

# 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    a__  :
			"""simple docstring"""







			def __init__(      self , lowercase , lowercase    )					->  Optional[Any]:
							'''simple docstring'''






							A__            =     key
							A__            =     freq


			def __str__(      self    )					->  Union[str, Any]:
							'''simple docstring'''






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



def    lowerCAmelCase__							(		SCREAMING_SNAKE_CASE_:							List[str]		,    SCREAMING_SNAKE_CASE_:							Optional[int]		,    SCREAMING_SNAKE_CASE_:							List[str]		,    SCREAMING_SNAKE_CASE_:							Tuple		,    SCREAMING_SNAKE_CASE_:							Optional[int]		,    SCREAMING_SNAKE_CASE_:							List[Any] )     -> Dict:





				if i > j or i < 0 or j > len(SCREAMING_SNAKE_CASE_ ) - 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(SCREAMING_SNAKE_CASE_		,    SCREAMING_SNAKE_CASE_		,    SCREAMING_SNAKE_CASE_		,    node - 1		,    key[node]		,    SCREAMING_SNAKE_CASE_ )
				print_binary_search_tree(SCREAMING_SNAKE_CASE_		,    SCREAMING_SNAKE_CASE_		,    node + 1		,    SCREAMING_SNAKE_CASE_		,    key[node]		,    SCREAMING_SNAKE_CASE_ )



def    lowerCAmelCase__							(		SCREAMING_SNAKE_CASE_:							Any )     -> Optional[Any]:
				nodes.sort(key=lambda SCREAMING_SNAKE_CASE_ : node.key )

				A__            =     len(SCREAMING_SNAKE_CASE_ )

				A__            =     [nodes[i].key for i in range(SCREAMING_SNAKE_CASE_ )]
				A__            =     [nodes[i].freq for i in range(SCREAMING_SNAKE_CASE_ )]

				# 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(SCREAMING_SNAKE_CASE_ )] for i in range(SCREAMING_SNAKE_CASE_ )]
				# 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(SCREAMING_SNAKE_CASE_ )] for i in range(SCREAMING_SNAKE_CASE_ )]
				# stores tree roots that will be used later for constructing binary search tree
				A__            =     [[i if i == j else 0 for j in range(SCREAMING_SNAKE_CASE_ )] for i in range(SCREAMING_SNAKE_CASE_ )]

				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(SCREAMING_SNAKE_CASE_ )

				print(F'\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}.' )
				print_binary_search_tree(SCREAMING_SNAKE_CASE_		,    SCREAMING_SNAKE_CASE_		,    0		,    n - 1		,    -1		,    SCREAMING_SNAKE_CASE_ )



def    lowerCAmelCase__							(		)     -> Dict:
				A__            =     [Node(SCREAMING_SNAKE_CASE_		,    randint(1		,    5_0 ) ) for i in range(1_0		,    0		,    -1 )]
				find_optimal_binary_search_tree(SCREAMING_SNAKE_CASE_ )


if __name__ == "__main__":
	main()




import math



def    lowerCAmelCase__							(		SCREAMING_SNAKE_CASE_:							int )     -> bool:




				'''simple docstring'''





				if 1 < number < 4:
								# 2 and 3 are primes
								return True
				elif number < 2 or number % 2 == 0 or number % 3 == 0:
								# Negatives, 0, 1, all even numbers, all multiples of 3 are not primes
								return False

				# All primes number are in format of 6k +/- 1
				for i in range(5		,    int(math.sqrt(SCREAMING_SNAKE_CASE_ ) + 1 )		,    6 ):
								if number % i == 0 or number % (i + 2) == 0:
												return False
				return True



def    lowerCAmelCase__							(		SCREAMING_SNAKE_CASE_:							int = 2_0_0_0_0_0_0 )     -> int:




				'''simple docstring'''





				return sum(num for num in range(3		,    SCREAMING_SNAKE_CASE_		,    2 ) if is_prime(SCREAMING_SNAKE_CASE_ ) ) + 2 if n > 2 else 0


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

