<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 i, int j, bool isRow){
        int n = board.size();
        int m = board[0].size();
        stack<pair<int, int>> st;
        st.push({i, j});
        while(!st.empty()){
            int u = st.top().first;
            int v = st.top().second;
            st.pop();
            board[u][v] = '.';
            if(u+1 < n && !isRow && board[u+1][v] == 'X') st.push({u+1, v});
            if(v+1 < m && isRow  && board[u][v+1] == 'X') st.push({u, v+1});
        }
    }

    int countBattleships(vector<vector<char>>& board) {
        int n = board.size();
        if(n == 0){
            return 0;
        }int m = board[0].size();
        int count = 0;
        for(int i = 0;i<n;i++){
            for(int j = 0;j<m;j++){
                if(board[i][j] == 'X'){
                    count++;
                    board[i][j] = '.';
                    if(i+1 <n && board[i+1][j] == 'X'){
                        dfs(board, i+1, j, false);
                    }else if(j+1 <m && board[i][j+1] == 'X'){
                        dfs(board, i, j+1, true);
                    }
                }
            }
        }return count;
    }
};
13000