<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 {
    int rows, cols; 
    int directions[4][2]  =  {{1,0}, {0,1}, {-1,0} , {0,-1}} ; 
    vector<vector<int>> visited ; 
    deque<vector<int>> bfs  ; 

    bool notValid(int row, int col ) {
        return row < 0 || col < 0 || row>= rows || col >=cols ; 
    }

    
    void searchFirstIsland(int row , int col, vector<vector<int>>& grid) {
        queue<vector<int>> q ; 
        visited[row][col] = 1 ;

        q.push({row,col}) ; 

        while(!q.empty()) {
            auto top = q.front() ; 
            q.pop() ; 
            bfs.push_back({top[0],top[1] , 0}) ; 
            for(auto d: directions) {
                int x = d[0] + top[0] , y = d[1] + top[1] ; 
                if(notValid(x, y) || visited[x][y] || grid[x][y] == 0){
                    continue ; 
                }
                q.push({x,y}) ; 
                visited[x][y] = 1 ;
            }
        }
    }


public:
    int shortestBridge(vector<vector<int>>& grid) {

        this->rows = grid.size();
        this->cols = grid[0].size(); 

        visited.resize(rows, vector<int>(cols, 0)) ; 
        int firstR = -1, firstC = -1, count =0 ;
        for(int row = 0 ; row < rows ; row++) { 
            for(int col = 0 ; col < cols ; col++) {
                if( grid[row][col] == 1) {
                    firstR = row;
                    firstC = col;
                    count ++ ; 
                    break ;
                }
            }
            if( firstR != -1) break ;
        }
        searchFirstIsland(firstR,firstC, grid) ; 
        //print(count);
        int minimumFlips = 0 ; 
        while(!bfs.empty()) { 
            auto top = bfs.front() ; 
            bfs.pop_front() ; 

            if(grid[top[0]][top[1]] == 1 && top[2] != 0 ) {
                return top[2]-1 ; 
            } 

            for( auto d : directions) {
                int nx = d[0] + top[0] , ny = d[1] + top[1] ; 
                if( notValid(nx, ny) || visited[nx][ny] ) continue ; 

                if( grid[nx][ny] == 1) bfs.push_front({nx,ny, top[2] + 1}) ; 
                else bfs.push_back({nx,ny, top[2] + 1}) ; 
                visited[nx][ny] = 1 ; 
            }
        }

        return minimumFlips ;

    }
};
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