<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>> dirs = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
    int n, m;
    queue<pair<int, int>> q;

    void dfs(vector<vector<int>>& grid, vector<vector<int>>& vis, int i,
             int j) {
        vis[i][j] = 1;
        q.push({i, j});

        for (auto dir : dirs) {
            int x = i + dir[0];
            int y = j + dir[1];

            if (x >= 0 && x < n && y >= 0 && y < m && !vis[x][y] && grid[x][y]) {
                dfs(grid, vis, x, y);
            }
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        vector<vector<int>> vis(n, vector<int>(m, 0));

        int found = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j]) {
                    found = 1;
                    dfs(grid, vis, i, j);
                    break;
                }
            }

            if (found) {
                break;
            }
        }


        // for (int i = 0; i < n; i++) {
        //     for (int j = 0; j < m; j++) {
        //        cout<<vis[i][j]<<" ";
        //     }
        //     cout<<"\n";
        // }

        int path = 0;

        while (!q.empty()) {
            int size = q.size();
            path++;

            while (size--) {
                auto top = q.front();
                q.pop();
                vis[top.first][top.second] = 1;

                for (auto dir : dirs) {
                    int x = top.first + dir[0];
                    int y = top.second + dir[1];

                    if (x >= 0 && x < n && y >= 0 && y < m) {
                        if (!vis[x][y] && grid[x][y]) {    
                            return path-1;
                        } else if (!vis[x][y]  && !grid[x][y]) {
                            vis[x][y] = 1;
                            q.push({x, y});
                        }
                    }
                }
            }
        }

        return -1;
    }
};
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