<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:
    vector<pair<int, int>> dirs = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
    void dfs(int i, int j, vector<vector<int>> &grid, set<vector<int>>& visited, int m, int n){
        visited.insert({i, j});
        for(auto [dx, dy]: dirs){
            int nx = i + dx;
            int ny = j + dy;
            if(nx >= 0 && ny >= 0 && nx < m && ny < n && grid[nx][ny] == 1 && visited.find({nx, ny}) == visited.end()){
                dfs(nx, ny, grid, visited, m, n);
            }
        }
    }
    int shortestBridge(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        set<vector<int>> visited;
        queue<vector<int>> qu;
        bool f = false;
        for(int i = 0; i < m; i++){
            if(f){
                break;
            }
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    dfs(i, j, grid, visited, m , n);
                    f = true;
                    break;
                }
            }
        }

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1 && visited.find({i, j}) == visited.end()){
                    qu.push({i, j});
                }
            }
        }
        int d = 0;
        while(!qu.empty()){
            int x = qu.size();
            while(x--){
                vector<int> tp = qu.front();
                qu.pop();
                for(auto [dx, dy]: dirs){
                    int nx = tp[0] + dx;
                    int ny = tp[1] + dy;

                    if(visited.find({nx, ny}) != visited.end()){
                        return d;
                    }
                    if(nx >= 0 && ny >= 0 && nx < m && ny < n && grid[nx][ny] == 0){
                        grid[nx][ny] = 1;
                        qu.push({nx, ny});
                    }
                }

            }
            d++;
        }
        return -1 ;
    }
};
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