302. Smallest Rectangle Enclosing Black Pixels

题目：

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.

链接： http://leetcode.com/problems/smallest-rectangle-enclosing-black-pixels/

题解：

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。

Time Complexity - O(mlogn + nlogm)， Space Complexity - O(1)

public class Solution {
    public int minArea(char[][] image, int x, int y) {
        if(image == null || image.length == 0) {
            return 0;
        }
        int rowNum = image.length, colNum = image[0].length;
        int left = binarySearch(image, 0, y, 0, rowNum, true, true);
        int right = binarySearch(image, y + 1, colNum, 0, rowNum, true, false);
        int top = binarySearch(image, 0, x, left, right, false, true);
        int bot = binarySearch(image, x + 1, rowNum, left, right, false, false);

        return (right - left) * (bot - top);
    }

    private int binarySearch(char[][] image, int lo, int hi, int min, int max, boolean searchHorizontal, boolean searchLo) {
        while(lo < hi) {
            int mid = lo + (hi - lo) / 2;
            boolean hasBlackPixel = false;
            for(int i = min; i < max; i++) {
                if((searchHorizontal ? image[i][mid] : image[mid][i]) == '1') {
                    hasBlackPixel = true;
                    break;
                }
            }
            if(hasBlackPixel == searchLo) {
                hi = mid;
            } else {
                lo = mid + 1;
            }
        }
        return lo;
    }
}

Reference:

https://leetcode.com/discuss/68246/c-java-python-binary-search-solution-with-explanation

https://leetcode.com/discuss/71898/java-concise-binary-search-4x-faster-than-dfs

https://leetcode.com/discuss/68407/clear-binary-search-python

https://leetcode.com/discuss/68233/java-dfs-solution-and-seeking-for-a-binary-search-solution

https://leetcode.com/discuss/68738/very-easy-dfs-java-solution-with-comments

https://leetcode.com/discuss/70670/dfs-bfs-binary-search-and-brute-force-interation