<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:
    void dfs(int row, int col, vector<vector<char>>& board, vector<vector<int>>& vis)
    {
        vis[row][col] = 1;
        int n = board.size();
        int m = board[0].size();
        int dr[] = {1,0};
        int dc[] = {0,1};
        for(int i=0;i<2;i++)
        {
            int nrow = row + dr[i];
            int ncol = col + dc[i];
            if(nrow < n && ncol < m && !vis[nrow][ncol] && board[nrow][ncol] == 'X')
                dfs(nrow,ncol,board,vis);
        }
    }
    int countBattleships(vector<vector<char>>& board) {
        int n = board.size();
        int m = board[0].size();
        vector<vector<int>> vis(n, vector<int> (m,0));
        int count  = 0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(!vis[i][j] && board[i][j] == 'X')
                {
                    dfs(i,j, board,vis);
                    count++;
                }
            }
        }
        return count;
    }
};
12200