
from __future__ import annotations






def        lowerCamelCase__    (    __lowerCAmelCase  :  list[int | float]    ):




   """simple docstring"""


   if len(__lowerCAmelCase    ) == 0:
      raise ValueError("find_min() arg is an empty sequence"    )
   lowerCAmelCase_      =     nums[0]
   for num in nums:
      lowerCAmelCase_      =     min(__lowerCAmelCase   ,							__lowerCAmelCase    )
   return min_num


if __name__ == "__main__":
   import doctest

   doctest.testmod(verbose=True)





# 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      _lowerCAmelCase :




				def __init__( self    ,    _UpperCamelCase    ,    _UpperCamelCase					)    ->				str:
							lowerCAmelCase_      =     key
							lowerCAmelCase_      =     freq




				def __str__( self					)    ->				Any:
							return f"""Node(key={self.key}, freq={self.freq})"""






def        lowerCamelCase__    (    __lowerCAmelCase  :  Optional[int]   ,							__lowerCAmelCase  :  Dict   ,							__lowerCAmelCase  :  Optional[Any]   ,							__lowerCAmelCase  :  Optional[int]   ,							__lowerCAmelCase  :  int   ,							__lowerCAmelCase  :  List[Any]    ):




			"""simple docstring"""


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

			lowerCAmelCase_      =     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(__lowerCAmelCase   ,							__lowerCAmelCase   ,							__lowerCAmelCase   ,							node - 1   ,							key[node]   ,							__lowerCAmelCase    )
			print_binary_search_tree(__lowerCAmelCase   ,							__lowerCAmelCase   ,							node + 1   ,							__lowerCAmelCase   ,							key[node]   ,							__lowerCAmelCase    )






def        lowerCamelCase__    (    __lowerCAmelCase  :  Union[str, Any]    ):




			"""simple docstring"""


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

			lowerCAmelCase_      =     len(__lowerCAmelCase    )

			lowerCAmelCase_      =     [nodes[i].key for i in range(__lowerCAmelCase    )]
			lowerCAmelCase_      =     [nodes[i].freq for i in range(__lowerCAmelCase    )]

			# 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.
			lowerCAmelCase_      =     [[freqs[i] if i == j else 0 for j in range(__lowerCAmelCase    )] for i in range(__lowerCAmelCase    )]
			# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
			# array
			lowerCAmelCase_      =     [[freqs[i] if i == j else 0 for j in range(__lowerCAmelCase    )] for i in range(__lowerCAmelCase    )]
			# stores tree roots that will be used later for constructing binary search tree
			lowerCAmelCase_      =     [[i if i == j else 0 for j in range(__lowerCAmelCase    )] for i in range(__lowerCAmelCase    )]

			for interval_length in range(2   ,							n + 1    ):
						for i in range(n - interval_length + 1    ):
									lowerCAmelCase_      =     i + interval_length - 1

									lowerCAmelCase_      =     sys.maxsize  # set the value to "infinity"
									lowerCAmelCase_      =     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
												lowerCAmelCase_      =     dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
												lowerCAmelCase_      =     dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
												lowerCAmelCase_      =     left + total[i][j] + right

												if dp[i][j] > cost:
															lowerCAmelCase_      =     cost
															lowerCAmelCase_      =     r

			print("Binary search tree nodes:"    )
			for node in nodes:
						print(__lowerCAmelCase    )

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






def        lowerCamelCase__    (    ):




			"""simple docstring"""


			lowerCAmelCase_      =     [Node(__lowerCAmelCase   ,							randint(1   ,							50    )    ) for i in range(10   ,							0   ,							-1    )]
			find_optimal_binary_search_tree(__lowerCAmelCase    )


if __name__ == "__main__":
			main()



