






# 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				,    lowerCamelCase__				,    lowerCamelCase__       ):




									"""simple docstring"""

									__UpperCamelCase :   Optional[int]									=key
									__UpperCamelCase :   Dict									=freq







				def __str__(					self       ):




									"""simple docstring"""

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






def 				A	(			a_				,a_				,a_				,a_				,a_				,a_   )	->		Tuple:
					if i > j or i < 0 or j > len(a_   ) - 1:
										return

					__UpperCamelCase :   Optional[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 				A	(			a_   )	->		List[Any]:
					nodes.sort(key=lambda a_   : node.key   )

					__UpperCamelCase :   Optional[Any]									=len(a_   )

					__UpperCamelCase :   Dict									=[nodes[i].key for i in range(a_   )]
					__UpperCamelCase :   str									=[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.
					__UpperCamelCase :   Tuple									=[[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
					__UpperCamelCase :   Optional[int]									=[[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
					__UpperCamelCase :   Tuple									=[[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   ):
															__UpperCamelCase :   List[str]									=i + interval_length - 1

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

																				if dp[i][j] > cost:
																									__UpperCamelCase :   Union[str, Any]									=cost
																									__UpperCamelCase :   int									=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	(			)	->		int:
					# A sample binary search tree
					__UpperCamelCase :   Optional[int]									=[Node(a_				,randint(1				,50   )   ) for i in range(10				,0				,-1   )]
					find_optimal_binary_search_tree(a_   )


if __name__ == "__main__":
							main()









from typing import TYPE_CHECKING

from ...utils import OptionalDependencyNotAvailable, _LazyModule, is_torch_available


A_			:Union[str, Any]	    =		{'''configuration_nat''': ['''NAT_PRETRAINED_CONFIG_ARCHIVE_MAP''', '''NatConfig''']}


try:
							if not is_torch_available():
														raise OptionalDependencyNotAvailable()
except OptionalDependencyNotAvailable:
							pass
else:
							A_			:Optional[Any]	    =		[
							    '''NAT_PRETRAINED_MODEL_ARCHIVE_LIST''',
							    '''NatForImageClassification''',
							    '''NatModel''',
							    '''NatPreTrainedModel''',
							    '''NatBackbone''',
							]

if TYPE_CHECKING:
							from .configuration_nat import NAT_PRETRAINED_CONFIG_ARCHIVE_MAP, NatConfig

							try:
														if not is_torch_available():
																					raise OptionalDependencyNotAvailable()
							except OptionalDependencyNotAvailable:
														pass
							else:
														from .modeling_nat import (
														    NAT_PRETRAINED_MODEL_ARCHIVE_LIST,
														    NatBackbone,
														    NatForImageClassification,
														    NatModel,
														    NatPreTrainedModel,
														)

else:
							import sys

							A_			:Tuple	    =		_LazyModule(__name__, globals()['''__file__'''], _import_structure, module_spec=__spec__)

