<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>> grid;
    int n;

    vector<vector<int>> dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};

    bool isValid(int i, int j) { return i >= 0 && j >= 0 && i < n && j < n; }

    void dfs(int i, int j, vector<vector<int>>& grid) {
        if (!isValid(i, j))
            return;

        for (auto d : dir) {
            int nx = i + d[0];
            int ny = j + d[1];
            if (isValid(nx, ny) && grid[nx][ny] == 1) {
                grid[nx][ny] = 2;
                dfs(nx, ny, grid);
            }
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        this->grid = grid;
        n = grid.size();
        int a = -1, b = -1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    a = i;
                    b = j;
                    break;
                }
            }
            if (a != -1 && b != -1)
                break;
        }
        grid[a][b] = 2;
        dfs(a, b, grid);

        vector<vector<int>> vis(n, vector<int>(n, 0));
        queue<pair<int, int>> q;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 2) {
                    q.push({i, j});
                    vis[i][j] = 1;
                }
            }
        }

        int steps = 0;
        while (!q.empty()) {
            int size = q.size();

            for (int i = 0; i < size; i++) {
                int x = q.front().first;
                int y = q.front().second;
                q.pop();
                for (auto d : dir) {
                    int nx = x + d[0];
                    int ny = y + d[1];
                    if (isValid(nx, ny) && !vis[nx][ny]) {
                        if (grid[nx][ny] == 1)
                            return steps;
                        q.push({nx, ny});
                        vis[nx][ny] = 1;
                    }
                }
            }
            steps++;
        }
        return -1;
    }
};
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