<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 n;
    int delta[5] = {0, 1, 0, -1, 0};
    void dfs(int i, int j, bool vis[], std::vector<std::vector<int>>& grid) {
        vis[i * n + j] = true;
        for (int d = 0; d < 4; ++d) {
            int di = i + delta[d], dj = j + delta[d + 1];
            if (di > -1 && di < n && dj > -1 && dj < n && grid[di][dj] && !vis[di * n + dj]) {
                dfs(di, dj, vis, grid);
            }
        }
    }
    int bfs(bool vis[], std::vector<std::vector<int>>& grid) {
        std::queue<int> q;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (vis[i * n + j]) q.push(i * n + j);
            }
        }
        int depth = 0;
        while(!q.empty()) {
            int sz = q.size();
            while (sz--) {
                int coord = q.front(); q.pop();
                for (int i = 0; i < 4; ++i) {
                    int di = coord / n + delta[i], dj = coord % n + delta[i + 1];
                    if (di > -1 && di < n && dj > -1 && dj < n && !vis[di * n + dj]) {
                        if (grid[di][dj]) return depth;
                        q.push(di * n + dj);
                        vis[di * n + dj] = true;
                    }
                }

            }
            ++depth;
        }
        return depth;
    }
public:
    int shortestBridge(vector<vector<int>>& grid) {
        n = grid.size();
        bool vis[n * n];
        std::memset(vis, false, sizeof vis);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j]) {
                    dfs(i, j, vis, grid);
                    return bfs(vis, grid);
                }
            }
        }
        return -1;
    }
};
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