<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 {
private:
    void dfs(int row, int col, vector<vector<int>>& grid, vector<vector<int>>& vis, queue<pair<int,pair<int,int>>> &q)
    {
        int n=grid.size();
        int m=grid[0].size();
        vis[row][col]=1;
        vector<int> drow={-1,0,1,0};
        vector<int> dcol={0,1,0,-1};

        q.push({0,{row,col}});

        for(int i=0;i<4;i++)
        {
            int nrow=row+drow[i];
            int ncol=col+dcol[i];

            if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and !vis[nrow][ncol] and grid[nrow][ncol]==1)
            {
                dfs(nrow,ncol,grid,vis,q);
            }
        }
    }

public:
    int shortestBridge(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();

        vector<vector<int>> vis(n,vector<int>(m,0));
        queue<pair<int,pair<int,int>>> q;
        bool found=false;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(!vis[i][j] and grid[i][j]==1)
                {
                    dfs(i,j,grid,vis,q);
                    found=true;
                    break;
                }
            }
            if(found) break;
        }
        int ans=INT_MAX;
        while(!q.empty())
        {
            int dist=q.front().first;
            int row=q.front().second.first;
            int col=q.front().second.second;
            q.pop();

            vector<int> drow={-1,0,1,0};
            vector<int> dcol={0,1,0,-1};

            for(int i=0;i<4;i++)
            {
                int nrow=row+drow[i];
                int ncol=col+dcol[i];

                if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and !vis[nrow][ncol])
                {
                    if(grid[nrow][ncol]==1)
                    {
                        ans=min(ans,dist);
                    }
                    else
                    {
                        vis[nrow][ncol]=1;
                        q.push({dist+1,{nrow,ncol}});
                    }
                }
            }
        }
        return ans;
    }
};
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