



# 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					,					a_					,					a_						):
											a__			 =						key
											a__			 =						freq







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




def      A_ (		__a       :			List[str]	,						__a       :			Optional[Any]	,						__a       :			Optional[Any]	,						__a       :			List[Any]	,						__a       :			Optional[Any]	,						__a       :			Any						):

				"""simple docstring"""
				if i > j or i < 0 or j > len(__a						) - 1:
								return

				a__			 =						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       :			Optional[Any]						):

				"""simple docstring"""
				nodes.sort(key=lambda __a						: node.key						)

				a__			 =						len(__a						)

				a__			 =						[nodes[i].key for i in range(__a						)]
				a__			 =						[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.
				a__			 =						[[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
				a__			 =						[[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
				a__			 =						[[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						):
												a__			 =						i + interval_length - 1

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

																if dp[i][j] > cost:
																				a__			 =						cost
																				a__			 =						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_ (		):

				"""simple docstring"""
				a__			 =						[Node(__a	,						randint(1	,						50						)						) for i in range(10	,						0	,						-1						)]
				find_optimal_binary_search_tree(__a						)


if __name__ == "__main__":
	main()






def      A_ (		__a       :			int						):

				"""simple docstring"""
				if number != int(__a						):
								raise ValueError("""factorial() only accepts integral values"""						)
				if number < 0:
								raise ValueError("""factorial() not defined for negative values"""						)
				a__			 =						1
				for i in range(1	,						number + 1						):
								value *= i
				return value

def      A_ (		__a       :			int						):

				"""simple docstring"""
				if not isinstance(__a	,						__a						):
								raise ValueError("""factorial() only accepts integral values"""						)
				if n < 0:
								raise ValueError("""factorial() not defined for negative values"""						)
				return 1 if n == 0 or n == 1 else n * factorial(n - 1						)


if __name__ == "__main__":
	import doctest

	doctest.testmod()

	UpperCAmelCase								=   int(input("""Enter a positive integer: """).strip() or 0)
	print(f"""factorial{n} is {factorial(n)}""")

