




"""simple docstring"""




# 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        UpperCAmelCase_  :
		def __init__(		self							,		UpperCamelCase_							,		UpperCamelCase_   )		->   Dict:
					__lowercase	:       Dict				=   key
					__lowercase	:       List[str]				=   freq
		def __str__(		self   )		->   List[Any]:
					return F"""Node(key={self.key}, freq={self.freq})"""







def       __UpperCAmelCase      (				__UpperCamelCase					,							__UpperCamelCase					,							__UpperCamelCase					,							__UpperCamelCase					,							__UpperCamelCase					,							__UpperCamelCase					):
			if i > j or i < 0 or j > len(__UpperCamelCase					) - 1:
						return

			__lowercase	:       int				=   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					):
			nodes.sort(key=lambda __UpperCamelCase					: node.key					)

			__lowercase	:       Any				=   len(__UpperCamelCase					)

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

									__lowercase	:       Union[str, Any]				=   sys.maxsize  # set the value to "infinity"
									__lowercase	:       Dict				=   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	:       Dict				=   dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
												__lowercase	:       Optional[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	:       int				=   cost
															__lowercase	:       Union[str, Any]				=   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      (				):
			# A sample binary search tree
			__lowercase	:       Any				=   [Node(__UpperCamelCase					,							randint(1					,							50					)					) for i in range(10					,							0					,							-1					)]
			find_optimal_binary_search_tree(__UpperCamelCase					)


if __name__ == "__main__":
			main()









"""simple docstring"""




from __future__ import annotations







def       __UpperCAmelCase      (				__UpperCamelCase					):
			__lowercase	:       List[Any]				=   [0] * len(__UpperCamelCase					)
			for i in range(1					,							len(__UpperCamelCase					)					):
						__lowercase	:       Optional[Any]				=   duration_times[i - 1] + waiting_times[i - 1]
			return waiting_times







def       __UpperCAmelCase      (				__UpperCamelCase					,							__UpperCamelCase					):
			return [
			    duration_time + waiting_times[i]
			    for i, duration_time in enumerate(__UpperCamelCase					)
			]







def       __UpperCAmelCase      (				__UpperCamelCase					):
			return sum(__UpperCamelCase					) / len(__UpperCamelCase					)







def       __UpperCAmelCase      (				__UpperCamelCase					):
			return sum(__UpperCamelCase					) / len(__UpperCamelCase					)


if __name__ == "__main__":
			# process id's
			a_            =   [1, 2, 3]

			# ensure that we actually have processes
			if len(processes) == 0:
						print('Zero amount of processes')
						raise SystemExit(0)

			# duration time of all processes
			a_            =   [1_9, 8, 9]

			# ensure we can match each id to a duration time
			if len(duration_times) != len(processes):
						print('Unable to match all id\'s with their duration time')
						raise SystemExit(0)

			# get the waiting times and the turnaround times
			a_            =   calculate_waiting_times(duration_times)
			a_            =   calculate_turnaround_times(duration_times, waiting_times)

			# get the average times
			a_            =   calculate_average_waiting_time(waiting_times)
			a_            =   calculate_average_turnaround_time(turnaround_times)

			# print all the results
			print('Process ID\tDuration Time\tWaiting Time\tTurnaround Time')
			for i, process in enumerate(processes):
						print(
						    F"{process}\t\t{duration_times[i]}\t\t{waiting_times[i]}\t\t"
						    F"{turnaround_times[i]}"
						)
			print(F"Average waiting time = {average_waiting_time}")
			print(F"Average turn around time = {average_turnaround_time}")



