<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,set<pair<int,int>> &vis){
         int n = grid.size();
         int m = grid[0].size();
         vis.insert({i,j});
         vector<pair<int,int>> directions{
            {0,1},
            {0,-1},
            {1,0},
            {-1,0}
         };
         for(auto& dir: directions){
            int newx = i+dir.first;
            int newy = j+dir.second;
            if(newx>=0 && newy>=0 && newx<n &&newy<m &&grid[newx][newy]==1&&vis.find({newx,newy})==vis.end()){
                vis.insert({newx,newy});
                dfs(newx,newy,grid,vis);
            }
         }
          
    }
    int bfs(vector<vector<int>>& grid,set<pair<int,int>> &vis){
        int res=0;
        int n = grid.size();
        int m = grid[0].size();
        queue<pair<int,int>> q(vis.begin(),vis.end());
         vector<pair<int,int>> directions{
            {0,1},
            {0,-1},
            {1,0},
            {-1,0}
         };
        while(!q.empty()){
            int n1 =q.size();
            while(n1){
                int i = q.front().first;
                int j = q.front().second;
                q.pop();
                for(auto& dir: directions){
                    int newx = i+dir.first;
                    int newy = j+dir.second;
                    if (newx >= 0 && newy >= 0 && newx < n && newy < m) {
                        if (grid[newx][newy] == 1 && vis.find({newx, newy}) == vis.end()) {
                            return res;
                        }
                        if (grid[newx][newy] == 0 && vis.find({newx, newy}) == vis.end()) {
                            q.push({newx, newy});
                            vis.insert({newx, newy});
                        }
                    }
                }
                n1--; 
            } 
            res++;
        }
        return res;
    }
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        bool found =false;
        set<pair<int,int>> vis;
        for(int i=0;i<n;i++){
            for(int j =0;j<m;j++){
                if(grid[i][j]==1){
                    dfs(i,j,grid,vis);
                    found =true;
                    break;
                }
            }
            if(found) break;
        }
        return bfs(grid, vis);
    }
};
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