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.

Tips:

找到包含所有black pixel的最小矩形。这里我们用二分查找。因为给定black pixel点(x，y)，并且所有black pixel都是联通的，以row search为例， 所有含有black pixel的column，映射到row x上时，必定是连续的。这样我们可以使用binary search，在0到y里面搜索最左边含有black pixel的一列。接下来可以继续搜索上下和右边界。搜索右边界和下边界的时候，其实我们要找的是第一个'0'，所以要传入一个boolean变量searchLo来判断。

Suppose we have a 2D array

"000000111000000"
"000000101000000"
"000000101100000"
"000001100100000"

Imagine we project the 2D array to the bottom axis with the rule "if a column has any black pixel it's projection is black otherwise white". The projected 1D array is

"000001111100000" Theorem

If there are only one black pixel region, then in a projected 1D array all the black pixels are connected.

Proof by contradiction

Assume to the contrary that there are disconnected black pixels at i and j where i < j in the 1D projection array. Thus there exists one column k, k in (i, j) and and the column k in the 2D array has no black pixel. Therefore in the 2D array there exists at least 2 black pixel regions separated by column k which contradicting the condition of "only one black pixel region".

Therefore we conclude that all the black pixels in the 1D projection array is connected. This means we can do a binary search in each half to find the boundaries, if we know one black pixel's position. And we do know that.

To find the left boundary, do the binary search in the [0, y) range and find the first column vector who has any black pixel.

To determine if a column vector has a black pixel is O(m) so the search in total is O(m log n)

We can do the same for the other boundaries. The area is then calculated by the boundaries. Thus the algorithm runs in O(m log n + n log m)

Code:

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;
    }