<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;
    int d[4][2] = {{0,1},{1,0},{-1,0},{0,-1}};
    bool check(vector<vector<int>>& nums,int i,int j){
        if(i<0||i>=nums.size()||j<0||j>=nums[0].size()) return 0;
        return 1;
    }
    void dfs(vector<vector<int>>& nums,vector<vector<int>>& vis,int i,int j){
        if(!check(nums,i,j) || vis[i][j]) return;
        if(nums[i][j]==0){
            vis[i][j]=1; q.push({i,j});
            return;
        }
        vis[i][j]=1;
        dfs(nums,vis,i+1,j); dfs(nums,vis,i-1,j);
        dfs(nums,vis,i,j+1); dfs(nums,vis,i,j-1);
    }
    int shortestBridge(vector<vector<int>>& nums) {
        vector<vector<int>> vis(nums.size(),vector<int>(nums[0].size(),0));
        bool st = 1;
        for(int i = 0; i<nums.size();++i){
            for(int j = 0; j<nums[0].size() && st;++j){
                if(nums[i][j]){
                    dfs(nums,vis,i,j); st=0;
                    break;
                }
            }
        }
        int l = 0;
        while(!q.empty()){
            int s = q.size();
            while(s--){
                auto[i,j] = q.front(); q.pop();
                if(nums[i][j]==1) return l;
                for(int k = 0; k<4;++k){
                    int x=i+d[k][0],y=j+d[k][1];
                    if(check(nums,x,y) && !vis[x][y]){
                        vis[x][y]=1; q.push({x,y});
                    }
                }
            }
            ++l;
        }
        return 0;
    }
};
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