




"""simple docstring"""

import math



def 				a (						__snake_case     :				Dict  ):



				'''simple docstring'''





				UpperCAmelCase_					:Optional[Any]      			=      math.log(2,    10  ) * number
				UpperCAmelCase_					:int      			=      round((log_number - int(_A  )),    15  )

				return difference



def 				a (						__snake_case     :				Tuple = 678910  ):



				'''simple docstring'''





				UpperCAmelCase_					:int      			=      90
				UpperCAmelCase_					:str      			=      0

				UpperCAmelCase_					:Any      			=      math.log(1.23,    10  )
				UpperCAmelCase_					:int      			=      math.log(1.24,    10  )
				UpperCAmelCase_					:List[Any]      			=      0

				while position < number:
								UpperCAmelCase_					:Dict      			=      log_difference(_A  )

								if difference >= upper_limit:
												power_iterator += 93

								elif difference < lower_limit:
												power_iterator += 196

								else:
												UpperCAmelCase_					:Optional[Any]      			=      power_iterator
												power_iterator += 196
												position += 1

				return previous_power


if __name__ == "__main__":
						import doctest

						doctest.testmod()

						print(f'''{solution() = }''')








"""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  _snake_case				:

				'''simple docstring'''




				def __init__(							self      :  str					,							snake_case      :  Optional[Any]					,							snake_case      :  str     ):
								UpperCAmelCase_					:Union[str, Any]      			=      key
								UpperCAmelCase_					:str      			=      freq





				def __str__(							self      :  Optional[Any]     ):
								return f'Node(key={self.key}, freq={self.freq})'



def 				a (						__snake_case     :				int,    __snake_case     :				Optional[int],    __snake_case     :				Tuple,    __snake_case     :				Any,    __snake_case     :				Dict,    __snake_case     :				List[str]  ):



				'''simple docstring'''





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

				UpperCAmelCase_					:List[str]      			=      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(__snake_case,    __snake_case,    __snake_case,    node - 1,    key[node],    __snake_case  )
				print_binary_search_tree(__snake_case,    __snake_case,    node + 1,    __snake_case,    key[node],    __snake_case  )



def 				a (						__snake_case     :				Optional[Any]  ):



				'''simple docstring'''





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

				UpperCAmelCase_					:List[str]      			=      len(__snake_case  )

				UpperCAmelCase_					:Any      			=      [nodes[i].key for i in range(__snake_case  )]
				UpperCAmelCase_					:int      			=      [nodes[i].freq for i in range(__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.
				UpperCAmelCase_					:Dict      			=      [[freqs[i] if i == j else 0 for j in range(__snake_case  )] for i in range(__snake_case  )]
				# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
				# array
				UpperCAmelCase_					:Optional[Any]      			=      [[freqs[i] if i == j else 0 for j in range(__snake_case  )] for i in range(__snake_case  )]
				# stores tree roots that will be used later for constructing binary search tree
				UpperCAmelCase_					:int      			=      [[i if i == j else 0 for j in range(__snake_case  )] for i in range(__snake_case  )]

				for interval_length in range(2,    n + 1  ):
								for i in range(n - interval_length + 1  ):
												UpperCAmelCase_					:Union[str, Any]      			=      i + interval_length - 1

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

																if dp[i][j] > cost:
																				UpperCAmelCase_					:Optional[Any]      			=      cost
																				UpperCAmelCase_					:Optional[int]      			=      r

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

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



def 				a (						):



				'''simple docstring'''





				UpperCAmelCase_					:Union[str, Any]      			=      [Node(__snake_case,    randint(1,    50  )  ) for i in range(10,    0,    -1  )]
				find_optimal_binary_search_tree(__snake_case  )


if __name__ == "__main__":
						main()


