





"""simple docstring"""






import heapq

def     __A      (			a_						:dict)		->		str:
				__a :						int			       =  []

				# 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(_UpperCamelCase   ,       [-1 * len(_UpperCamelCase), (key, value)])

				# chosen_vertices = set of chosen vertices
				__a :						Optional[Any]			       =  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
								__a :						Optional[Any]			       =  heapq.heappop(_UpperCamelCase)[1][0]
								chosen_vertices.add(_UpperCamelCase)

								# 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]:
																__a :						List[str]			       =  elem[1][1].index(_UpperCamelCase)
																del elem[1][1][index]
																elem[0] += 1
        # re-order the queue
								heapq.heapify(_UpperCamelCase)
				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)}')






"""simple docstring"""






from __future__ import annotations

import typing
from collections import Counter

def     __A      (			a_						:int)		->		typing.Counter[int]:
				__a :						typing.Counter[int]			       =  Counter()
				for base in range(1   ,       max_perimeter + 1):
								for perpendicular in range(a_   ,       max_perimeter + 1):
												__a :						Any			       =  (base * base + perpendicular * perpendicular) ** 0.5
												if hypotenuse == int(a_):
																__a :						List[Any]			       =  int(base + perpendicular + hypotenuse)
																if perimeter > max_perimeter:
																				continue
																triplets[perimeter] += 1
				return triplets

def     __A      (			a_						:int = 10_00)		->		int:
				__a :						Dict			       =  pythagorean_triple(a_)
				return triplets.most_common(1)[0][0]


if __name__ == "__main__":
		print(F'Perimeter {solution()} has maximum solutions')