<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:
    void dfs(vector<vector<int>>& grid, int i, int j, set<pair<int, int>>& st) {
        int n = grid.size();
        if (i < 0 || j < 0 || i >= n || j >= n) {
            return;
        }
        if (grid[i][j] == 0) {
            return;
        }
        if (st.find({i, j}) != st.end()) {
            return;
        }
        st.insert({i, j});
        dfs(grid, i + 1, j, st);
        dfs(grid, i, j + 1, st);
        dfs(grid, i - 1, j, st);
        dfs(grid, i, j - 1, st);
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        set<pair<int, int>> st;

        bool found = true;
        for (int i = 0; i < n && found; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    dfs(grid, i, j, st);
                    st.insert({i, j});
                    found = false;
                    break;
                }
            }
        }

        // BFS
        queue<pair<int, int>> q;
        for (auto& s : st) {
            q.push(s);
        }

        int level = 0;
        while (!q.empty()) {

            int l = q.size();

            while (l--) {
                int r = q.front().first;
                int c = q.front().second;
                q.pop();

                vector<int> dirRow = {1, 0, -1, 0};
                vector<int> dirCol = {0, 1, 0, -1};

                for (int i = 0; i < 4; i++) {
                    int newR = r + dirRow[i];
                    int newC = c + dirCol[i];

                    if (newR < 0 || newC < 0 || newR >= n || newC >= n || st.find({newR, newC}) != st.end()) {
                        continue;
                    }


                    if (grid[newR][newC] == 1) {
                        return level;
                    }
                    q.push({newR, newC});
                    st.insert({newR, newC});

                }
            }
            level++;
        }
        return -1;
    }
};
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