(mn))
        # Space (O(mn))
    
        # Smarter solution

        if not board or len(board) == 0 or len(board[0]) == 0:
            return 0
        
        count = 0
        for row in range(len(board)):
            for col in range(len(board[0])):
                if board[row][col] == 'X':
                    if (row == 0 or board[row - 1][col] != "X") and (col == 0 or board[row][col - 1] != "X"):
                        count += 1
        return count
17200