_302.java

package com.fishercoder.solutions;

/**
 * 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.
 */
public class _302 {
    class Solution {
        private char[][] image;

        public int minArea(char[][] iImage, int x, int y) {
            image = iImage;
            int m = image.length;
            int 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;
                int 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;
                int mid = (i + j) / 2;
                while (k < right && image[mid][k] == '0') {
                    ++k;
                }
                if (k < right == opt) {
                    j = mid;
                } else {
                    i = mid + 1;
                }
            }
            return i;
        }
    }
}