Find largest rectangle containing only zeros in an N×N binary matrix

Question

Given an NxN binary matrix (containing only 0\'s or 1\'s), how can we go about finding largest rectangle containing all 0\'s?

Example:

      I
    0          


        

Answer 1

Here is a version of jfs' solution, which also delivers the position of the largest rectangle:

from collections import namedtuple
from operator import mul

Info = namedtuple('Info', 'start height')

def max_rect(mat, value=0):
    """returns (height, width, left_column, bottom_row) of the largest rectangle 
    containing all `value`'s.

    Example:
    [[0, 0, 0, 0, 0, 0, 0, 0, 3, 2],
     [0, 4, 0, 2, 4, 0, 0, 1, 0, 0],
     [1, 0, 1, 0, 0, 0, 3, 0, 0, 4],
     [0, 0, 0, 0, 4, 2, 0, 0, 0, 0],
     [0, 0, 0, 2, 0, 0, 0, 0, 0, 0],
     [4, 3, 0, 0, 1, 2, 0, 0, 0, 0],
     [3, 0, 0, 0, 2, 0, 0, 0, 0, 4],
     [0, 0, 0, 1, 0, 3, 2, 4, 3, 2],
     [0, 3, 0, 0, 0, 2, 0, 1, 0, 0]]
     gives: (3, 4, 6, 5)
    """
    it = iter(mat)
    hist = [(el==value) for el in next(it, [])]
    max_rect = max_rectangle_size(hist) + (0,)
    for irow,row in enumerate(it):
        hist = [(1+h) if el == value else 0 for h, el in zip(hist, row)]
        max_rect = max(max_rect, max_rectangle_size(hist) + (irow+1,), key=area)
        # irow+1, because we already used one row for initializing max_rect
    return max_rect

def max_rectangle_size(histogram):
    stack = []
    top = lambda: stack[-1]
    max_size = (0, 0, 0) # height, width and start position 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), 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), start), key=area)

    return max_size

def area(size):
    return size[0] * size[1]