<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 {
private:
    queue<vector<int>> que;
    vector<vector<bool>> visited;

    vector<vector<int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    void dfs(vector<vector<int>>& grid, int p, int q, int n, int m, vector<vector<bool>>& visited) {
        grid[p][q] = 2;
        visited[p][q] = true;
        bool water = false;

        for(auto d : directions) {
            int row = p + d[0];
            int col = q + d[1];
            if(row >= 0 && col >= 0 && row < n && col < m) {
                if(!visited[row][col] && grid[row][col] == 1)
                    dfs(grid, row, col, n, m, visited);
                if(grid[row][col] == 0)
                    water = true;
            }
        }

        if(water) {
            que.push({p, q});
            visited[p][q] = true;
        }
    }

    int findShortestBridge(vector<vector<int>>& grid, vector<vector<bool>>& visited) {
        int n = grid.size(), m = grid[0].size();
        queue<vector<int>> q;

        // for(auto coast : coastline) {
        //     q.push({coast.first, coast.second});
        //     visited[coast.first][coast.second] = true;
        // }

        int dist = 0;

        while(!que.empty()) {
            int qSize = que.size();
            dist++;
            
            for(int i=0; i<qSize; i++) {
                vector<int> cell = que.front();
                que.pop();

                for(auto d : directions) {
                    int row = cell[0] + d[0];
                    int col = cell[1] + d[1];
                    if(row >= 0 && col >= 0 && row < n && col < m && !visited[row][col] && grid[row][col] != 2) {
                        if(grid[row][col] == 1)
                            return dist;
                        else {
                            que.push({row, col});
                            visited[row][col] = true;
                        }
                    }
                }
            }
        }

        return -1;
    }
public:
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        bool found = false;
        visited = vector<vector<bool>>(n, vector<bool>(m, false));

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                if(grid[i][j]) {
                    dfs(grid, i, j, n, m, visited);
                    found = true;
                    break;
                }
            }
            if(found)   break;
        }

        return findShortestBridge(grid, visited) - 1;
    }
};
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