Smallest Rectangle Enclosing Black Pixels

Source

• leetcode: 302

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.

Java

public int minArea(char[][] image, int x, int y) {
    int m = image.length, n = image[0].length;
    int colMin = binarySearch(image, true, 0, y, 0, m, true);
    int colMax = binarySearch(image, true, y + 1, n, 0, m, false);
    int rowMin = binarySearch(image, false, 0, x, colMin, colMax, true);
    int rowMax = binarySearch(image, false, x + 1, m, colMin, colMax, false);
    return (rowMax - rowMin) * (colMax - colMin);
}

public int binarySearch(char[][] image, boolean horizontal, int lower, int upper, int min, int max, boolean goLower) {
    while(lower < upper) {
        int mid = lower + (upper - lower) / 2;
        boolean inside = false;
        for(int i = min; i < max; i++) {
            if((horizontal ? image[i][mid] : image[mid][i]) == '1') {
                inside = true; 
                break;
            }
        }
        if(inside == goLower) {
            upper = mid;
        } else {
            lower = mid + 1;
        }
    }
    return lower;
}