<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(int row, int column, vector<vector<int>>& grid, int n, int m) {
        // Base cases
        if (row < 0 || row >= n || column < 0 || column >= m || grid[row][column] != 1) {
            return;
        }
        // Mark the cell as part of the first island
        grid[row][column] = 2;
        // Spread in all four directions
        dfs(row + 1, column, grid, n, m);
        dfs(row - 1, column, grid, n, m);
        dfs(row, column + 1, grid, n, m);
        dfs(row, column - 1, grid, n, m);
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        // Find the first island
        bool first_found = false;
        for (int i = 0; i < n && !first_found; ++i) {
            for (int j = 0; j < m && !first_found; ++j) {
                if (grid[i][j] == 1) {
                    dfs(i, j, grid, n, m);
                    first_found = true;
                }
            }
        }

        // Add all cells marked as part of the first island to the queue
        queue<pair<int, int>> q;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (grid[i][j] == 2) {
                    q.push({i, j});
                }
            }
        }

        // BFS to find the shortest path to the second island
        vector<int> dr{1, -1, 0, 0};
        vector<int> dc{0, 0, 1, -1};
        int count = 0;

        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                auto [r, c] = q.front();
                q.pop();
                for (int k = 0; k < 4; ++k) {
                    int nr = r + dr[k];
                    int nc = c + dc[k];
                    if (nr >= 0 && nr < n && nc >= 0 && nc < m) {
                        if (grid[nr][nc] == 1) {
                            return count;
                        }
                        if (grid[nr][nc] == 0) {
                            grid[nr][nc] = 2;
                            q.push({nr, nc});
                        }
                    }
                }
            }
            ++count;
        }
        return -1;
    }
};

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