<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){
        grid[i][j] = 2;
        if(i+1<grid.size() && grid[i+1][j]==1)dfs(i+1,j,grid);
        if(i-1>=0 && grid[i-1][j]==1)dfs(i-1,j,grid);
        if(j+1<grid[0].size() && grid[i][j+1]==1)dfs(i,j+1,grid);
        if(j-1>=0 && grid[i][j-1]==1)dfs(i,j-1,grid);
    }
    int shortestBridge(vector<vector<int>>& grid) {
        bool check = 1;
        for(int i=0;i<grid.size();++i){
            for(int j=0;j<grid[0].size();++j){
                if(grid[i][j]==1){
                    dfs(i,j,grid);
                    check = 0;
                    break;
                }
            }
            if(!check)break;
        }
        queue<vector<int>>q;
        vector<vector<bool>>vis(grid.size(), vector<bool>(grid[0].size(),0));
        for(int i=0;i<grid.size();++i){
            for(int j=0;j<grid[0].size();++j){
                if(grid[i][j]==1){
                    // grid[i][j] = 1;
                    q.push({i,j,0});
                }
            }
        }
        int ans = INT_MAX;
        while(!q.empty()){
            int x = q.front()[0];
            int y = q.front()[1];
            int z = q.front()[2];
            // vis[x][y] = 1;
            q.pop();
            if(grid[x][y]==2){
                ans=min(ans,z-1);
                continue;
            }
            if(x+1<grid.size() && grid[x+1][y]!=1 && !vis[x+1][y]){
                vis[x+1][y] = 1;
                q.push({x+1,y,z+1});
            }
            if(x-1>=0 && grid[x-1][y]!=1 && !vis[x-1][y]){
                vis[x-1][y] = 1;
                q.push({x-1,y,z+1});
            }
            if(y+1<grid[0].size() && grid[x][y+1]!=1 && !vis[x][y+1]){
                vis[x][y+1] = 1;
                q.push({x,y+1,z+1});
            }
            if(y-1>=0 && grid[x][y-1]!=1 && !vis[x][y-1]){
                vis[x][y-1] = 1;
                q.push({x,y-1,z+1});
            }
        }
        return ans;
    }
};
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