<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 row;
    int col;
    int delrow[4] = {0, 1, 0, -1};
    int delcol[4] = {-1, 0, 1, 0};
    bool isSafe(int i, int j) {
        return (i < row && i >= 0 && j < col && j >= 0);
    }
    void dfs(int i, int j, vector<vector<int>>& grid,
             set<pair<int, int>>& visited) {
        if (!isSafe(i, j) || visited.find({i, j}) != visited.end() ||
            grid[i][j] == 0) {
            return;
        }
        visited.insert({i, j});
        for (int p = 0; p < 4; p++) {
            int nrow = i + delrow[p];
            int ncol = j + delcol[p];
            dfs(nrow, ncol, grid, visited);
        }
    }
    int bfs(vector<vector<int>>& grid, set<pair<int, int>>& visited) {
        int level = 0;
        queue<pair<int, int>> que;
        for (auto& it : visited) {
            que.push({it});
        }
        while (!que.empty()) {
            int n = que.size();
     
            for (int i = 0; i < n; i++) {

                int row = que.front().first;
                int col = que.front().second;
                que.pop();

                for (int p = 0; p < 4; p++) {
                    int nrow = row + delrow[p];
                    int ncol = col + delcol[p];
                    if (isSafe(nrow, ncol) and
                        visited.find({nrow, ncol}) == visited.end()) {
                        if (grid[nrow][ncol] == 1) {
                            return level;
                        }
                        que.push({nrow, ncol});
                        visited.insert({nrow, ncol});
                    }
                }
            }
                   level++;
        }
        return level;
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int count = 0;
        row = grid.size();
        col = grid[0].size();
        set<pair<int, int>> visited;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == 1) {
                    dfs(i, j, grid, visited);
                    return bfs(grid, visited);
                }
            }
        }

        return -1;
    }
};
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