<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 {
public:
    int countBattleships(vector<vector<char>>& board) {
        int ans = 0;
         
        for(int i = 0; i < board.size(); i++)
        {
            bool continu = false;
            int cnt = 0; // prevent one duplicate
            for(int j = 0; j < board[0].size(); j++)
            {
                if(board[i][j] == 'X')
                {
                   continu = check1(board, i, j);
                   if(continu) cnt++;
                }
                else
                { 
                    if(continu) ans++;
                    continu = false;
                    if(cnt == 1 && j - 1 >= 0) board[i][j - 1] = '.';
                    cnt = 0;
                }
                if(j == board[0].size() - 1 && continu) ans++;
                if(j == board[0].size() - 1 && cnt == 1) board[i][j] = '.';
            }
            
        }

        //if(board.size() == 1 && board[0].size() == 1) return ans;

        for(int j = 0; j < board[0].size(); j++)
        {
            bool continu = false;
            int cnt = 0;
            for(int i = 0; i < board.size(); i++)
            {
                if(board[i][j] == 'X')
                {
                   continu = check2(board, i, j);
                }
                else
                {
                    if(continu) ans++;
                    continu = false;
                    cnt = 0;
                }
                if(i == board.size() - 1 && continu) ans++;
            }
        }

        return ans;
    }


    bool check1(vector<vector<char>>& board, int i, int j)
    {
        if(i - 1 >= 0 && board[i - 1][j] == 'X') return false; // up
        if(i + 1 < board.size() && board[i + 1][j] == 'X') return false; // down

        return true;
    }

    bool check2(vector<vector<char>>& board, int i, int j)
    {
        if(j - 1 >= 0 && board[i][j - 1] == 'X') return false; // left
        if(j + 1 < board[0].size() && board[i][j + 1] == 'X') return false; // right

        return true;
    }
};
11800