
# 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							:   Dict			,  snake_case							:   str			,  snake_case							:   Tuple						)   -> List[str]:
							_lowercase					:  str    			=      key
							_lowercase					:  Any    			=      freq






	def __str__(self							:   int						)   -> Dict:
							return F"""Node(key={self.key}, freq={self.freq})"""



def 		UpperCamelCase    (					_UpperCAmelCase     :      List[Any]   , _UpperCAmelCase     :      Union[str, Any]   , _UpperCAmelCase     :      Any   , _UpperCAmelCase     :      str   , _UpperCAmelCase     :      Dict   , _UpperCAmelCase     :      Optional[int]     )      -> Optional[int]:






						'''simple docstring'''

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

						_lowercase					:  Optional[Any]    			=      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     :      Dict     )      -> Optional[Any]:






						'''simple docstring'''

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

						_lowercase					:  Any    			=      len(_UpperCAmelCase     )

						_lowercase					:  Optional[int]    			=      [nodes[i].key for i in range(_UpperCAmelCase     )]
						_lowercase					:  str    			=      [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					:  Optional[int]    			=      [[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					:  Optional[int]    			=      [[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					:  List[str]    			=      [[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					:  int    			=      i + interval_length - 1

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

																								if dp[i][j] > cost:
																														_lowercase					:  List[Any]    			=      cost
																														_lowercase					:  Tuple    			=      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    (					)      -> Optional[Any]:






						'''simple docstring'''

						_lowercase					:  Dict    			=      [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



def 		UpperCamelCase    (					_UpperCAmelCase     :      list[int]   , _UpperCAmelCase     :      int     )      -> int:






						'''simple docstring'''

						_lowercase			,						_lowercase					:  Any    			=      0, len(_UpperCAmelCase     ) - 1
						while start_ind <= end_ind:
												if array[start_ind] == search_item:
																		return start_ind
												elif array[end_ind] == search_item:
																		return end_ind
												else:
																		start_ind += 1
																		end_ind -= 1
    # returns -1 if search_item is not found in array
						return -1


if __name__ == "__main__":
							print(double_linear_search(list(range(100)), 40))




