<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>

#include <vector>
#include <queue>
#include <utility>

using namespace std;

class Solution {
public:
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, -1, 0, 1};
    vector<vector<bool>> visited;

    int shortestBridge(vector<vector<int>>& grid) {
        queue<vector<int>> q; // {row, col, distance}
        visited.resize(grid.size(), vector<bool>(grid[0].size(), false));
        bool flag = false;

        // Find and mark the first island
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) {
                    dfs(i, j, grid, q);
                    flag = true;
                    break;
                }
            }
            if (flag) break;
        }

        // Perform BFS to find the shortest bridge
        while (!q.empty()) {
            vector<int> cell = q.front();
            q.pop();
            int x = cell[0], y = cell[1], dist = cell[2];

            for (int k = 0; k < 4; k++) {
                int nx = x + dx[k];
                int ny = y + dy[k];
                
                if (nx >= 0 && nx < grid.size() && ny >= 0 && ny < grid[0].size() && !visited[nx][ny]) {
                    if (grid[nx][ny] == 1) {
                        return dist;
                    }
                    q.push({nx, ny, dist + 1});
                    visited[nx][ny] = true;
                }
            }
        }

        return -1; // Should not reach here if there are two islands
    }

    void dfs(int i, int j, vector<vector<int>>& grid, queue<vector<int>>& q) {
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || visited[i][j] || grid[i][j] == 0) {
            return;
        }
        visited[i][j] = true;
        q.push({i, j, 0});
        for (int k = 0; k < 4; k++) {
            int x = i + dx[k];
            int y = j + dy[k];
            dfs(x, y, grid, q);
        }
    }
};

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