<p>Given a 2D&nbsp;<code>grid</code> consists of <code>0s</code> (land)&nbsp;and <code>1s</code> (water).&nbsp; An <em>island</em> is a maximal 4-directionally connected group of <code><font face="monospace">0</font>s</code> and a <em>closed island</em>&nbsp;is an island <strong>totally</strong>&nbsp;(all left, top, right, bottom) surrounded by <code>1s.</code></p>

<p>Return the number of <em>closed islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png" style="width: 240px; height: 120px;" /></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
Islands in gray are closed because they are completely surrounded by water (group of 1s).</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png" style="width: 160px; height: 80px;" /></p>

<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1],
&nbsp;              [1,0,0,0,0,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,1,0,1,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length, grid[0].length &lt;= 100</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;=1</code></li>
</ul>

class Solution {
public:
    vector<pair<int, int>> GetNeighbors(vector<vector<int>> &grid, int sr, int sc) {
        vector<pair<int, int>> neighbors;
        int delta[] = {-1, 0, 1, 0, -1};

        for (int i = 0; i < 4; i++) {
            int nr = sr + delta[i];
            int nc = sc + delta[i + 1];
            if (0 <= nr && nr < grid.size() && 0 <= nc && nc < grid[0].size()) {
                neighbors.emplace_back(make_pair(nr, nc));
            }
        }

        return neighbors;
    }

    bool IsClosedIsland(vector<vector<int>> &grid, int sr, int sc) {
        deque<pair<int, int>> q;
        bool is_closed = true;

        if (sr == 0 || sr == grid.size() - 1 || sc == 0 || sc == grid[0].size() - 1) {
            is_closed = false;
        }
        q.push_back({sr, sc});
        grid[sr][sc] = 1;

        while (!q.empty()) {
            vector<pair<int, int>> neighbors = GetNeighbors(grid, q.front().first, q.front().second);
            for (auto neighbor : neighbors) {
                if (grid[neighbor.first][neighbor.second] == 1) {
                    continue;
                }
                if (neighbor.first == 0 || neighbor.first == grid.size() - 1 || neighbor.second == 0 || neighbor.second == grid[0].size() - 1) {
                    is_closed = false;
                }
                q.push_back({neighbor.first, neighbor.second});
                grid[neighbor.first][neighbor.second] = 1;
            }
            q.pop_front();
        }

        return is_closed;
    }

    int closedIsland(vector<vector<int>> &grid) {
        int res = 0;

        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) {
                    continue;
                }
                if (IsClosedIsland(grid, i, j)) {
                    res++;
                }
            }
        }

        return res;
    }
};

// Topics: Graph, BFS

// Time Complexity: O(r * c)
22138