Smallest Rectangle Enclosing Black Pixels

An image is represented by a binary matrix with 0 as a white pixel and 1 as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y) of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.

For example, given the following image:

[
  "0010",
  "0110",
  "0100"
]

and x = 0, y = 2,Return 6.

A BFS approach. Find the min and max of x and y during the BFS. Mark visited cell as ‘2’ to avoid duplicate visit.

//Can I manipulate this image?
class Solution {
public:
    int minArea(vector<vector<char>>& image, int x, int y) {
        if(image.size() == 0 || image[0].size() == 0) return 0;
        queue<pair<int, int>> q;
        q.push(make_pair(x, y));
        image[x][y] = '2';
        int left, right;
        left = right = x;
        int top, bottom;
        top = bottom = y;
        while(!q.empty()){
            int x = q.front().first;
            int y = q.front().second;
            q.pop();
            left = min(left, x);
            right = max(right, x);
            top = min(top, y);
            bottom = max(bottom, y);
            int dict[][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
            for(int i = 0; i < 4; i++){
                int nx = x + dict[i][0];
                int ny = y + dict[i][1];
                if(nx < 0 || ny < 0 || nx >= image.size() || ny >= image[0].size()){
                    continue; // not valid
                }
                if(image[nx][ny] != '1'){
                    continue; // not a black pixel
                }
                
                q.push(make_pair(nx, ny));
                image[nx][ny] = '2';
            }
        }
        int width = right - left + 1;
        int height = bottom - top + 1;
        return width * height;
    }
};