<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:
    int dfs(vector<vector<int>>& grid , int i , int j){

        // Directions for up, down, left, right
        vector<pair<int , int>> directions= {{1,0},{-1,0},{0,1},{0,-1}};

        // Check for boundary conditions and if the cell is water (0)
        if(i<0 || i>=grid.size() || j<0 || j>=grid[0].size() || grid[i][j]==0){
            return 0;
        }

        // Mark the current cell as visited (set to 0)
        grid[i][j]=0;

        // Initialize the current area as 1 (current cell is part of the island)
        int ans = 1;

        // Explore all four directions
        for(auto &dir : directions){
            int newx = i + dir.first;
            int newy = j + dir.second;

            // Recursively explore the neighbors and add to the current area
            ans += dfs(grid , newx , newy);
        }

        return ans; // Return the total area of the island
    }

    int maxAreaOfIsland(vector<vector<int>>& grid) {

        if(grid.empty()){
            return 0;
        }

        int n = grid.size();
        int m = grid[0].size();
        int maxArea = 0;

        // Traverse the grid to find all islands
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                // If it's land (1), perform a DFS and calculate the area
                if(grid[i][j] == 1){
                    maxArea = max(maxArea, dfs(grid, i, j));
                }
            }
        }
        
        return maxArea; 
    }
};
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