85. Maximal Rectangle
Description
Given a rows x cols
binary matrix
filled with 0
's and 1
's, find the largest rectangle containing only 1
's and return its area.
Example 1:
Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]] Output: 6 Explanation: The maximal rectangle is shown in the above picture.
Example 2:
Input: matrix = [["0"]] Output: 0
Example 3:
Input: matrix = [["1"]] Output: 1
Constraints:
rows == matrix.length
cols == matrix[i].length
1 <= row, cols <= 200
matrix[i][j]
is'0'
or'1'
.
Solution
maximal-rectangle.py
class Solution:
def maximalRectangle(self, matrix: List[List[str]]) -> int:
if not matrix: return 0
rows, cols = len(matrix), len(matrix[0])
left = [0] * cols
right = [cols] * cols
heights = [0] * cols
res = 0
for i in range(rows):
currLeft, currRight = 0, cols
for j in range(cols):
if matrix[i][j] == "1":
heights[j] += 1
else:
heights[j] = 0
for j in range(cols):
if matrix[i][j] == "1":
left[j] = max(left[j], currLeft)
else:
left[j] = 0
currLeft = j + 1
for j in range(cols - 1, -1, -1):
if matrix[i][j] == "1":
right[j] = min(right[j], currRight)
else:
right[j] = cols
currRight = j
for j in range(cols):
res = max(res, (right[j] - left[j]) * heights[j])
return res