



def 						_lowerCAmelCase    (					__lowerCamelCase :  str  ):



 """simple docstring"""





 __SCREAMING_SNAKE_CASE							:     List[str]							=						str(__lowerCamelCase  ).strip()
 if not bin_string:
  raise ValueError("Empty string was passed to the function"  )
 __SCREAMING_SNAKE_CASE							:     Union[str, Any]							=						bin_string[0] == "-"
 if is_negative:
  __SCREAMING_SNAKE_CASE							:     str							=						bin_string[1:]
 if not all(char in "01" for char in bin_string  ):
  raise ValueError("Non-binary value was passed to the function"  )
 __SCREAMING_SNAKE_CASE							:     Dict							=						0
 for char in bin_string:
  __SCREAMING_SNAKE_CASE							:     List[str]							=						2 * decimal_number + int(__lowerCamelCase  )
 return -decimal_number if is_negative else decimal_number


if __name__ == "__main__":
      from doctest import testmod

      testmod()







'''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 	__SCREAMING_SNAKE_CASE		:







  '''simple docstring'''






  def __init__(					self ,						snake_case_ ,						snake_case_		):




       '''simple docstring'''

       UpperCAmelCase_				:  Union[str, Any]  = key
       UpperCAmelCase_				:  Tuple  = freq






  def __str__(					self		):




       '''simple docstring'''

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







def    _lowerCamelCase       ( lowerCamelCase_   :				str      ,	lowerCamelCase_   :				List[str]      ,	lowerCamelCase_   :				Dict      ,	lowerCamelCase_   :				List[Any]      ,	lowerCamelCase_   :				int      ,	lowerCamelCase_   :				Optional[int]  ):



     """simple docstring"""





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

     UpperCAmelCase_				:  Optional[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(lowerCamelCase_      ,	lowerCamelCase_      ,	lowerCamelCase_      ,	node - 1      ,	key[node]      ,	lowerCamelCase_  )
     print_binary_search_tree(lowerCamelCase_      ,	lowerCamelCase_      ,	node + 1      ,	lowerCamelCase_      ,	key[node]      ,	lowerCamelCase_  )


def    _lowerCamelCase       ( lowerCamelCase_   :				Optional[int]  ):



     """simple docstring"""





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

     UpperCAmelCase_				:  int  = len(lowerCamelCase_  )

     UpperCAmelCase_				:  Dict  = [nodes[i].key for i in range(lowerCamelCase_  )]
     UpperCAmelCase_				:  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.
     UpperCAmelCase_				:  Optional[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
     UpperCAmelCase_				:  Any  = [[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
     UpperCAmelCase_				:  Optional[int]  = [[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  ):
               UpperCAmelCase_				:  Optional[Any]  = i + interval_length - 1

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

                    if dp[i][j] > cost:
                         UpperCAmelCase_				:  Dict  = cost
                         UpperCAmelCase_				:  Tuple  = 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"""





     UpperCAmelCase_				:  Any  = [Node(lowerCamelCase_      ,	randint(1      ,	50  )  ) for i in range(10      ,	0      ,	-1  )]
     find_optimal_binary_search_tree(lowerCamelCase_  )


if __name__ == "__main__":
 main()






