Leetcode #302: Smallest Rectangle Enclosing Black Pixels
In this guide, we solve Leetcode #302 Smallest Rectangle Enclosing Black Pixels in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Depth-First Search, Breadth-First Search, Array, Binary Search, Matrix
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: image = [["0","0","1","0"],["0","1","1","0"],["0","1","0","0"]], x = 0, y = 2
Output: 6
Python Solution
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)
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.