<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 maxI;
    int maxJ;
    vector<vector<int>> dirs {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    
    // This function marks the entire first island
    void markIsland(vector<vector<int>>& grid, int i, int j, queue<pair<int, int>>& q) {
        if (i < 0 || i >= maxI || j < 0 || j >= maxJ || grid[i][j] != 1)
            return;
        
        grid[i][j] = 2; // Mark the cell as part of the first island
        q.push({i, j}); // Add this cell to the queue for BFS
        
        // Recursively mark the neighboring cells of the island
        for (auto dir : dirs) {
            markIsland(grid, i + dir[0], j + dir[1], q);
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        maxI = grid.size();
        maxJ = grid[0].size();
        
        queue<pair<int, int>> q;
        bool found = false;
        
        // Find and mark the first island
        for (int i = 0; i < maxI; i++) {
            for (int j = 0; j < maxJ; j++) {
                if (grid[i][j] == 1) {
                    markIsland(grid, i, j, q);  // Mark all parts of the first island
                    found = true;
                    break;
                }
            }
            if (found) break;
        }
        
        // Perform BFS to expand from the first island and find the second island
        int level = 0;
        while (!q.empty()) {
            int n = q.size();
            while (n--) {
                auto [x, y] = q.front();
                q.pop();
                
                // Explore all 4 directions
                for (auto dir : dirs) {
                    int newX = x + dir[0];
                    int newY = y + dir[1];
                    
                    if (newX < 0 || newX >= maxI || newY < 0 || newY >= maxJ)
                        continue;

                    if (grid[newX][newY] == 1) {  // Found the second island
                        return level;
                    }
                    
                    if (grid[newX][newY] == 0) {  // Expand the BFS to water
                        grid[newX][newY] = -1;  // Mark as visited
                        q.push({newX, newY});
                    }
                }
            }
            level++;  // Increment the distance level for the next layer of BFS
        }
        
        return -1;  // Should not reach here
    }
};

39835