My intuition is to use DFS. Similar to problem of "Number to Islands". The mistake I made is in red. I need to mark the visited position otherwise, I'll get infinite recursive DFS call. Also the rectangle corner points should be updated in the beginning of DFS.

1:  class Solution {  
2:  private:  
3:    int maxI, minI, maxJ, minJ;  
4:    int row, col;  
5:    vector<pair<int,int>> dir = {{-1,0},{1,0},{0,-1},{0,1}};  
6:  public:  
7:    int minArea(vector<vector<char>>& image, int x, int y) {  
8:      maxI = maxJ = 0;  
9:      minI = minJ = INT_MAX;  
10:      row = image.size();  
11:      if (row == 0) return 0;  
12:      col = image[0].size();  
13:      dfs(image, x, y);  
14:      return (maxI - minI + 1) * (maxJ - minJ + 1);  
15:    }  
16:    void dfs(vector<vector<char>> &image, int i, int j) {  
17:      image[i][j] = '0';  
18:      minI = min(minI, i);  
19:      minJ = min(minJ, j);  
20:      maxI = max(maxI, i);  
21:      maxJ = max(maxJ, j);  
22:      for (int k = 0; k < dir.size(); k++) {  
23:        int ii = i + dir[k].first;  
24:        int jj = j + dir[k].second;  
25:        if (ii < 0 || ii == row || jj < 0 || jj == col || image[ii][jj] == '0') continue;  
26:        dfs(image, ii, jj);  
27:      }  
28:    }  
29:  };  

There is a binary search solution in the top rated solutions but I don't quite understand. Need to revisit.