<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:
    int n, start = 0, end = 0;
    vector<int> list;
    void dfs(vector<vector<int>>& grid, int x, int y){
        if(x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1){
            grid[x][y] = 2;
            list[end++] = x * n + y;
            dfs(grid, x + 1, y);
            dfs(grid, x - 1, y);
            dfs(grid, x, y + 1);
            dfs(grid, x, y - 1);
        }
    }
    int bfs(vector<vector<int>>& grid){
        int d[5] = {0, 1, 0, -1, 0};
        while(start != end){
            int x = list[start] / n;
            int y = list[start] % n;
            start++;
            for(int k = 0; k < 4; k++){
                int x1 = x + d[k];
                int y1 = y + d[k + 1];
                if(x1 < 0 || x1 >= n || y1 < 0 || y1 >= n || grid[x1][y1] > 1){
                    continue;
                }
                if(grid[x1][y1] == 1){
                    return grid[x][y] - 2;
                }
                grid[x1][y1] = grid[x][y] + 1;
                list[end++] = x1 * n + y1;
            }
        }
        return -1;
    }
    int shortestBridge(vector<vector<int>>& grid) {
        n = grid.size();
        list.assign(n * n, 0);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    dfs(grid, i, j);
                    return bfs(grid);
                }
            }
        }
        return -1;
    }
};
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