"""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 ,				_SCREAMING_SNAKE_CASE ,				_SCREAMING_SNAKE_CASE	):


      '''simple docstring'''

      __lowerCamelCase    		=					key
      __lowerCamelCase    		=					freq







   def __str__(	self	):


      '''simple docstring'''

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



def       A__    (	lowercase_     :	Tuple      , lowercase_     :	Dict      , lowercase_     :	int      , lowercase_     :	Dict      , lowercase_     :	int      , lowercase_     :	Union[str, Any]						)  ->							Optional[Any]:
   if i > j or i < 0 or j > len(A__						) - 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(A__      , A__      , A__      , node - 1      , key[node]      , A__						)
   print_binary_search_tree(A__      , A__      , node + 1      , A__      , key[node]      , A__						)



def       A__    (	lowercase_     :	Optional[Any]						)  ->							Union[str, Any]:
   nodes.sort(key=lambda lowercase_						: node.key						)

   __lowerCamelCase    		=					len(A__						)

   __lowerCamelCase    		=					[nodes[i].key for i in range(A__						)]
   __lowerCamelCase    		=					[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.
   __lowerCamelCase    		=					[[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
   __lowerCamelCase    		=					[[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
   __lowerCamelCase    		=					[[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						):
         __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(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__    (	)  ->							Optional[Any]:
   # A sample binary search tree
   __lowerCamelCase    		=					[Node(A__      , randint(1      , 50						)						) for i in range(10      , 0      , -1						)]
   find_optimal_binary_search_tree(A__						)


if __name__ == "__main__":
     main()







from ...utils import is_transformers_available
from .pipeline_latent_diffusion_superresolution import LDMSuperResolutionPipeline


if is_transformers_available():
				from .pipeline_latent_diffusion import LDMBertModel, LDMTextToImagePipeline


