<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:
    void dfs(vector<vector<int>>& grid, int i, int j, vector<vector<int>>& visited){
        int n = grid.size();
        if (i < 0 || i >= n || j < 0 || j >= n || visited[i][j] 
        || grid[i][j] == 0) return;

        visited[i][j] = 1;
        
        vector<pair<int,int>> vp{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        for (int k = 0; k<4; k++){
            int ni = i + vp[k].first, nj = j + vp[k].second;
            dfs(grid, ni,nj, visited);
        }
    }
    int bfs(vector<vector<int>>& grid, vector<vector<int>>& visited){
        int n = visited.size();
        queue<vector<int>> q;
        for (int i = 0; i<n; i++){
            for (int j = 0; j<n; j++){
                if (visited[i][j]){
                    cout << "Q:I: J: "<<i <<" and "<<j<<endl;
                    q.push({i, j, 0});
                }
            }
        }
        cout << "Q: SIXE: "<<q.size()<<endl;
        while (!q.empty()){
            auto front = q.front();
            q.pop();
            vector<pair<int,int>> vp{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
                   
                    
            for (int k = 0; k<4; k++){
                int ni = front[0] + vp[k].first, 
                nj = front[1] + vp[k].second;
                if (ni >= 0 && ni < n && nj >= 0 && nj < n){
                    if (visited[ni][nj]) continue;
                    visited[ni][nj] = 1;
                    if (grid[ni][nj] == 1) return front[2];
                    q.push({ni, nj, front[2]+1});
                }
            }
        }
        return -1;
    }
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> visited(n, vector<int>(n, 0));
        for (int i = 0; i<n; i++){
            for (int j = 0; j<n; j++){
                if (grid[i][j]) {
                    dfs(grid, i, j, visited);
                    return bfs(grid, visited);
                }
            }
        }
        cout << "XX"<<endl;
        return -1;
    }
};
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