




'''simple docstring'''



from transformers import ProcessorMixin





class     lowerCAmelCase   ( __lowerCAmelCase):
       __lowercase  :      Union[str, Any]        =			'''AutoFeatureExtractor'''
       __lowercase  :      List[Any]        =			'''AutoTokenizer'''












'''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  )      ->  Optional[int]:



            '''simple docstring'''




            __snake_case    					=  key
            __snake_case    					=  freq




       def __str__(   self  )      ->  List[Any]:



            '''simple docstring'''




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







def 			_UpperCamelCase (_lowerCamelCase							:     str   ,	_lowerCamelCase							:     List[str]   ,	_lowerCamelCase							:     Union[str, Any]   ,	_lowerCamelCase							:     int   ,	_lowerCamelCase							:     List[Any]   ,	_lowerCamelCase							:     Tuple	)->   Any:




     '''simple docstring'''





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

     __snake_case    					=  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 			_UpperCamelCase (_lowerCamelCase							:     List[Any]	)->   Union[str, Any]:




     '''simple docstring'''





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

     __snake_case    					=  len(_lowerCamelCase	)

     __snake_case    					=  [nodes[i].key for i in range(_lowerCamelCase	)]
     __snake_case    					=  [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.
     __snake_case    					=  [[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
     __snake_case    					=  [[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
     __snake_case    					=  [[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	):
               __snake_case    					=  i + interval_length - 1

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

                    if dp[i][j] > cost:
                         __snake_case    					=  cost
                         __snake_case    					=  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 			_UpperCamelCase ()->   Union[str, Any]:




     '''simple docstring'''





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


if __name__ == "__main__":
     main()






