
"""simple docstring"""





_a								=						{}



def  lowerCamelCase__						(     __snake_case,	__snake_case,	__snake_case      )				->						int:


					"""simple docstring"""





					if late == 3 or absent == 2:
										return 0

					# if we have no days left, and have not failed any other rules,
					# we have a prize string
					if days == 0:
										return 1

					# No easy solution, so now we need to do the recursive calculation

					# First, check if the combination is already in the cache, and
					# if yes, return the stored value from there since we already
					# know the number of possible prize strings from this point on
					_UpperCamelCase									=    (days, absent, late)
					if key in cache:
										return cache[key]

					# now we calculate the three possible ways that can unfold from
					# this point on, depending on our attendance today

					# 1) if we are late (but not absent), the "absent" counter stays as
					# it is, but the "late" counter increases by one
					_UpperCamelCase									=    _calculate(days - 1,	__snake_case,	late + 1      )

					# 2) if we are absent, the "absent" counter increases by 1, and the
					# "late" counter resets to 0
					_UpperCamelCase									=    _calculate(days - 1,	absent + 1,	0      )

					# 3) if we are on time, this resets the "late" counter and keeps the
					# absent counter
					_UpperCamelCase									=    _calculate(days - 1,	__snake_case,	0      )

					_UpperCamelCase									=    state_late + state_absent + state_ontime

					_UpperCamelCase									=    prizestrings
					return prizestrings



def  lowerCamelCase__						(     __snake_case = 30      )				->						int:


					"""simple docstring"""





					return _calculate(__snake_case,	absent=0,	late=0      )


if __name__ == "__main__":
				print(solution())








"""simple docstring"""





import heapq



def  lowerCamelCase__						(     __snake_case      )				->						set[int]:


					"""simple docstring"""





					_UpperCamelCase									=    []

					# for each node and his adjacency list add them and the rank of the node to queue
					# using heapq module the queue will be filled like a Priority Queue
					# heapq works with a min priority queue, so I used -1*len(v) to build it
					for key, value in graph.items():
										# O(log(n))
										heapq.heappush(__snake_case,	[-1 * len(__snake_case      ), (key, value)]      )

					# chosen_vertices = set of chosen vertices
					_UpperCamelCase									=    set()

					# while queue isn't empty and there are still edges
					#   (queue[0][0] is the rank of the node with max rank)
					while queue and queue[0][0] != 0:
										# extract vertex with max rank from queue and add it to chosen_vertices
										_UpperCamelCase									=    heapq.heappop(__snake_case      )[1][0]
										chosen_vertices.add(__snake_case      )

										# Remove all arcs adjacent to argmax
										for elem in queue:
															# if v haven't adjacent node, skip
															if elem[0] == 0:
																				continue
															# if argmax is reachable from elem
															# remove argmax from elem's adjacent list and update his rank
															if argmax in elem[1][1]:
																				_UpperCamelCase									=    elem[1][1].index(__snake_case      )
																				del elem[1][1][index]
																				elem[0] += 1
        # re-order the queue
										heapq.heapify(__snake_case      )
					return chosen_vertices


if __name__ == "__main__":
				import doctest

				doctest.testmod()

				_a								=						{0: [1, 3], 1: [0, 3], 2: [0, 3, 4], 3: [0, 1, 2], 4: [2, 3]}
				print(F"""Minimum vertex cover:\n{greedy_min_vertex_cover(graph)}""")






