<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 {
public:
    
    vector<vector<int>> dir = {{1,0}, {-1,0}, {0,-1}, {0,1}};
    
    bool isValid(vector<vector<int>> &grid, int x, int y, int islandName)
    {
        if (x < 0 || x>=grid.size() || y< 0 || y>=grid[0].size() || grid[x][y] == islandName || grid[x][y] == 0)
            return false;
        return true;
    }
    
    void dfs(vector<vector<int>> &grid, int x, int y, int islandName)
    {
        if (!isValid(grid, x, y, islandName))
            return;
        
        grid[x][y] = islandName;
        
        for(vector<int>d: dir)
        {
            int newX = x+d[0];
            int newY = y+d[1];
            
            dfs(grid, newX, newY, islandName);
        }
    }
    
    void groupIslands(vector<vector<int>> &grid)
    {
        int islandName = 2;
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if (grid[i][j] == 1)
                {
                    dfs(grid, i, j, islandName);
                    islandName++;
                }
            }
        }
    }
    
    bool isAtleastOneSizeZero(vector<vector<int>> &grid, int x, int y)
    {
        for(vector<int> d: dir)
        {
            int newX = x+d[0];
            int newY = y+d[1];
            if (newX < 0 || newX>=grid.size() || newY< 0 || newY>=grid[0].size())
                continue;
            if(grid[newX][newY] == 0)
                return true;
        }
        return false;
    }
    
    int shortestBridge(vector<vector<int>>& grid) {
        groupIslands(grid);
        
        queue<pair<pair<int,int> ,int>> q;
        set<pair<int,int>> vis;
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if (grid[i][j] == 2)
                {
                    if (isAtleastOneSizeZero(grid, i, j))
                    {
                        q.push({{i,j}, 0});
                        vis.insert({i,j});
                    }
                }
            }
        }
        
        while(q.size() > 0)
        {
            auto front = q.front();
            q.pop();
            if (grid[front.first.first][front.first.second] == 3)
                return front.second - 1;
            for(vector<int> d: dir)
            {
                int newX = front.first.first+d[0];
                int newY = front.first.second+d[1];
                if (newX < 0 || newX>=grid.size() || newY< 0 || newY>=grid[0].size())
                    continue;
                if ((grid[newX][newY] == 0 || grid[newX][newY] == 3) && vis.find({newX, newY}) == vis.end())
                {
                    q.push({{newX, newY}, front.second+1});
                    vis.insert({newX, newY});
                }
            }
        }
        return -1;
        
    }
};
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