<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>

const vector<vector<int>> dirs = {{1,0}, {-1,0}, {0,1},{0,-1}};

class Solution {
public:
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<pair<int,int>> bfs;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                if(grid[i][j] == 0) continue;
                markAsIsland2(i,j,grid,bfs);
                return distIsland1(grid,bfs);
            }
        }
        return -1;
    }

private:
    void markAsIsland2(int i, int j, vector<vector<int>>& grid, queue<pair<int,int>> &bfs){
        int n = grid.size();
        if(i < 0 || i >= n || j < 0 || j >= n) return;
        if(grid[i][j] == 0 || grid[i][j] == 2) return;
        grid[i][j] = 2;
        bfs.push({i,j});
        for(auto dir : dirs) markAsIsland2(i + dir[0], j + dir[1], grid, bfs);
    }

    // min distance -> BFS
    int distIsland1(vector<vector<int>>& grid, queue<pair<int,int>> &bfs){
        int n = grid.size(), breadth = 0;
        while(!bfs.empty()){
            for(int breadthSize = bfs.size(); breadthSize > 0; --breadthSize){
                auto curr = bfs.front();
                bfs.pop();
                if(curr.first < 0 || curr.first >= n || curr.second < 0 || curr.second >= n) continue;
                if(grid[curr.first][curr.second] == 1) return breadth-1;
                if(grid[curr.first][curr.second] == -1) continue;
                grid[curr.first][curr.second] = -1;
                for(auto dir : dirs) bfs.push({curr.first+dir[0], curr.second+dir[1]});
            }
            breadth++;
        }
        return -1;
    }
};
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