Is Smallest Rectangle Enclosing Black Pixels easy or hard?

LeetCode classifies this problem as Hard. The challenge is recognizing that rows and columns containing black pixels form contiguous regions, which allows binary search on boundaries instead of scanning the entire matrix.

Smallest Rectangle Enclosing Black Pixels Python/Java solution

Implementations typically scan rows or columns to verify whether a black pixel exists and then use binary search to narrow the boundary indices. The same algorithm works in Python, Java, C++, and Go because it relies only on matrix traversal and index-based binary search.

What is the best approach for Smallest Rectangle Enclosing Black Pixels?

The most efficient approach uses binary search to locate the top, bottom, left, and right boundaries that contain black pixels. Because rows and columns with black pixels form contiguous ranges, binary search can shrink the search space quickly. This produces a time complexity of O(m log n + n log m) with constant extra space.

Is Smallest Rectangle Enclosing Black Pixels asked at Google/Amazon/Meta?

This problem is commonly associated with Google interview preparation and appears frequently in discussions of advanced matrix search techniques. It tests binary search over a grid and understanding of monotonic boundaries in matrices.

What data structure is used in Smallest Rectangle Enclosing Black Pixels?

The problem mainly uses a 2D array (matrix). Depending on the solution, you may also apply BFS or DFS with a queue or recursion stack, or binary search over row and column indices to find rectangle boundaries.

What is the time complexity of Smallest Rectangle Enclosing Black Pixels?

The optimal binary search solution runs in O(m log n + n log m) time because each boundary search performs log steps and scans a row or column. Brute force and BFS/DFS approaches take O(m*n) time since they may inspect every cell in the matrix.

How to solve Smallest Rectangle Enclosing Black Pixels in O(m log n + n log m)?

Perform binary search to locate the first and last rows and columns that contain at least one black pixel. For each candidate row or column during search, scan across it to check whether a '1' exists. Finding four boundaries this way yields the smallest rectangle containing all black pixels.

302. Smallest Rectangle Enclosing Black Pixels - Practice Online

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.