600. Smallest Rectangle Enclosing Black Pixels [LintCode]

An image is represented by a binary matrix with 0as a white pixel and 1as 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.

Example

For example, given the following image:

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

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

public class Solution {
    /**
     * @param image a binary matrix with '0' and '1'
     * @param x, y the location of one of the black pixels
     * @return an integer
     */
    public int minArea(char[][] image, int x, int y) {
        // Write your code here

        if(image == null || image.length == 0) return 0;
        if(image[0] == null || image[0].length == 0) return 0;

        int M = image[0].length;
        int N = image.length;
        int left = 0;
        int right = 0;
        int up = 0;
        int bottom = 0;

        // find the left bound
        int start = 0;
        int end = y;
        int mid;
        while(start + 1 < end)
        {
            mid = start + (end - start) / 2;

            if(colHasBlackPixel(image, mid))
            {
                end = mid;
            }
            else
            {
                start = mid;
            }
        }
        if(colHasBlackPixel(image, start)) left = start;
        else if(colHasBlackPixel(image, end)) left = end;
        else return 0;

        // find the right bound
        start = y;
        end = M - 1;
        while(start + 1 < end)
        {
            mid = start + (end - start) / 2;

            if(colHasBlackPixel(image, mid))
            {
                start = mid;
            }
            else
            {
                end = mid;
            }
        }
        if(colHasBlackPixel(image, end)) right = end;
        else if(colHasBlackPixel(image, start)) right = start;
        else return 0;

        // find the up bound
        start = 0;
        end = x;
        while(start + 1 < end)
        {
            mid = start + (end - start) / 2;

            if(rowHasBlackPixel(image, mid))
            {
                end = mid;
            }
            else
            {
                start = mid;
            }
        }
        if(rowHasBlackPixel(image, start)) up = start;
        else if(rowHasBlackPixel(image, end)) up = end;
        else return 0;

        // find the bottom bound
        start = x;
        end = N - 1;
        while(start + 1 < end)
        {
            mid = start + (end - start) / 2;

            if(rowHasBlackPixel(image, mid))
            {
                start = mid;
            }
            else
            {
                end = mid;
            }
        }
        if(rowHasBlackPixel(image, end)) bottom = end;
        else if(rowHasBlackPixel(image, start)) bottom = start;
        else return 0;

        return (right - left + 1) * (bottom - up + 1);

    }

    private boolean colHasBlackPixel(char[][] image, int col)
    {   
        if(image == null || col < 0) return false;

        for(int i = 0; i < image.length; i++)
        {
            if(image[i][col] == '1') return true;
        }
        return false;
    }

    private boolean rowHasBlackPixel(char[][] image, int row)
    {
        if(image == null || row < 0) return false;

        for(int i = 0; i < image[0].length; i++)
        {
            if(image[row][i] == '1') return true;
        }
        return false;
    }
}