<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:
    set<pair<int, int>> visited;

    void dfs(vector<vector<int>>& arr, int i, int j, int n, int m) {
        if (i < 0 || j < 0 || i >= n || j >= m) return;
        if (arr[i][j] == 0) return;
        if (visited.find({i, j}) != visited.end()) return;
        visited.insert({i, j});
        arr[i][j] = -1;
        dfs(arr, i + 1, j, n, m);
        dfs(arr, i - 1, j, n, m);
        dfs(arr, i, j + 1, n, m);
        dfs(arr, i, j - 1, n, m);
    }

    int shortestBridge(vector<vector<int>>& arr) {
        int a = -3;
        int n = arr.size();
        int m = arr[0].size();

        // Find the first island and mark it with -1
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (arr[i][j] == 1) {
                    dfs(arr, i, j, n, m);
                    a = 0;
                    break;
                }
            }
            if (a == 0) break;
        }

        set<pair<int, int>> s;
        set<int> st;
        queue<pair<int, int>> q;

        // Push all the cells of the second island into the queue
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (arr[i][j] == 1 && s.find({i, j}) == s.end()) {
                    q.push({i, j});
                    s.insert({i, j});
                }
            }
        }

        q.push({-2, -2});
        int count = 0;

        while (!q.empty()) {
            int row = q.front().first;
            int col = q.front().second;
            q.pop();

            if (row == -2 && col == -2) {
                count++;
                if (!q.empty()) q.push({-2, -2});
                continue;
            }

            if (st.find(-1) != st.end()) break;

            // Check all four directions
            if (row + 1 < n && s.find({row + 1, col}) == s.end()) {
                q.push({row + 1, col});
                s.insert({row + 1, col});
                st.insert(arr[row + 1][col]);
            }
            if (row - 1 >= 0 && s.find({row - 1, col}) == s.end()) {
                q.push({row - 1, col});
                s.insert({row - 1, col});
                st.insert(arr[row - 1][col]);
            }
            if (col + 1 < m && s.find({row, col + 1}) == s.end()) {
                q.push({row, col + 1});
                s.insert({row, col + 1});
                st.insert(arr[row][col + 1]);
            }
            if (col - 1 >= 0 && s.find({row, col - 1}) == s.end()) {
                q.push({row, col - 1});
                s.insert({row, col - 1});
                st.insert(arr[row][col - 1]);
            }
        }

        return count ;
    }
};

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