<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<pair<int,int>,int>> q;
    vector<vector<int>> direction ={{-1,0}, {1,0},{0,-1},{0,1}};
    void dfs( vector<vector<int>>& grid, int i, int j ){
        q.push( pair(pair(i,j), 0) );
        grid[i][j] = 2;
        for( auto d: direction ){
            int r = d[0] + i;
            int c = d[1] + j;
            if(r >=0 && c>=0 && r< grid.size() && c< grid[0].size()&& grid[r][c] ==1){
                dfs(grid, r,c);
            }
        }

    }

    int shortestBridge(vector<vector<int>>& grid) {
        for(int i=0; i< grid.size();i++){
            bool found = false;
            for(int j=0; j < grid[0].size();j++){
                if(grid[i][j] == 1){
                    dfs(grid,i,j );
                    found=true;
                    break;
                } 

            }
            if(found) break;
        }
        while(!q.empty()){
            auto [p, dist] = q.front();
            int row = p.first;
            int col = p.second;
            q.pop();
            for( auto d: direction ){
                int r = d[0] + row;
                int c = d[1] + col;
                if(r >= 0 && c>= 0 && r< grid.size() && c< grid[0].size() ){
                    if(grid[r][c] == 1) return dist;
                    if(grid[r][c] ==0){
                        grid[r][c] = -1;
                        q.push( pair( pair(r,c),dist+1 ) );
                    }
                }
            }
            
        }
        return 0;
    }
};
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