<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:
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        bool found = false;
        queue<pair<int, int>> q;
        vector<vector<bool>> visited(n, vector<bool>(n, false));

        // Step 1: Identify the first island and mark all its positions
        for (int i = 0; i < n && !found; ++i) {
            for (int j = 0; j < n && !found; ++j) {
                if (grid[i][j] == 1) {
                    dfs(grid, visited, q, i, j, n);
                    found = true;
                }
            }
        }

        // Step 2: Expand from the first island using BFS to find the shortest bridge to the second island
        int steps = 0;
        vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

        while (!q.empty()) {
            int size = q.size();
            while (size--) {
                auto [x, y] = q.front();
                q.pop();

                for (auto& dir : directions) {
                    int nx = x + dir[0], ny = y + dir[1];
                    if (nx >= 0 && nx < n && ny >= 0 && ny < n && !visited[nx][ny]) {
                        if (grid[nx][ny] == 1) {
                            return steps;
                        }
                        if (grid[nx][ny] == 0) {
                            visited[nx][ny] = true;
                            q.push({nx, ny});
                        }
                    }
                }
            }
            steps++;
        }

        return -1;  // This line should never be reached since there are exactly two islands.
    }

private:
    void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, queue<pair<int, int>>& q, int i, int j, int n) {
        if (i < 0 || j < 0 || i >= n || j >= n || visited[i][j] || grid[i][j] == 0) {
            return;
        }
        visited[i][j] = true;
        q.push({i, j});
        vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        for (auto& dir : directions) {
            dfs(grid, visited, q, i + dir[0], j + dir[1], n);
        }
    }
};
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