<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:
    vector<vector<bool>> visited; // Keeps track of visited nodes
    int temp = 0;

    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();

        // Initialize the visited vector to be the same size as the grid, filled with 'false'
        visited.resize(rows, vector<bool>(cols, false));
        
        int ans = 0; // Variable to store the maximum area of an island

        // Iterate over each cell in the grid
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (!visited[i][j] && grid[i][j] == 1) { 
                    // If the cell is unvisited and part of an island (grid[i][j] == 1)
                    temp = 0; // Reset the temporary area count
                    bfs(grid, i, j); // Perform BFS to explore the entire island
                    ans = max(ans, temp); // Update the maximum area found so far
                }
            }
        }
        return ans; // Return the maximum area of an island
    }

    void bfs(vector<vector<int>>& grid, int i, int j) {
        visited[i][j] = true; // Mark the current cell as visited
        temp++; // Increment the area count for the current island

        int rows = grid.size();
        int cols = grid[0].size();

        // Check all four possible directions (up, down, left, right)
        if (i - 1 >= 0 && grid[i - 1][j] == 1 && !visited[i - 1][j]) {
            bfs(grid, i - 1, j);
        }
        if (i + 1 < rows && grid[i + 1][j] == 1 && !visited[i + 1][j]) {
            bfs(grid, i + 1, j);
        }
        if (j - 1 >= 0 && grid[i][j - 1] == 1 && !visited[i][j - 1]) {
            bfs(grid, i, j - 1);
        }
        if (j + 1 < cols && grid[i][j + 1] == 1 && !visited[i][j + 1]) {
            bfs(grid, i, j + 1);
        }
    }
};

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