<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>> &nums, int ir, int jc) {
        stack<pair<int, int>> q;
        vector<int> s = {-1, 0, 1, 0};
        vector<int> s1 = {0, 1 ,0, -1};
        q.push({ir, jc});

        while(q.size() > 0) {
            int row = q.top().first;
            int col = q.top().second;
            q.pop();
            nums[row][col] = '.';
            for(int i=0; i<4; i++)
            {
                int r = row + s[i];
                int c = col + s1[i];
                if(r>=0 && c>=0 && r<nums.size() && c<nums[0].size() && nums[r][c] == 'X')
                    q.push({r, c});
            }
        }
    }

    int countBattleships(vector<vector<char>>& nums) {
        int count = 0;
        for(int i=0; i<nums.size(); i++)
            for(int j=0; j<nums[0].size(); j++)
                if(nums[i][j] == 'X') {
                    dfs(nums, i, j);
                    count++;
                }
        return count;
    }
};
13300