Shortest Bridge
<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>
Depth-First Search; Breadth-First Search; Array; Matrix
class Solution {
public:
    const static inline vector<int> dirs = {-1, 0, 1, 0, -1};

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<pair<int, int>> q;
        function<void(int, int)> dfs = [&](int i, int j) {
            grid[i][j] = 2;
            q.emplace(i, j);
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1) {
                    dfs(x, y);
                }
            }
        };
        for (int i = 0, x = 1; i < n && x; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j]) {
                    dfs(i, j);
                    x = 0;
                    break;
                }
            }
        }
        int ans = 0;
        while (1) {
            for (int h = q.size(); h; --h) {
                auto [i, j] = q.front();
                q.pop();
                for (int k = 0; k < 4; ++k) {
                    int x = i + dirs[k], y = j + dirs[k + 1];
                    if (x >= 0 && x < n && y >= 0 && y < n) {
                        if (grid[x][y] == 1) return ans;
                        if (grid[x][y] == 0) {
                            grid[x][y] = 2;
                            q.emplace(x, y);
                        }
                    }
                }
            }
            ++ans;
        }
    }
};