<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 dir[5] = {0, 1, 0, -1, 0};
    
    void paint(vector<vector<int>>& A, int i, int j, vector<pair<int, int>> &q) {
        // paint island
        // i, j is the original point
        if (min(i, j) >= 0 && max(i, j) < A.size() && A[i][j] == 1) {
            A[i][j] = 2;
            q.push_back({i, j});
            for (int d = 0; d < 4; ++d)
                paint(A, i + dir[d], j + dir[d + 1], q);
        }
    }
    
    int shortestBridge(vector<vector<int>>& A) {
        
        vector<pair<int, int>> q;
        // find one of the island first. put all parcels into queue.
        // make their values to 2
        int flag = -9;
        for (int i = 0; q.size() == 0 && i < A.size(); ++i)
            for (int j = 0; q.size() == 0 && j < A[0].size(); ++j){
                if(A[i][j] == 1){
                    paint(A, i, j, q);
                    flag = -99;
                    break;
                }
                if(flag == -99) break;
            }
        while (!q.empty()) {
            vector<pair<int, int>> q1;
            for (auto [i, j] : q) {
                for (int d = 0; d < 4; ++d) {
                    int x = i + dir[d], y = j + dir[d + 1];
                    if (min(x, y) >= 0 && max(x, y) < A.size()) {
                        // if met another island colored by 1, the original label
                        // then current value - 2 will be number of zeroes met b4.
                        if (A[x][y] == 1)
                            return A[i][j] - 2;
                        if (A[x][y] == 0) {
                            A[x][y] = A[i][j] + 1;
                            q1.push_back({x, y});
                        } // other values means met before so skip
                    }
                }
            }
            swap(q, q1);
        }
        return 0;
    }
};





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