<p>You are given an <code>n x n</code> binary matrix <code>grid</code>. You are allowed to change <strong>at most one</strong> <code>0</code> to be <code>1</code>.</p>

<p>Return <em>the size of the largest <strong>island</strong> in</em> <code>grid</code> <em>after applying this operation</em>.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,0],[0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1],[1,0]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1],[1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    
    void dfs(int i,int j,vector<vector<int>>& grid,int &sz,int x){
   if(i<0||i>=grid.size()||j<0||j>=grid[0].size()||grid[i][j]==x||grid[i][j]==0){return ;}
       
        sz++;
        grid[i][j]=x;
        dfs(i+1,j,grid,sz,x);
        dfs(i-1,j,grid,sz,x);
        dfs(i,j+1,grid,sz,x);
        dfs(i,j-1,grid,sz,x);
        
        
        
    }
    
    
    int largestIsland(vector<vector<int>>& grid) {
        int ans=-1; int n=grid.size();
        for(int i=0;i<grid.size();i++){
            
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j]!=0)continue;
                grid[i][j]=1;
                int sz=0;
                dfs(i,j,grid,sz,-1);
                ans=max(ans,sz);
                dfs(i,j,grid,sz,1);
                grid[i][j]=0;
                
                
            }
            
            
        }
        return ans<0 ? n*n : ans;
        
    }
};
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