<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) {
        queue<vector<int>> bfsq;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[0].size(); ++j) {
                if (grid[i][j] == 1) {
                    markIsland(grid, i, j, bfsq);
                    break;
                }
            }
            if (!bfsq.empty()) {
                break;
            }
        }
        while (!bfsq.empty()) {
            auto& top = bfsq.front();
            for (int k = 0; k < 4; ++k) {
                int ik = top[0] + DIRS[k][0], jk = top[1] + DIRS[k][1];
                if (ik >= 0 && ik < grid.size() && jk >= 0 && jk < grid[0].size()) {
                    if (grid[ik][jk] == 0) {
                        grid[ik][jk] = 2;
                        bfsq.push({ik, jk, top[2]+1});
                    } else if (grid[ik][jk] == 1) {
                        return top[2];
                    }
                }
            }
            bfsq.pop();
        }
        return -1;
    }

private:
    const int DIRS[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

    void markIsland(vector<vector<int>>& grid, int i, int j, queue<vector<int>>& bfsq) {
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] != 1) {
            return;
        }
        bfsq.push({i, j, 0});
        grid[i][j] = 2;
        for (int k = 0; k < 4; ++k) {
            markIsland(grid, i+DIRS[k][0], j+DIRS[k][1], bfsq);
        }
    }
};

// [0 0 0 0 0 0]
// [0 1 0 0 0 0]
// [1 1 0 0 0 0]
// [1 1 0 0 0 0]
// [0 0 0 0 0 0]
// [0 0 1 1 0 0]
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