1 public class Solution { 2     private char[][] image; 3     public int minArea(char[][] image, int x, int y) { 4         if (image.length == 0 || image[0].length == 0) { 5             return 0; 6         } 7         this.image = image; 8         int n = image.length, m = image[0].length; 9         int left = searchRow(0, x, 0, m, true);10         int right = searchRow(x + 1, n, 0, m, false);11         int top = searchCols(0, y, 0, n, true);12         int bottom = searchCols(y + 1, m, 0, n, false);13         return (right - left) * (bottom - top);14     }15     16     private int searchCols(int i, int j, int top, int bottom, boolean bound) {17         while (i != j) {18             int k = top;19             int mid = (i + j) / 2;20             while (k < bottom && image[k][mid] == '0') {21                 k++;22             }23             if(k < bottom == bound) {24                 j = mid;25             } else {26                 i = mid + 1;27             }28         }29         return i;30     }   31     32     private int searchRow(int i, int j, int left, int right, boolean bound) {33         while (i != j) {34             int k = left;35             int mid = (i + j) / 2;36             while (k < right && image[mid][k] == '0') {37                 k++;38             }39             if(k < right == bound) {40                 j = mid;41             } else {42                 i = mid + 1;43             }44         }45         return i;46     }47 }

Binary search:

1. The bound is defined to search "0" or "1". When search 0 - x, we want to get the most left "1". So k < right mean there is a one, j = mid. But when we do x - n search, we want to get most left "0". So when k < right, we still need to keep searching mid + 1.