<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:
    vector<pair<int,int>>dir = {{0,1},{1,0},{0,-1},{-1,0}};
    int n,m;
    void bfs(int i, int j, vector<vector<int>>&vis, vector<vector<char>>& board){
        queue<pair<int,int>>q;
        q.push({i,j});
        vis[i][j] = 1;
        while(!q.empty()){
            auto [x,y] = q.front();
            q.pop();
            for(auto [dx,dy] : dir){
                int nx = x + dx;
                int ny = y + dy;
                if(nx >= 0 && nx < n && ny >= 0 && ny < m && board[nx][ny] == 'X' && !vis[nx][ny]){
                    q.push({nx,ny});
                    vis[nx][ny] = 1;
                }
            }
        }
    }
    void dfs(int x, int y, vector<vector<int>>&vis, vector<vector<char>>& board){
        vis[x][y] = 1;
        for(auto [dx,dy] : dir){
            int nx = x + dx;
            int ny = y + dy;
            if(nx >= 0 && nx < n && ny >= 0 && ny < m && board[nx][ny] == 'X' && !vis[nx][ny]){
                dfs(nx,ny,vis,board);
            }
        }
    }
    int countBattleships(vector<vector<char>>& board) {
        n = board.size(), m = board[0].size();
        int ans = 0;
        vector<vector<int>>vis(n,vector<int>(m,0));
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(!vis[i][j] && board[i][j] == 'X'){
                    bfs(i,j,vis,board);
                    ans++;
                }
            }
        }
        return ans;
    }
};
13500