<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(vector<vector<char>>& board, int r, int c,int nr[],int nc[],vector<vector<int>>&vis)
    {   
        int n=board.size();
        int m=board[0].size();
        vis[r][c]=1;
        for(int i=0;i<4;i++)
        {
            int nrr=r+nr[i];
            int ncc=c+nc[i];
            if(nrr>=0 && nrr<n && ncc>=0 && ncc<m && board[nrr][ncc]=='X' && !vis[nrr][ncc])
            {
                dfs(board, nrr, ncc,nr,nc,vis);
            }
        }
    }
    int countBattleships(vector<vector<char>>& board)
    {
        int n=board.size();
        int m=board[0].size();
       if (n==0)
       {
            return 0;
       }
        int c = 0;
        int nr[4]={-1,0,1,0};
        int nc[4]={0,1,0,-1};
        vector<vector<int>>vis(n,vector<int>(m,0));
        for (int i=0; i<n; i++)
        {
            for (int j=0;j<m;j++)
            {
                if(board[i][j]=='X' && !vis[i][j]) 
                {
                    c++;
                    dfs(board, i, j,nr,nc,vis);
                }
            }
        }
        return c;
    }     
};
12300