Here’s a solution based on the “Largest Rectangle in a Histogram” problem suggested by @j_random_hacker in the comments:
[Algorithm] works by iterating through
rows from top to bottom, for each row
solving this problem, where the
“bars” in the “histogram” consist of
all unbroken upward trails of zeros
that start at the current row (a
column has height 0 if it has a 1 in
the current row).
The input matrix mat may be an arbitrary iterable e.g., a file or a network stream. Only one row is required to be available at a time.
#!/usr/bin/env python
from collections import namedtuple
from operator import mul
Info = namedtuple('Info', 'start height')
def max_size(mat, value=0):
"""Find height, width of the largest rectangle containing all `value`'s."""
it = iter(mat)
hist = [(el==value) for el in next(it, [])]
max_size = max_rectangle_size(hist)
for row in it:
hist = [(1+h) if el == value else 0 for h, el in zip(hist, row)]
max_size = max(max_size, max_rectangle_size(hist), key=area)
return max_size
def max_rectangle_size(histogram):
"""Find height, width of the largest rectangle that fits entirely under
the histogram.
"""
stack = []
top = lambda: stack[-1]
max_size = (0, 0) # height, width of the largest rectangle
pos = 0 # current position in the histogram
for pos, height in enumerate(histogram):
start = pos # position where rectangle starts
while True:
if not stack or height > top().height:
stack.append(Info(start, height)) # push
elif stack and height < top().height:
max_size = max(max_size, (top().height, (pos - top().start)),
key=area)
start, _ = stack.pop()
continue
break # height == top().height goes here
pos += 1
for start, height in stack:
max_size = max(max_size, (height, (pos - start)), key=area)
return max_size
def area(size):
return reduce(mul, size)
The solution is O(N), where N is the number of elements in a matrix. It requires O(ncols) additional memory, where ncols is the number of columns in a matrix.
Latest version with tests is at https://gist.github.com/776423