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302. Smallest Rectangle Enclosing Black Pixels πŸ”’

Description

You are given an m x n binary matrix image where 0 represents a white pixel and 1 represents a black pixel.

The black pixels are connected (i.e., there is only one black region). Pixels are connected horizontally and vertically.

Given two integers x and y that represents the location of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.

You must write an algorithm with less than O(mn) runtime complexity

 

Example 1:

Input: image = [["0","0","1","0"],["0","1","1","0"],["0","1","0","0"]], x = 0, y = 2
Output: 6

Example 2:

Input: image = [["1"]], x = 0, y = 0
Output: 1

 

Constraints:

  • m == image.length
  • n == image[i].length
  • 1 <= m, n <= 100
  • image[i][j] is either '0' or '1'.
  • 0 <= x < m
  • 0 <= y < n
  • image[x][y] == '1'.
  • The black pixels in the image only form one component.

Solutions

Solution 1

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class Solution:
    def minArea(self, image: List[List[str]], x: int, y: int) -> int:
        m, n = len(image), len(image[0])
        left, right = 0, x
        while left < right:
            mid = (left + right) >> 1
            c = 0
            while c < n and image[mid][c] == '0':
                c += 1
            if c < n:
                right = mid
            else:
                left = mid + 1
        u = left
        left, right = x, m - 1
        while left < right:
            mid = (left + right + 1) >> 1
            c = 0
            while c < n and image[mid][c] == '0':
                c += 1
            if c < n:
                left = mid
            else:
                right = mid - 1
        d = left
        left, right = 0, y
        while left < right:
            mid = (left + right) >> 1
            r = 0
            while r < m and image[r][mid] == '0':
                r += 1
            if r < m:
                right = mid
            else:
                left = mid + 1
        l = left
        left, right = y, n - 1
        while left < right:
            mid = (left + right + 1) >> 1
            r = 0
            while r < m and image[r][mid] == '0':
                r += 1
            if r < m:
                left = mid
            else:
                right = mid - 1
        r = left
        return (d - u + 1) * (r - l + 1)

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class Solution {

    public int minArea(char[][] image, int x, int y) {
        int m = image.length, n = image[0].length;
        int left = 0, right = x;
        while (left < right) {
            int mid = (left + right) >> 1;
            int c = 0;
            while (c < n && image[mid][c] == '0') {
                ++c;
            }
            if (c < n) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        int u = left;
        left = x;
        right = m - 1;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            int c = 0;
            while (c < n && image[mid][c] == '0') {
                ++c;
            }
            if (c < n) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        int d = left;
        left = 0;
        right = y;
        while (left < right) {
            int mid = (left + right) >> 1;
            int r = 0;
            while (r < m && image[r][mid] == '0') {
                ++r;
            }
            if (r