<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 bfs(int i, int j, vector<vector<int>>& grid, queue<pair<int,int>>& q) {
        int dx[] = {0, 0, -1, 1};
        int dy[] = {-1, 1, 0, 0};
        int n = grid.size();
        vector<vector<int>> vis(n, vector<int>(n, 0));
        
        queue<pair<int,int>> tempQ;
        grid[i][j] = 2;
        tempQ.push({i, j});
        q.push({i, j});
        
        while (!tempQ.empty()) {
            int r = tempQ.front().first;
            int c = tempQ.front().second;
            tempQ.pop();

            for (int k = 0; k < 4; k++) {
                int nr = r + dx[k];
                int nc = c + dy[k];

                if (nr >= 0 && nr < n && nc >= 0 && nc < n && grid[nr][nc] == 1) {
                    grid[nr][nc] = 2;
                    tempQ.push({nr, nc});
                    q.push({nr, nc}); 
                }
            }
        }
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        int dx[] = {0, 0, -1, 1};
        int dy[] = {-1, 1, 0, 0};

        queue<pair<int, int>> q;

        for (int i = 0; i < n; i++) {
            bool found = false;
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    bfs(i, j, grid, q);
                    found = true;
                    break;
                }
            }
            if (found) break;
        }

        vector<vector<int>> vis(n, vector<int>(n, 0));

        int steps = 0;
        while (!q.empty()) {
            int sz = q.size();
            while (sz--) {
                auto [r, c] = q.front();
                q.pop();

                for (int k = 0; k < 4; k++) {
                    int nr = r + dx[k];
                    int nc = c + dy[k];

                    if (nr >= 0 && nc >= 0 && nr < n && nc < n && !vis[nr][nc]) {
                        if (grid[nr][nc] == 1) {
                            return steps;
                        }
                        if (grid[nr][nc] == 0) {
                            q.push({nr, nc});
                        }
                        vis[nr][nc] = 1;
                    }
                }
            }
            steps++;
        }

        return -1;
    }
};

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