<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) {
        const vector<vector<int>> dirs{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        queue<pair<int, int>> mq;
        int n = grid.size();

        function<void(int, int)> Dfs = [&](int row, int col) {
            grid[row][col] = 2;
            mq.push({row, col});
            for (auto dir: dirs) {
                int r = row + dir[0];
                int c = col + dir[1];
                if (r>=0 && r<n && c>=0 && c<n && grid[r][c] == 1) Dfs(r, c);
            }
        };

        for (int i=0; i<n; i++) {
            for (int j=0; j<n; j++) {
                if (grid[i][j]) {
                    Dfs(i, j);
                    i=j=n;
                }
            }
        }

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

        int distance = 0;
        while (!mq.empty()) {
            int num_mq = mq.size();
            for (int i=0; i<num_mq; i++) {
                auto pos = mq.front();
                mq.pop();
                for (auto dir: dirs) {
                    int r = pos.first + dir[0];
                    int c = pos.second + dir[1];
                    if (r>=0 && r<n && c>=0 && c<n) {
                        if (grid[r][c] == 1) return distance;
                        else if (grid[r][c] == 0) {
                            mq.push({r, c});
                            grid[r][c] = 2;
                        }
                    }
                }
            }
            distance++;
        }

        return -1;
    }
};
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