<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(int i , int j ,vector<vector<int>>& grid ,queue<vector<int>>&q ){
        if(i<0||i>=grid.size() || j<0 || j>= grid[0].size() || grid[i][j] == 2 || grid[i][j] == 0)return;
        grid[i][j] = 2;
        q.push({i,j,0});
        dfs(i , j-1 , grid ,q);
        dfs(i , j+1 , grid ,q);
        dfs(i+1 , j , grid ,q);
        dfs(i-1 , j , grid ,q);
    }
    int shortestBridge(vector<vector<int>>& grid) {
       
        queue<vector<int>>q;
        for(int i=0;i<grid.size();i++){
            int f=0;
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j]){
                    f=1;
                    dfs(i,j,grid,q);
                    break;
                }
            }
            if(f)break;
        }
        
        cout<<q.size()<<endl;
        int dr[4] = {0,0,-1,1};
        int dc[4] = {-1,1,0,0};
        while(!q.empty()){
            auto v = q.front();
            q.pop();
            int i = v[0] , j = v[1] , dis = v[2];
            for(int k=0;k<4;k++){
                int r = i+dr[k];
                int c = j+dc[k];
                if(r<0 || c<0 || r>= grid.size() || c>=grid.size())continue;
                if(grid[r][c] == 2)continue;
                if(grid[r][c] == 1)return dis;
                grid[r][c] = 2; 
                q.push({r,c,dis+1});
            }
        }
        return 0;

    }
};
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