<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>> g;
    void dfs(int r,int c,vector<vector<int>> &grid,vector<vector<int>> &vis){
        int m=grid.size(),n=grid[0].size();
        if(r<0 || r>=m || c<0 || c>=n || grid[r][c]==0 || vis[r][c]) return;
        vis[r][c]=1;

        g.push_back({r,c});

        int dr[]={0,1,0,-1};
        int dc[]={1,0,-1,0};


        for(int i=0;i<4;i++){
            int nr=r+dr[i],nc=c+dc[i];
            dfs(nr,nc,grid,vis);
        }

    }
    int shortestBridge(vector<vector<int>>& grid) {
        int m=grid.size(),n=grid[0].size(),ans=1e9;

        vector<vector<int>> vis(m,vector<int>(n,0));
        bool once=false;

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1 && !vis[i][j] && !once){
                    once=true;
                    dfs(i,j,grid,vis);
                }
                else if(grid[i][j]==1 && !vis[i][j]){
                    for(auto p:g){
                        int dist=abs(p.first-i)+abs(p.second-j);
                        ans=min(ans,dist);
                    }
                } 
            }
        }

        return ans-1;
    }
};
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