<p>You are given an <code>m x n</code> binary matrix <code>grid</code>. An island is a group of <code>1</code>&#39;s (representing land) connected <strong>4-directionally</strong> (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.</p>

<p>The <strong>area</strong> of an island is the number of cells with a value <code>1</code> in the island.</p>

<p>Return <em>the maximum <strong>area</strong> of an island in </em><code>grid</code>. If there is no island, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/01/maxarea1-grid.jpg" style="width: 500px; height: 310px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> The answer is not 11, because the island must be connected 4-directionally.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,0,0,0,0,0,0,0]]
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    set<pair<int,int>> visited;
    int count = 0;
    void bfs(int x, int y, vector<vector<int>>& grid){
        count = 1;
        visited.insert({x,y});
        queue<pair<int,int>> q;
        q.push({x,y});
        int a[4] = {-1,0,1,0};
        int b[4] = {0,1,0,-1};
        while(!q.empty()){
            int X = q.front().first;
            int Y = q.front().second;
            q.pop();
            for(int i = 0; i<4; i++){
                int g = X + a[i];
                int h = Y + b[i];
                if(g<0 || g>=grid.size() || h<0 || h>=grid[0].size()){
                    continue;
                }
                if(!visited.contains({g,h}) && grid[g][h] == 1){
                    q.push({g,h});
                    visited.insert({g,h});
                    count++;
                }
            }
        }
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int ans = 0;
        for(int i = 0; i<m; i++){
            for(int j = 0; j<n; j++){
                if(!visited.contains({i,j}) && grid[i][j] == 1){
                    bfs(i,j,grid);
                    ans = max(ans,count);
                }
            }
        }
        return ans;
    }
};
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