<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>>& grid, int x, int y, vector<pair<int,int>>& res) {
       vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        res.push_back({x, y});
        // visited[x][y] = true;
        grid[x][y]='.';
          
        for (auto& dir : directions) {
            int nx = x + dir.first;
            int ny = y + dir.second;

            if (nx >= 0 && nx < grid.size() && ny >= 0 && ny < grid[0].size()  && grid[nx][ny] == 'X') {
                dfs(grid, nx, ny,  res);
            }
        }
    }
    bool check(vector<pair<int, int>>& indexes) {
        int flagr=1,r=indexes[0].first,flagc=1,c=indexes[0].second;
        for (auto index : indexes) {
            if(index.first!=r)flagr=false;
            if(index.second!=c)flagc=false;
        }
        return flagr||flagc;
    }

    int countBattleships(vector<vector<char>>& grid) {
        if (grid.empty()) return 0;
    
        int rows = grid.size();
        int cols = grid[0].size();
    
        // vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        int count = 0;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if ( grid[i][j] == 'X') {
                    vector<pair<int, int>> tmp;
                    dfs(grid, i, j, tmp);
                        count+=check(tmp);
                }
            }
        }

        return count;
    }
};

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