<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:
    vector<vector<int>> directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    
    void dfs(int i, int j, vector<vector<int>>& grid, queue<pair<int, int>>& q) {
        int n = grid.size(), m = grid[0].size();
        if (i < 0 || j < 0 || i >= n || j >= m || grid[i][j] != 1) return;
        
        grid[i][j] = 2;
        q.push({i, j});
        
        for (auto dir : directions) {
            int newI = i + dir[0];
            int newJ = j + dir[1];
            dfs(newI, newJ, grid, q);
        }
    }
    
    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        queue<pair<int, int>> q;
        bool found = false;
        
        // Step 1: Find and mark the first island using DFS
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (grid[i][j] == 1) {
                    dfs(i, j, grid, q);
                    found = true;
                    break;
                }
            }
            if (found) break;
        }
        
        // Step 2: BFS to find the shortest path to the second island
        int steps = 0;
        while (!q.empty()) {
            int size = q.size();
            while (size--) {
                auto [i, j] = q.front(); q.pop();
                
                for (auto dir : directions) {
                    int newI = i + dir[0];
                    int newJ = j + dir[1];
                    
                    if (newI < 0 || newJ < 0 || newI >= n || newJ >= m) continue;
                    
                    if (grid[newI][newJ] == 1) return steps;
                    
                    if (grid[newI][newJ] == 0) {
                        grid[newI][newJ] = 2;
                        q.push({newI, newJ});
                    }
                }
            }
            steps++;
        }
        
        return -1;
    }
};

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