




# 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 		lowercase_       :




					def __init__(   self:		Optional[int],     _lowercase:		Any,     _lowercase:		Any):


									'''simple docstring'''


									__lowerCAmelCase					=  key
									__lowerCAmelCase					=  freq







					def __str__(   self:		List[str]):


									'''simple docstring'''


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





def      UpperCAmelCase  (						UpperCamelCase__						,			UpperCamelCase__						,			UpperCamelCase__						,			UpperCamelCase__						,			UpperCamelCase__						,			UpperCamelCase__   )     ->     int:







				'''simple docstring'''


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

				__lowerCAmelCase					=  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      UpperCAmelCase  (						UpperCamelCase__   )     ->     str:







				'''simple docstring'''


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

				__lowerCAmelCase					=  len(UpperCamelCase__   )

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

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

																if dp[i][j] > cost:
																				__lowerCAmelCase					=  cost
																				__lowerCAmelCase					=  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      UpperCAmelCase  (						)     ->     int:







				'''simple docstring'''


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


if __name__ == "__main__":
	main()










'''simple docstring'''

def     __UpperCamelCase					(				_lowercase = 1000       )			->      int:
	_lowercase       :							Dict  	=		0
	_lowercase       :							int  	=		0
	while 1:
		num += 3
		if num >= n:
			break
		total += num
		num += 2
		if num >= n:
			break
		total += num
		num += 1
		if num >= n:
			break
		total += num
		num += 3
		if num >= n:
			break
		total += num
		num += 1
		if num >= n:
			break
		total += num
		num += 2
		if num >= n:
			break
		total += num
		num += 3
		if num >= n:
			break
		total += num
	return total


if __name__ == "__main__":
		print(F'''{solution() = }''')
