




"""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     lowercase :







				def __init__(							self				:   List[Any] ,       _lowerCamelCase				:   List[str] ,       _lowerCamelCase				:   List[str]    ):



					"""simple docstring"""

					A_					:			Optional[int]						=						key
					A_					:			Optional[Any]						=						freq







				def __str__(							self				:   List[str]    ):



					"""simple docstring"""

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

def 						lowercase_ (						_UpperCAmelCase					, _UpperCAmelCase					, _UpperCAmelCase					, _UpperCAmelCase					, _UpperCAmelCase					, _UpperCAmelCase		):




	"""simple docstring"""






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

	A_					:			Dict						=						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 						lowercase_ (						_UpperCAmelCase		):




	"""simple docstring"""






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

	A_					:			Any						=						len(_UpperCAmelCase		)

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

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

				if dp[i][j] > cost:
					A_					:			List[Any]						=						cost
					A_					:			List[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 						lowercase_ (						):




	"""simple docstring"""






	A_					:			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"""

def 						lowercase_ (						_UpperCAmelCase		):




	"""simple docstring"""






	A_					:			Optional[int]						=						9

	while num > 0:
		if num % 10 != digit:
			return False
		num //= 100
		digit -= 1

	return True

def 						lowercase_ (						):




	"""simple docstring"""






	A_					:			Dict						=						1_3890_2663

	while not is_square_form(num * num		):
		if num % 10 == 3:
			num -= 6  # (3 - 6) % 10 = 7
		else:
			num -= 4  # (7 - 4) % 10 = 3

	return num * 10


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





