code: 
  
  
  
  
  """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 							lowerCamelCase		:
  
  
  
  
  
  
  
     def __init__(							self    ,      SCREAMING_SNAKE_CASE_    ,      SCREAMING_SNAKE_CASE_ ):
         UpperCamelCase				:   Tuple          =							key
         UpperCamelCase				:   Dict          =							freq
  
  
  
  
  
  
  
     def __str__(							self ):
         return f'Node(key={self.key}, freq={self.freq})'
  
  
  
  
  
  
  
  def 					A_			(		snake_case_     :       Optional[int]						,snake_case_     :       Optional[int]						,snake_case_     :       str						,snake_case_     :       List[str]						,snake_case_     :       str						,snake_case_     :       List[str]    ):
  
  
  
  
  
  
  
      '''simple docstring'''
  
  
  
  
  
  
  
      if i > j or i < 0 or j > len(_A    ) - 1:
          return
  
      UpperCamelCase				:   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(_A						,_A						,_A						,node - 1						,key[node]						,_A    )
      print_binary_search_tree(_A						,_A						,node + 1						,_A						,key[node]						,_A    )
  def 					A_			(		snake_case_     :       Optional[Any]    ):
  
  
  
  
  
  
  
      '''simple docstring'''
      nodes.sort(key=lambda snake_case_    : node.key    )
  
      UpperCamelCase				:   List[Any]          =							len(_A    )
  
      UpperCamelCase				:   str          =							[nodes[i].key for i in range(_A    )]
      UpperCamelCase				:   Dict          =							[nodes[i].freq for i in range(_A    )]
  
      # 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				:   List[str]          =							[[freqs[i] if i == j else 0 for j in range(_A    )] for i in range(_A    )]
      # sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
      # array
      UpperCamelCase				:   Optional[Any]          =							[[freqs[i] if i == j else 0 for j in range(_A    )] for i in range(_A    )]
      # stores tree roots that will be used later for constructing binary search tree
      UpperCamelCase				:   Union[str, Any]          =							[[i if i == j else 0 for j in range(_A    )] for i in range(_A    )]
  
      for interval_length in range(2						,n + 1    ):
          for i in range(n - interval_length + 1    ):
              UpperCamelCase				:   Optional[Any]          =							i + interval_length - 1
  
              UpperCamelCase				:   Any          =							sys.maxsize  # set the value to "infinity"
              UpperCamelCase				:   List[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
                  UpperCamelCase				:   Optional[int]          =							dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
                  UpperCamelCase				:   List[str]          =							dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
                  UpperCamelCase				:   int          =							left + total[i][j] + right
  
                  if dp[i][j] > cost:
                      UpperCamelCase				:   List[Any]          =							cost
                      UpperCamelCase				:   Any          =							r
  
      print("""Binary search tree nodes:"""    )
      for node in nodes:
          print(_A    )
  
      print(f'\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}.'    )
      print_binary_search_tree(_A						,_A						,0						,n - 1						,-1						,_A    )
  def 					A_			(		):
  
  
  
  
  
  
  
      '''simple docstring'''
      UpperCamelCase				:   Dict          =							[Node(_A						,randint(1						,5_0    )    ) for i in range(1_0						,0						,-1    )]
      find_optimal_binary_search_tree(_A    )
  
  
  if __name__ == "__main__":
         main()
  
  
  
  
  
style_context: 
  
  
  
  
  """simple docstring"""
  
  
  
  
  import numpy as np
  import qiskit
  from qiskit import Aer, ClassicalRegister, QuantumCircuit, QuantumRegister, execute
  def 					A_			(		snake_case_     :       float = np.pi / 2						,snake_case_     :       float = np.pi / 2						,snake_case_     :       float = np.pi / 2    ):
  
  
  
  
  
  
  
      '''simple docstring'''
      UpperCamelCase				:   Union[str, Any]          =							QuantumRegister(3						,"""qr"""    )  # Define the number of quantum bits
      UpperCamelCase				:   Any          =							ClassicalRegister(1						,"""cr"""    )  # Define the number of classical bits
  
      UpperCamelCase				:   Any          =							QuantumCircuit(snake_case_						,snake_case_    )  # Define the quantum circuit.
  
      # Build the circuit
      quantum_circuit.u(snake_case_						,snake_case_						,snake_case_						,0    )  # Quantum State to teleport
      quantum_circuit.h(1    )  # add hadamard gate
      quantum_circuit.cx(
          1						,2    )  # add control gate with qubit 1 as control and 2 as target.
      quantum_circuit.cx(0						,1    )
      quantum_circuit.h(0    )
      quantum_circuit.cz(0						,2    )  # add control z gate.
      quantum_circuit.cx(1						,2    )
  
      quantum_circuit.measure([2]						,[0]    )  # measure the qubit.
  
      # Simulate the circuit using qasm simulator
      UpperCamelCase				:   Tuple          =							Aer.get_backend("""aer_simulator"""    )
      UpperCamelCase				:   Any          =							execute(snake_case_						,snake_case_						,shots=1_0_0_0    )
  
      return job.result().get_counts(snake_case_    )
  
  
  if __name__ == "__main__":
    print(
        '''Total count for teleported state is: '''
        F'''{quantum_teleportation(np.pi/2, np.pi/2, np.pi/2)}'''
    )
  
  
  
  
  