<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 {
    bool isValid(int i, int j, vector<vector<int>>& grid ){
        int n = grid.size();
        int m = grid[0].size();

        if(i>=0 && i<n && j>=0 && j<m){
            return true;
        }
        return false;
    }
    void dfs(int i,int j,vector<vector<int>>& grid, queue<vector<int>>& q){
        if(!isValid(i,j,grid)){
            return;
        }
        if(grid[i][j]==0 || grid[i][j]==2)
        return;
        grid[i][j] = 2;
        q.push({i,j,0});

        vector<pair<int,int>> dir = {{0,1},{1,0},{0,-1},{-1,0}};

        for(auto d:dir){
            int x = i+d.first;
            int y = j+d.second;

            dfs(x,y,grid,q);
        }
    }
public:
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        queue<vector<int>> q;
        int flag = 0;

        for(int i=0;i<n;i++){
            if(flag) break;
            for(int j=0;j<m;j++){
                if(grid[i][j]){
                    dfs(i,j,grid,q);
                    flag=1;
                    break;
                }
            }
        }
        int ans = -1;

        vector<pair<int,int>> dir = {{0,1},{1,0},{0,-1},{-1,0}};
        cout << q.size()<<endl;

        while(!q.empty()){
            vector<int> top = q.front();
            q.pop();
            int i = top[0];
            int j = top[1];
            int dist = top[2];

            for(auto d:dir){
                int x = i + d.first;
                int y = j + d.second;

                if(!isValid(x,y,grid) || grid[x][y]==2){
                    continue;
                }

                if(grid[x][y]==0){
                    grid[x][y]=2;
                    q.push({x,y,dist+1});
                }
                else{
                    return dist;
                }

            }
        }

        return ans;

    }
};
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