Python: solving Water Area problem in most bizaare and inefficient way

result.py

heights = [0, 8, 0, 0, 5, 0, 0, 10, 0, 0, 1, 1, 0, 3]

"""Result
(0 for void, 1 for a pillar, 2 for water)

[[0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 2, 2, 2, 2, 2, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 2, 2, 2, 2, 2, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 2, 2, 2, 2, 2, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 2, 2, 1, 2, 2, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 2, 2, 1, 2, 2, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1],
 [0, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 1],
 [0, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1]]
 
 """

waterarea.py

# why not trying with visual
def waterArea(heights):
	if not heights:
		return 0
  
	from pprint import pprint
  
	Max = max(heights)
	graphic = [[0 for width in range(len(heights))] for height in range(Max)]
	area = 0
  
	for index, height in enumerate(heights):
		i = len(graphic) - 1  # from the bottom
		while height > 0:
			graphic[i][index] = 1  # set a pillar
			i -= 1
			height -= 1  # building higher

	for i, line in enumerate(graphic):
		indices = [i for i, x in enumerate(line) if x == 1]  # get all pillars' indices
		if len(indices) <= 1:  # only 1 pillars around
			continue

		first, second = indices.pop(0), indices.pop(0)
		while True:
			for between in range(first + 1, second):  # change 0 to 2 between 1 (pillars)
				graphic[i][between] = 2
				area += 1  # add water area
				
			if len(indices) == 0:
				break
				
			first, second = second, indices.pop(0)
			
	pprint(graphic)
	return area