<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:
    queue<pair<int,int>>q;
    void dfs(vector<vector<int>>& grid, int i, int j , int n, int m){
        if(i>=n||j>=m||i<0||j<0||grid[i][j]==2||grid[i][j]==0) return;
        grid[i][j]=2;
        q.push({i,j});
        dfs(grid, i+1,j,n,m);
        dfs(grid, i,j+1,n,m);
        dfs(grid, i,j-1,n,m);
        dfs(grid, i-1,j,n,m);
    }
    int bfs(vector<vector<int>>&grid){
        int n=grid.size();
        int m=grid[0].size();
        int d=0;
        int mini=INT_MAX;
        int dr[]={0,1,0,-1};
        int dc[]={1,0,-1,0};
        while(!q.empty()){
            int size=q.size();
            while(size--){
                int row=q.front().first;
                int col=q.front().second;
                q.pop();
                for(int i=0;i<4;i++){
                    int nrow=dr[i]+row;
                    int ncol=dc[i]+col;
                    if(nrow<n&&nrow>=0&&ncol<m&&ncol>=0&&grid[nrow][ncol]==1){
                        mini=min(mini,d);
                    }
                    if(nrow<n&&nrow>=0&&ncol<m&&ncol>=0&&grid[nrow][ncol]==0){
                        q.push({nrow,ncol});
                        grid[nrow][ncol]=2;
                    }
                }
            }
            d++;
        }
        return mini;
    }
    int shortestBridge(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        bool flag=false;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1&&flag==false){
                    dfs(grid,i,j,n,m);
                    flag=true;
                    break;
                }
            }
            if(flag==true) break;
        }
        return bfs(grid);
    }
};
26065