

# 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    A__:
						"""simple docstring"""






						def __init__(  self       ,     _lowercase       ,     _lowercase  )      ->				str:
									a_		:		str				    =				key
									a_		:		Optional[Any]				    =				freq





						def __str__(  self  )      ->				List[Any]:
									return F'''Node(key={self.key}, freq={self.freq})'''







def 			_UpperCAmelCase							(		a__       ,					a__       ,					a__       ,					a__       ,					a__       ,					a__) ->    Optional[int]:

			'''simple docstring'''



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

			a_		:		List[Any]				    =				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 			_UpperCAmelCase							(		a__) ->    Optional[int]:

			'''simple docstring'''



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

			a_		:		List[Any]				    =				len(a__)

			a_		:		Optional[Any]				    =				[nodes[i].key for i in range(a__)]
			a_		:		Any				    =				[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_		:		str				    =				[[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_		:		List[str]				    =				[[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_		:		Any				    =				[[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_		:		List[Any]				    =				i + interval_length - 1

									a_		:		Optional[int]				    =				sys.maxsize  # set the value to "infinity"
									a_		:		Union[str, Any]				    =				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_		:		List[str]				    =				dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
												a_		:		Optional[Any]				    =				dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
												a_		:		Optional[Any]				    =				left + total[i][j] + right

												if dp[i][j] > cost:
															a_		:		Optional[Any]				    =				cost
															a_		:		List[str]				    =				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 			_UpperCAmelCase							(		) ->    Dict:

			'''simple docstring'''



			a_		:		Optional[int]				    =				[Node(a__       ,					randint(1       ,					5_0)) for i in range(1_0       ,					0       ,					-1)]
			find_optimal_binary_search_tree(a__)


if __name__ == "__main__":
	main()





import sys

from transformers import AutoTokenizer


__snake_case							:     int 						=	sys.argv[1]
__snake_case							:     Dict 						=	sys.argv[2]
__snake_case							:     Dict 						=	int(sys.argv[3])

__snake_case							:     Union[str, Any] 						=	0

__snake_case							:     Any 						=	AutoTokenizer.from_pretrained(model_name_or_path)
max_len -= tokenizer.num_special_tokens_to_add()

with open(dataset, """rt""") as f_p:
	for line in f_p:
		__snake_case							:     str 						=	line.rstrip()

		if not line:
			print(line)
			__snake_case							:     Optional[int] 						=	0
			continue

		__snake_case							:     Dict 						=	line.split()[0]

		__snake_case							:     str 						=	len(tokenizer.tokenize(token))

		# Token contains strange control characters like \x96 or \x95
		# Just filter out the complete line
		if current_subwords_len == 0:
			continue

		if (subword_len_counter + current_subwords_len) > max_len:
			print("""""")
			print(line)
			__snake_case							:     Optional[int] 						=	current_subwords_len
			continue

		subword_len_counter += current_subwords_len

		print(line)


