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

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

    void dfs(vector<vector<int>>& grid, int i, int j, set<pair<int, int>>& vis) {
        if (!isSafe(i, j) || grid[i][j] == 0 || vis.find({i, j}) != vis.end()) return;
        vis.insert({i, j});

        for (auto& it : dir) {
            int _i = i + it[0];
            int _j = j + it[1];
            
            dfs(grid, _i, _j, vis);
        }
    }

    int bfs(vector<vector<int>>& grid, set<pair<int, int>>& vis) {
        queue<pair<int, int>> q;
        for (auto &it : vis) {
            q.push(it);
        }
        int level = 0;
        while (!q.empty()) {
            int s = q.size();

            while (s--) {
                pair<int, int> p = q.front();
                q.pop();

                int x = p.first;
                int y = p.second;

                for (auto d : dir) {
                    int _x = x + d[0];
                    int _y = y + d[1];

                    if (isSafe(_x, _y)) {
                        if (grid[_x][_y] == 1 && vis.find({_x, _y}) == vis.end()) {
                            return level;
                        }
                        if (vis.find({_x, _y}) == vis.end()) {
                            vis.insert({_x, _y});
                            q.push({_x, _y});
                        }
                    }
                }
            }
            level++;
        }
        return -1;
    }

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

        set<pair<int, int>> vis;

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

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