





# 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 					__magic_name__  :





     """simple docstring"""



     def __init__(							self       :Dict			,      snake_case       :Tuple			,      snake_case       :Union[str, Any]  ):





        '''simple docstring'''




        A_					:				List[Any]          =			key
        A_					:				Any          =			freq





     def __str__(							self       :str  ):





        '''simple docstring'''




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




def      __snake_case						(					_lowerCAmelCase : Union[str, Any]				,				_lowerCAmelCase : List[str]				,				_lowerCAmelCase : Any				,				_lowerCAmelCase : Any				,				_lowerCAmelCase : Tuple				,				_lowerCAmelCase : Optional[Any]	) ->						Optional[int]:
   if i > j or i < 0 or j > len(_lowerCAmelCase	) - 1:
      return

   A_					:				Union[str, Any]          =			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      __snake_case						(					_lowerCAmelCase : int	) ->						List[Any]:
   nodes.sort(key=lambda _lowerCAmelCase	: node.key	)

   A_					:				Optional[Any]          =			len(_lowerCAmelCase	)

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

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

            if dp[i][j] > cost:
               A_					:				List[str]          =			cost
               A_					:				Optional[int]          =			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      __snake_case						(					) ->						Union[str, Any]:
   # A sample binary search tree
   A_					:				Optional[int]          =			[Node(_lowerCAmelCase				,				randint(1				,				50	)	) for i in range(10				,				0				,				-1	)]
   find_optimal_binary_search_tree(_lowerCAmelCase	)


if __name__ == "__main__":
 main()








from __future__ import annotations

def      __snake_case						(					_lowerCAmelCase : int	) ->						list[int]:
   assert (
       isinstance(_lowerCAmelCase				,				_lowerCAmelCase	) and mask > 0
   ), f"mask needs to be positive integer, your input {mask}"
   A_					:				Optional[Any]          =			[]
   A_					:				Optional[int]          =			mask

   while submask:
      all_submasks.append(_lowerCAmelCase	)
      A_					:				List[str]          =			(submask - 1) & mask

   return all_submasks


if __name__ == "__main__":
 import doctest

 doctest.testmod()
