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Solutions for "Building Visibility" #1 solution in Clear category for Building Visibility by zeixes
# https://py.checkio.org/mission/buildings-visibility/publications/yoichi/python-3/first/share/0f1fd5cf14674becd006491919cfda30/
from collections import namedtuple
def intersect(build, builds) -> bool:
# check height and weight for every building and subtract from current
for build_1 in builds:
if build_1.height >= build.height:
if (build.x[1] <= build_1.x[0] and build.x[0] < build_1.x[0]) or (
build.x[0] >= build_1.x[1] and build.x[1] > build_1.x[1]):
continue
elif build_1.x[1] > build.x[0] >= build_1.x[0]:
build.x[0] = build_1.x[1]
elif build_1.x[0] < build.x[1] and build.x[1] >= build_1.x[1]:
build.x[1] = build_1.x[0]
return build.x[0] < build.x[1]
def checkio(buildings):
buildings.sort(key=lambda item: item[1])
build_lst = []
for params in buildings:
building = namedtuple('Build', 'x,y,height')
building.x, building.y, building.height = [params[0], params[2]], [params[1], params[3]], params[4]
if not build_lst or intersect(building, build_lst):
build_lst.append(building)
for b in build_lst:
print(b.x)
print()
return len(build_lst)
if __name__ == '__main__':
assert checkio([
[1, 1, 4, 5, 3.5],
[2, 6, 4, 8, 5],
[5, 1, 9, 3, 6],
[5, 5, 6, 6, 8],
[7, 4, 10, 6, 4],
[5, 7, 10, 8, 3]
]) == 5, "First"
assert checkio([
[1, 1, 11, 2, 2],
[2, 3, 10, 4, 1],
[3, 5, 9, 6, 3],
[4, 7, 8, 8, 2]
]) == 2, "Second"
assert checkio([
[1, 1, 3, 3, 6],
[5, 1, 7, 3, 6],
[9, 1, 11, 3, 6],
[1, 4, 3, 6, 6],
[5, 4, 7, 6, 6],
[9, 4, 11, 6, 6],
[1, 7, 11, 8, 3.25]
]) == 4, "Third"
assert checkio([
[0, 0, 1, 1, 10]
]) == 1, "Alone"
assert checkio([
[2, 2, 3, 3, 4],
[2, 5, 3, 6, 4]
]) == 1, "Shadow"
May 23, 2024
