302. 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.

Solution：Binary Search

思路：
Time Complexity: O(logn + logm) Space Complexity: O(1)

Solution Code：

class Solution {
    private char[][] image;
    public int minArea(char[][] iImage, int x, int y) {
        image = iImage;
        int m = image.length, n = image[0].length;
        int left = searchColumns(0, y, 0, m, true);
        int right = searchColumns(y + 1, n, 0, m, false);
        int top = searchRows(0, x, left, right, true);
        int bottom = searchRows(x + 1, m, left, right, false);
        return (right - left) * (bottom - top);
    }
    private int searchColumns(int i, int j, int top, int bottom, boolean opt) {
        while (i != j) {
            int k = top, mid = (i + j) / 2;
            while (k < bottom && image[k][mid] == '0') ++k;
            if (k < bottom == opt)
                j = mid;
            else
                i = mid + 1;
        }
        return i;
    }
    private int searchRows(int i, int j, int left, int right, boolean opt) {
        while (i != j) {
            int k = left, mid = (i + j) / 2;
            while (k < right && image[mid][k] == '0') ++k;
            if (k < right == opt)
                j = mid;
            else
                i = mid + 1;
        }
        return i;
    }
    //  Runtime: 1 ms
}