<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 helper(vector<vector<char>>& grid, int i, int j){
        // Base case: check for invalid indices and if the current cell is water ('0')
        if(i<0 || j<0 || i>=grid.size() || j>=grid[i].size() || grid[i][j]=='.')
          return;
        
        // Mark the current cell as visited by setting it to '0' (water)
        grid[i][j]='.';

        // Recursively call helper function for all adjacent cells
        helper(grid, i+1, j); // Down
        helper(grid, i, j+1); // Right
        helper(grid, i, j-1); // Left
        helper(grid, i-1, j); // Up
    }
    int countBattleships(vector<vector<char>>& grid) {
        int ans = 0; // Initialize count of islands to 0
        // Iterate over each cell in the grid
        for(int i = 0; i < grid.size(); i++){
            for(int j = 0; j < grid[i].size(); j++){
                // If the cell contains '1', it is part of an island
                if(grid[i][j] == 'X') {
                    ans++; // Increment the island count
                    // Call the helper function to mark the entire island as visited
                    helper(grid, i, j);
                }
            }
        }
        return ans;
    }
};
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