<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 i, int j, int n, int m, vector<vector<int>>& grid) {
        if(i < 0 || i >= n || j < 0 || j >= m || grid[i][j] == 0 || grid[i][j] == 2) return;
        grid[i][j] = 2;
        dfs(i - 1, j, n, m, grid);
        dfs(i, j + 1, n, m, grid);
        dfs(i + 1, j, n, m, grid);
        dfs(i, j - 1, n, m, grid);
    }

    int bfs(queue<vector<int>>& q, vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        int x[4] = {0, 1, 0, -1};
        int y[4] = {1, 0, -1, 0};
        while(!q.empty()) {
            vector<int> node = q.front(); q.pop();
            int i = node[0], j = node[1], w = node[2];
            if(grid[i][j] == 2) return w - 1;
            for(int k = 0; k < 4; ++k) {
                int r = i + x[k], c = j + y[k];
                if(r >= 0 && r < n && c >= 0 && c < m && grid[r][c] != 1) {
                    grid[r][c] = (grid[r][c] == 2 ? 2 : 1);
                    q.push({r, c, w + 1});
                }
            }
        }
        return 0;
    }

    int shortestBridge(vector<vector<int>>& grid) {
        // set one island to another color let's say 2
        // find the shortest path from color 1 to color 2
        int n = grid.size(), m = grid[0].size();
        bool islandFound = false;
        queue<vector<int>> q;
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                if(grid[i][j] == 1) {
                    // set island to color 2
                    dfs(i, j, n, m, grid);
                    islandFound = true;
                    break;
                }
            }
            if(islandFound) break;
        }

        // add the island
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                if(grid[i][j] == 1) q.push({i, j, 0});
            }
        }

        return bfs(q, grid);
    }
};
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