



# 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     :Tuple      ,  lowercase__     :str      ,  lowercase__     :str    ):
										lowercase						   =       key
										lowercase						   =       freq



					def __str__(		self     :Any    ):
										return F"""Node(key={self.key}, freq={self.freq})"""





def 		__snake_case						(    _UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase				):
					"""simple docstring"""






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

					lowercase						   =       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(_UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase       ,			node - 1       ,			key[node]       ,			_UpperCAmelCase				)
					print_binary_search_tree(_UpperCAmelCase       ,			_UpperCAmelCase       ,			node + 1       ,			_UpperCAmelCase       ,			key[node]       ,			_UpperCAmelCase				)





def 		__snake_case						(    _UpperCAmelCase				):
					"""simple docstring"""






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

					lowercase						   =       len(_UpperCAmelCase				)

					lowercase						   =       [nodes[i].key for i in range(_UpperCAmelCase				)]
					lowercase						   =       [nodes[i].freq for i in range(_UpperCAmelCase				)]

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

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

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

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

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

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





def 		__snake_case						(    ):
					"""simple docstring"""






					lowercase						   =       [Node(_UpperCAmelCase       ,			randint(1       ,			50				)				) for i in range(10       ,			0       ,			-1				)]
					find_optimal_binary_search_tree(_UpperCAmelCase				)


if __name__ == "__main__":
		main()











from __future__ import annotations

# Define the Gravitational Constant G and the function
__magic_name__       =     6.6743E-11  # unit of G : m^3 * kg^-1 * s^-2





def 		__snake_case						(    _UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase       ,			_UpperCAmelCase				):
					"""simple docstring"""






					lowercase						   =       mass_a * mass_a

					if (force, mass_a, mass_a, distance).count(0				) != 1:
										raise ValueError('One and only one argument must be 0'				)
					if force < 0:
										raise ValueError('Gravitational force can not be negative'				)
					if distance < 0:
										raise ValueError('Distance can not be negative'				)
					if mass_a < 0 or mass_a < 0:
										raise ValueError('Mass can not be negative'				)
					if force == 0:
										lowercase						   =       GRAVITATIONAL_CONSTANT * product_of_mass / (distance**2)
										return {"force": force}
					elif mass_a == 0:
										lowercase						   =       (force) * (distance**2) / (GRAVITATIONAL_CONSTANT * mass_a)
										return {"mass_1": mass_a}
					elif mass_a == 0:
										lowercase						   =       (force) * (distance**2) / (GRAVITATIONAL_CONSTANT * mass_a)
										return {"mass_2": mass_a}
					elif distance == 0:
										lowercase						   =       (GRAVITATIONAL_CONSTANT * product_of_mass / (force)) ** 0.5
										return {"distance": distance}
					raise ValueError('One and only one argument must be 0'				)


# Run doctest
if __name__ == "__main__":
		import doctest

		doctest.testmod()






