



# This Python program implements an optimal binary search tree (abbreviated BST)
# building dynamic programming algorithm that delivers O(n^2) performance.
#
# The goal of the optimal BST problem is to build a low-cost BST for a
# given set of nodes, each with its own key and frequency. The frequency
# of the node is defined as how many time the node is being searched.
# The search cost of binary search tree is given by this formula:
#
# cost(1, n) = sum{i = 1 to n}((depth(node_i) + 1) * node_i_freq)
#
# where n is number of nodes in the BST. The characteristic of low-cost
# BSTs is having a faster overall search time than other implementations.
# The reason for their fast search time is that the nodes with high
# frequencies will be placed near the root of the tree while the nodes
# with low frequencies will be placed near the leaves of the tree thus
# reducing search time in the most frequent instances.
import sys
from random import randint



class       _lowercase				:





	"""simple docstring"""




	def __init__(self						,			lowerCamelCase_						,			lowerCamelCase_						):
						"""simple docstring"""

						a   	=  key
						a   	=  freq




	def __str__(self						):
						"""simple docstring"""

						return F'''Node(key={self.key}, freq={self.freq})'''






def        a( A			:       Tuple					,	A			:       int					,	A			:       Tuple					,	A			:       Optional[int]					,	A			:       Union[str, Any]					,	A			:       int			)						->			Union[str, Any]:
					"""simple docstring"""





					if i > j or i < 0 or j > len(A			) - 1:
										return

					a   	=  root[i][j]
					if parent == -1:  # root does not have a parent
										print(f'''{key[node]} is the root of the binary search tree.'''			)
					elif is_left:
										print(f'''{key[node]} is the left child of key {parent}.'''			)
					else:
										print(f'''{key[node]} is the right child of key {parent}.'''			)

					print_binary_search_tree(A					,	A					,	A					,	node - 1					,	key[node]					,	A			)
					print_binary_search_tree(A					,	A					,	node + 1					,	A					,	key[node]					,	A			)






def        a( A			:       List[Any]			)						->			Optional[int]:
					"""simple docstring"""





					nodes.sort(key=lambda A			: node.key			)

					a   	=  len(A			)

					a   	=  [nodes[i].key for i in range(A			)]
					a   	=  [nodes[i].freq for i in range(A			)]

					# This 2D array stores the overall tree cost (which's as minimized as possible);
					# for a single key, cost is equal to frequency of the key.
					a   	=  [[freqs[i] if i == j else 0 for j in range(A			)] for i in range(A			)]
					# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
					# array
					a   	=  [[freqs[i] if i == j else 0 for j in range(A			)] for i in range(A			)]
					# stores tree roots that will be used later for constructing binary search tree
					a   	=  [[i if i == j else 0 for j in range(A			)] for i in range(A			)]

					for interval_length in range(2					,	n + 1			):
										for i in range(n - interval_length + 1			):
															a   	=  i + interval_length - 1

															a   	=  sys.maxsize  # set the value to "infinity"
															a   	=  total[i][j - 1] + freqs[j]

															# Apply Knuth's optimization
															# Loop without optimization: for r in range(i, j + 1):
															for r in range(root[i][j - 1]					,	root[i + 1][j] + 1			):  # r is a temporal root
																				a   	=  dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
																				a   	=  dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
																				a   	=  left + total[i][j] + right

																				if dp[i][j] > cost:
																									a   	=  cost
																									a   	=  r

					print("Binary search tree nodes:"			)
					for node in nodes:
										print(A			)

					print(f'''\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}.'''			)
					print_binary_search_tree(A					,	A					,	0					,	n - 1					,	-1					,	A			)






def        a( )						->			Optional[int]:
					"""simple docstring"""





					a   	=  [Node(A					,	randint(1					,	50			)			) for i in range(10					,	0					,	-1			)]
					find_optimal_binary_search_tree(A			)


if __name__ == "__main__":
				main()





import platform
from argparse import ArgumentParser

import huggingface_hub
import pandas
import pyarrow

from datasets import __version__ as version
from datasets.commands import BaseDatasetsCLICommand






def        a( A			:       Dict			)						->			Optional[int]:
					"""simple docstring"""





					return EnvironmentCommand()







class       _lowercase				(  lowerCAmelCase      ):





	"""simple docstring"""




	@staticmethod
	def 	UpperCamelCase_			(lowerCamelCase_						):
						"""simple docstring"""

						a   	=  parser.add_parser("env"						,			help="Print relevant system environment info."						)
						download_parser.set_defaults(func=lowerCamelCase_						)




	def 	UpperCamelCase_			(self						):
						"""simple docstring"""

						a   	=  {
						    "`datasets` version": version,
						    "Platform": platform.platform(),
						    "Python version": platform.python_version(),
						    "Huggingface_hub version": huggingface_hub.__version__,
						    "PyArrow version": pyarrow.__version__,
						    "Pandas version": pandas.__version__,
						}

						print("\nCopy-and-paste the text below in your GitHub issue.\n"						)
						print(self.format_dict(lowerCamelCase_						)						)

						return info




	@staticmethod
	def 	UpperCamelCase_			(lowerCamelCase_						):
						"""simple docstring"""

						return "\n".join([F'''- {prop}: {val}''' for prop, val in d.items()]						) + "\n"
