<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(vector<vector<int>>& g, vector<vector<int>>& v, queue<pair<int, int>>& q, int x, int y, int n) {
        if (x < 0 || y < 0 || x >= n || y >= n || v[x][y] || g[x][y] != 1) return;
        v[x][y] = 1;
        q.push({x, y});
        vector<int> dx = {-1, 0, 1, 0};
        vector<int> dy = {0, 1, 0, -1};
        for (int i = 0; i < 4; i++) {
            dfs(g, v, q, x + dx[i], y + dy[i], n);
        }
    }
    int shortestBridge(vector<vector<int>>& g) {
        int n = g.size();
        vector<vector<int>> v(n, vector<int>(n, 0));
        queue<pair<int, int>> q;
        bool f = false;
        vector<int> dx = {-1, 0, 1, 0};
        vector<int> dy = {0, 1, 0, -1};
        for (int i = 0; i < n && !f; i++) {
            for (int j = 0; j < n && !f; j++) {
                if (g[i][j] == 1) {
                    dfs(g, v, q, i, j, n);
                    f = true;
                }
            }
        }
        int s = 0;
        while (!q.empty()) {
            int sz = q.size();
            while (sz--) {
                auto [x, y] = q.front();
                q.pop();
                for (int i = 0; i < 4; i++) {
                    int nx = x + dx[i];
                    int ny = y + dy[i];
                    if (nx >= 0 && ny >= 0 && nx < n && ny < n && !v[nx][ny]) {
                        if (g[nx][ny] == 1) return s;
                        v[nx][ny] = 1;
                        q.push({nx, ny});
                    }
                }
            }
            s++;
        }
        return -1;
    }
};
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