

# 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 			snake_case_     :







			'''simple docstring'''

			def __init__(  self				:       Optional[Any]							,     _UpperCamelCase				:       Tuple							,     _UpperCamelCase				:       Dict							)   ->Union[str, Any]:
									snake_case_     =			key
									snake_case_     =			freq







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






def 						__SCREAMING_SNAKE_CASE      (SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__ ):
						if i > j or i < 0 or j > len(SCREAMING_SNAKE_CASE__ ) - 1:
												return

						snake_case_     =			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(SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	node - 1						,	key[node]						,	SCREAMING_SNAKE_CASE__ )
						print_binary_search_tree(SCREAMING_SNAKE_CASE__						,	SCREAMING_SNAKE_CASE__						,	node + 1						,	SCREAMING_SNAKE_CASE__						,	key[node]						,	SCREAMING_SNAKE_CASE__ )


def 						__SCREAMING_SNAKE_CASE      (SCREAMING_SNAKE_CASE__ ):
						nodes.sort(key=lambda SCREAMING_SNAKE_CASE__ : node.key )

						snake_case_     =			len(SCREAMING_SNAKE_CASE__ )

						snake_case_     =			[nodes[i].key for i in range(SCREAMING_SNAKE_CASE__ )]
						snake_case_     =			[nodes[i].freq for i in range(SCREAMING_SNAKE_CASE__ )]

						# 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.
						snake_case_     =			[[freqs[i] if i == j else 0 for j in range(SCREAMING_SNAKE_CASE__ )] for i in range(SCREAMING_SNAKE_CASE__ )]
						# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
						# array
						snake_case_     =			[[freqs[i] if i == j else 0 for j in range(SCREAMING_SNAKE_CASE__ )] for i in range(SCREAMING_SNAKE_CASE__ )]
						# stores tree roots that will be used later for constructing binary search tree
						snake_case_     =			[[i if i == j else 0 for j in range(SCREAMING_SNAKE_CASE__ )] for i in range(SCREAMING_SNAKE_CASE__ )]

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

																		snake_case_     =			sys.maxsize  # set the value to "infinity"
																		snake_case_     =			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
																								snake_case_     =			dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
																								snake_case_     =			dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
																								snake_case_     =			left + total[i][j] + right

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

						print('''Binary search tree nodes:''' )
						for node in nodes:
												print(SCREAMING_SNAKE_CASE__ )

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


def 						__SCREAMING_SNAKE_CASE      ():
						# A sample binary search tree
						snake_case_     =			[Node(SCREAMING_SNAKE_CASE__						,	randint(1						,	50 ) ) for i in range(10						,	0						,	-1 )]
						find_optimal_binary_search_tree(SCREAMING_SNAKE_CASE__ )


if __name__ == "__main__":
					main()


import warnings

from .generation import GenerationMixin





class 			snake_case_     (	__A      ):







			'''simple docstring'''

			warnings.warn(
			    "Importing `GenerationMixin` from `src/transformers/generation_utils.py` is deprecated and will "
			    "be removed in Transformers v5. Import as `from transformers import GenerationMixin` instead."			, __A			, )