<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:

    void dfs(int row, int col, vector<vector<int>>& grid, vector<vector<int>>& vis)
    {
        int n = grid.size();
        int m = grid[0].size();
        vis[row][col] = 1;

        int dr[] = {0, 1, 0, -1};
        int dc[] = {-1, 0, 1, 0};

        for(int i = 0; i < 4; i++)
        {
            int nr = row + dr[i];
            int nc = col + dc[i];

            if(nr >= 0 && nr < n && nc >= 0 && nc < m && grid[nr][nc] == 1 && !vis[nr][nc])
            {
                dfs(nr, nc, grid, vis);
            }
        }
    }

    int bfs(vector<vector<int>>& grid, vector<vector<int>>& vis)
    {
        queue<pair<int, int>> q;
        int n = grid.size();
        int m = grid[0].size();

        int dr[] = {0, 1, 0, -1};
        int dc[] = {-1, 0, 1, 0};

        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(vis[i][j] == 1)
                    q.push({i,j});
            }
        }

        int level = 0;
        while(!q.empty())
        {
            int size = q.size();
            for(int i = 0; i < size; i++)
            {
                auto node = q.front();
                q.pop();
                int u = node.first;
                int v = node.second;

                for(int i = 0; i < 4; i++)
                {
                    int x = u + dr[i];
                    int y = v + dc[i];

                    if(x >= 0 && x<n && y >= 0 && y < m && !vis[x][y])
                    {
                        vis[x][y] = 1;

                        if(grid[x][y] == 1) return level;
                        q.push({x, y});
                    }
                }
            }
            level++;
        }
        return level;
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<int>> vis(n, vector<int>(m, 0));

        //first perform a dfs to mark all the lands in island 1
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(grid[i][j] == 1 && !vis[i][j])
                {
                    dfs(i, j, grid, vis);
                    return bfs(grid, vis);
                }
            }
        }
        return -1;
    }
};
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