<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:
    int r,c;
    queue<vector<int>>q;
    bool valid(int i,int j){
        if(i<0 || i>=r || j<0 || j>=c) return false;
        return true;
    }
    vector<vector<int>>dir={{-1,0},{1,0},{0,-1},{0,1}};//t,b,l,r
    void dfs(vector<vector<int>>&grid, int i,int j, int& colour, vector<vector<bool>>&visited){
        grid[i][j]=colour;
        if(colour==2){
            q.push({i,j});
        }
        visited[i][j]=true;
        for(auto x: dir){
            if(valid(i+x[0],j+x[1])&&grid[i+x[0]][j+x[1]]==1 && !visited[i+x[0]][j+x[1]]){
                dfs(grid, i+x[0],j+x[1],colour, visited);
            }
        }
    }
    int shortestBridge(vector<vector<int>>& grid) {
       r = grid.size();
       c = grid[0].size();
       vector<vector<bool>>visited(r,vector<bool>(c,false));
       int island=2;
       int i,j;
       bool sig=false;
       for(i=0;i<r;i++){
        for( j=0;j<c;j++){
            if(grid[i][j]==1 && !visited[i][j]){
                sig = true;
                dfs(grid,i,j,island,visited); 
                break;
            }
        }
        if(sig) break;
       }
       int bridge=0;
       int t;
       while(!q.empty()){
            t = q.size();
            bridge++;
            while(t--){
                auto f = q.front();
                q.pop();
                for(auto x: dir){
                    if(valid(f[0]+x[0],f[1]+x[1])&&grid[f[0]+x[0]][f[1]+x[1]]!=2 && !visited[f[0]+x[0]][f[1]+x[1]]){
                        if(grid[f[0]+x[0]][f[1]+x[1]]==1) {
                           return bridge-1;
                        }
                        visited[f[0]+x[0]][f[1]+x[1]]=true;
                        q.push({f[0]+x[0],f[1]+x[1]});
                    }
                }
            }
       }
       return bridge;
    }
};
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