<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}};
    int shortestBridge(vector<vector<int>>& grid) {
        for(int i=0; i<grid.size(); i++)
        {
            for(int j=0; j<grid.size(); j++)
            {
                if(grid[i][j] == 1)
                {
                    dfs(grid, i, j);
                    return bfs(grid);
                }
            }
        }
        return -1;
    }

    void dfs(vector<vector<int>>& grid, int r, int c)
    {
        if(r < 0 || c < 0 || r >= grid.size() || c >= grid.size())
            return;

        if(grid[r][c] != 1)
            return;

        grid[r][c] = -1;

        for(int i=0; i<dir.size(); i++)
        {
            dfs(grid, r + dir[i][0], c + dir[i][1]);   
        }
    }

    int bfs(vector<vector<int>>& grid)
    {
        queue<pair<int, int>> q;
        for(int i=0; i<grid.size(); i++)
        {
            for(int j=0; j<grid.size(); j++)
            {
                if(grid[i][j] == -1)
                {
                    q.push({i, j});
                }
            }
        }

        int ans = 0;
        while(!q.empty())
        {
            int sz = q.size();
            while(sz-- > 0)
            {
                auto idx = q.front();
                q.pop();

                int i = idx.first;
                int j = idx.second;

                for(auto d: dir)
                {
                    int r = i + d[0];
                    int c = j + d[1];

                    if(r < 0 || c < 0 || r >= grid.size() || c >= grid.size())
                        continue;
                    
                    if(grid[r][c] < 0)
                        continue;

                    if(grid[r][c] == 1)
                        return ans;
                    
                    grid[r][c] = -1;

                    q.push({r, c});
                }
            }
            ans++;
        }
        return ans;
    }
};
31675