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

    int findShortest(int n, vector<vector<int>>& grid){
        queue<vector<int>> q;
        vector<vector<bool>> vis(n, vector<bool>(n, false));

        // we are going from 2 to 1;
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == 2){
                    if((i-1>0 && grid[i-1][j]==0) || (j-1>0 && grid[i][j-1]==0) ||
                        (i+1<n && grid[i+1][j]==0) || (j+1<n && grid[i][j+1]==0) ){
                            q.push({i,j,0});
                            vis[i][j] = true;
                        }
                }
            }
        }

        int ans = INT_MAX;
        vector<pair<int, int>> dirs = {{0,1}, {1,0}, {0,-1}, {-1,0}};
        while(!q.empty()){
            vector<int> u = q.front();
            q.pop();

            for(auto [dx,dy]: dirs){
                int x = u[0] + dx;
                int y = u[1] + dy;

                if(x<0 || y<0 || x>=n || y>=n || vis[x][y] || grid[x][y] == 2){
                    continue;
                }

                if(grid[x][y] == 0){
                    q.push({x,y,u[2]+1});
                    vis[x][y] = true;
                }else if(grid[x][y] == 1){
                    ans = min(ans, u[2]);
                }
            }
        }

        return ans;
    }

    int shortestBridge(vector<vector<int>>& grid) {
        int n = grid.size();
        bool isColored = false;
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == 1){
                    color(i,j, n, grid);
                    isColored = true;
                    break;
                }
            }
            if(isColored)
                break;
        }
        
        int ans = findShortest(n, grid);

        return ans;
    }
};
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