<p>You are given an <code>n x n</code> binary matrix <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>&#39;s not connected to any other <code>1</code>&#39;s. There are <strong>exactly two islands</strong> in <code>grid</code>.</p>

<p>You may change <code>0</code>&#39;s to <code>1</code>&#39;s to connect the two islands to form <strong>one island</strong>.</p>

<p>Return <em>the smallest number of </em><code>0</code><em>&#39;s you must flip to connect the two islands</em>.</p>

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

<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> grid = [[0,1,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There are exactly two islands in <code>grid</code>.</li>
</ul>

class Solution {
    vector<vector<int>>visited;
    void DFS(int i, int j, int n, vector<vector<int>>& grid, queue<vector<int>>&q1 )
    {
        if(i<0 || j<0 || i>=n || j>=n || grid[i][j]==0 || visited[i][j]==1 )
        {
            return;
        }
        visited[i][j]=1;
        q1.push({i,j,0});

        vector<int>X={0,0,-1,1}; 
        vector<int>Y={-1,1,0,0}; 
        for(int k=0; k<4; k++)
        {
            int x=i+X[k], y=j+Y[k];
            DFS(x,y,n,grid,q1);
        }
    }

public:
    int shortestBridge(vector<vector<int>>& grid) {
        int n=grid.size();
        queue<vector<int>>q1;
        visited.resize(n, vector<int>(n,0));
        bool b1=false;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n ; j++)
            {
                if(grid[i][j]==1)
                {
                    b1=true;
                    DFS(i, j,n, grid,q1 );
                    break;
                }
            }
            if(b1)
            {
                break;
            }
        }    
        // visited.resize(n, vector<int>(n,0));
        while(!q1.empty())
        {
            auto it=q1.front(); 
            q1.pop();
            int i=it[0], j=it[1];
            vector<int>X={0,0,-1,1}; 
            vector<int>Y={-1,1,0,0}; 
            for(int k=0; k<4; k++)
            {
                int x=i+X[k], y=j+Y[k];
                if(x>=0 && y>=0 && x<n && y<n)
                {
                    if(visited[x][y]==1)
                    {
                        continue;
                    }
                    else if(grid[x][y]==1){
                        return it[2];
                    }
                    else{
                        visited[x][y]=1;
                        q1.push({x,y, it[2]+1});
                    }
                }
            }

        }
        return 0;
    }
};
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