<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 {
private:
    int dx[4] = {1, -1, 0, 0}; // direction for x
    int dy[4] = {0, 0, -1, 1}; // direction for y

    bool check(int n, int x, int y)
    {
        return x >= 0 and x < n and y >= 0 and y < n; 
    }

public:
    void dfs(vector<vector<int>> &grid, vector<vector<bool>> &visited, queue<pair<int, pair<int, int>>> &q, int cx, int cy)
    {
        q.push({0, {cx, cy}});
        visited[cx][cy] = true;

        for(int i = 0; i < 4; ++i)
        {
            int nx = cx + dx[i];
            int ny = cy + dy[i];

            if(check(int(grid.size()), nx, ny) and grid[nx][ny] == 1 and !visited[nx][ny])
            {
                dfs(grid, visited, q, nx, ny);
            }
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = int(grid.size());
        queue<pair<int, pair<int, int>>> q; // dist - x - y
        vector<vector<bool>> visited(n, vector<bool> (n));

        bool found = false;
        for(int x = 0; x < n; ++x) 
        {
            for(int y = 0; y < n; ++y)
            {    
                if (grid[x][y] == 1 and !visited[x][y])
                {
                    found = true;
                    dfs(grid, visited, q, x, y);
                    break;
                }
            }

            if (found) break;
        }

        int ans = 10001;
        while(!q.empty())
        {
            int dist = q.front().first; // current dist
            int cx = q.front().second.first; // current x
            int cy = q.front().second.second; // current y
            q.pop();

            for(int i = 0; i < 4; ++i)
            {
                int nx = cx + dx[i]; // next x
                int ny = cy + dy[i]; // next y

                if (check(n, nx, ny) and grid[nx][ny] == 0 and !visited[nx][ny])
                {
                    q.push({dist + 1, {nx, ny}});
                    visited[nx][ny] = true;
                }
                else if (check(n, nx, ny) and grid[nx][ny] == 1 and !visited[nx][ny])
                {
                    ans = min(ans, dist);
                }
            }
        }

        return ans;
    }
};
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