
'''simple docstring'''
# 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        SCREAMING_SNAKE_CASE__	:







				def __init__(    self		,							A_		,							A_			)->						Union[str, Any]:







											'''simple docstring'''






											UpperCamelCase					       =			key
											UpperCamelCase					       =			freq







				def __str__(    self			)->						str:







											'''simple docstring'''






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







def 						A_(			A		:						int						,			A		:						Optional[int]						,			A		:						Optional[int]						,			A		:						str						,			A		:						Union[str, Any]						,			A		:						Optional[int]):
							if i > j or i < 0 or j > len(A) - 1:
														return

							UpperCamelCase					       =			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					       =			len(A)

							UpperCamelCase					       =			[nodes[i].key for i in range(A)]
							UpperCamelCase					       =			[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					       =			[[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					       =			[[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					       =			[[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					       =			i + interval_length - 1

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

																												if dp[i][j] > cost:
																																			UpperCamelCase					       =			cost
																																			UpperCamelCase					       =			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_(			):
							# A sample binary search tree
							UpperCamelCase					       =			[Node(A						,			randint(1						,			50)) for i in range(10						,			0						,			-1)]
							find_optimal_binary_search_tree(A)


if __name__ == "__main__":
						main()






"""simple docstring"""


from .hash_table import HashTable







class 		lowerCamelCase__					(					__magic_name__				):





			'''simple docstring'''



			def __init__(							self     ,     *__UpperCAmelCase     ,     **__UpperCAmelCase  ) ->     List[str]:
				super().__init__(*__UpperCAmelCase     ,     **__UpperCAmelCase  )



			def 					_lowerCAmelCase		(							self     ,     __UpperCAmelCase     ,     __UpperCAmelCase=None  ) ->     Union[str, Any]:
				_lowerCAmelCase								=1
				_lowerCAmelCase								=self.hash_function(key + i * i  )

				while self.values[new_key] is not None and self.values[new_key] != key:
					i += 1
					_lowerCAmelCase								=(
					    self.hash_function(key + i * i  )
					    if not self.balanced_factor() >= self.lim_charge
					    else None
					)

					if new_key is None:
						break

				return new_key



