



"""simple docstring"""







import sys
from random import randint






class   UpperCAmelCase__		:



		def __init__(			self	:     Optional[int]   ,				snake_case	:     List[Any]   ,				snake_case	:     Dict					)      ->    Any:






									'''simple docstring'''

									A    =       key
									A    =       freq



		def __str__(			self	:     Union[str, Any]					)      ->    Optional[Any]:






									'''simple docstring'''

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



def 	lowerCAmelCase__      (  lowerCamelCase__					,   lowerCamelCase__					,   lowerCamelCase__					,   lowerCamelCase__					,   lowerCamelCase__					,   lowerCamelCase__      )			->				int:
							if i > j or i < 0 or j > len(UpperCamelCase__      ) - 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(UpperCamelCase__					,   UpperCamelCase__					,   UpperCamelCase__					,   node - 1					,   key[node]					,   UpperCamelCase__      )
							print_binary_search_tree(UpperCamelCase__					,   UpperCamelCase__					,   node + 1					,   UpperCamelCase__					,   key[node]					,   UpperCamelCase__      )

def 	lowerCAmelCase__      (  lowerCamelCase__      )			->				Tuple:
							nodes.sort(key=lambda lowerCamelCase__      : node.key      )

							A    =       len(UpperCamelCase__      )

							A    =       [nodes[i].key for i in range(UpperCamelCase__      )]
							A    =       [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.
							A    =       [[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
							A    =       [[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
							A    =       [[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      ):
																					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(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 	lowerCAmelCase__      (  )			->				int:
							# A sample binary search tree
							A    =       [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				(				UpperCamelCase__				:int   ,				UpperCamelCase__				:list			)		->						list:
				snake_case__      :   List[str]      =    sorted(UpperCamelCase__   ,				key=lambda UpperCamelCase__			: value[2]   ,				reverse=UpperCamelCase__			)

				# Create a list of size equal to the maximum deadline
				# and initialize it with -1
				snake_case__      :   Any      =    max(UpperCamelCase__   ,				key=lambda UpperCamelCase__			: value[1]			)[1]
				snake_case__      :   str      =    [-1] * max_deadline

				# Finding the maximum profit and the count of jobs
				snake_case__      :   List[str]      =    0
				snake_case__      :   Optional[Any]      =    0
				for job in jobs:
								# Find a free time slot for this job
								# (Note that we start from the last possible slot)
								for i in range(job[1] - 1   ,				-1   ,				-1			):
												if time_slots[i] == -1:
																snake_case__      :   Optional[Any]      =    job[0]
																count += 1
																max_profit += job[2]
																break
				return [count, max_profit]


if __name__ == "__main__":
				import doctest

				doctest.testmod()




