<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:
        # Solution 1 - using DFS with visit set. 
        # Time - O(m*n)
        # Space - O(m*n) 

        # visit = set()
        # ROWS, COLS = len(board), len(board[0])

        # directions = [(0, 1), (1, 0), (-1, 0), (0, -1)]

        # def dfs(r, c):
        #     visit.add((r, c))
            
        #     for d in directions:
        #         new_r, new_c = d[0] + r, d[1] + c

        #         if new_r >= 0 and new_r < ROWS and new_c >=0 \
        #             and new_c < COLS and (new_r, new_c) not in visit and board[new_r][new_c] == 'X':
        #             dfs(new_r, new_c)

        # res = 0

        # for r in range(ROWS):
        #     for c in range(COLS):
        #         if (r, c) not in visit and board[r][c] == 'X':
        #             dfs(r, c)
        #             res += 1

        # return res


        # Solution 2 - 
        # Time - O(m*n)
        # Space - O(1)

        res = 0

        ROWS, COLS = len(board), len(board[0])

        for r in range(ROWS):
            for c in range(COLS):

                prev_r, prev_c = r-1, c-1

                if (prev_r < 0 or board[prev_r][c] == '.') and (prev_c < 0 or board[r][prev_c] == '.') and board[r][c] == 'X':
                    res += 1

        return res
        
17000