

# 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      _a				:




      """simple docstring"""

      def __init__(      self       :							List[Any]	, UpperCAmelCase       :							List[str]	, UpperCAmelCase       :							Optional[int]  ):
            A_	     =							key
            A_	     =							freq







      def __str__(      self       :							Optional[Any]  ):
            return f'''Node(key={self.key}, freq={self.freq})'''




def 			__snake_case	(   __UpperCamelCase  :   List[str]							,__UpperCamelCase  :   Dict							,__UpperCamelCase  :   Dict							,__UpperCamelCase  :   Optional[int]							,__UpperCamelCase  :   List[str]							,__UpperCamelCase  :   Any						):



      """simple docstring"""

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

      A_	     =							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(__UpperCamelCase							,__UpperCamelCase							,__UpperCamelCase							,node - 1							,key[node]							,__UpperCamelCase						)
      print_binary_search_tree(__UpperCamelCase							,__UpperCamelCase							,node + 1							,__UpperCamelCase							,key[node]							,__UpperCamelCase						)

def 			__snake_case	(   __UpperCamelCase  :   Dict						):



      """simple docstring"""

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

      A_	     =							len(__UpperCamelCase						)

      A_	     =							[nodes[i].key for i in range(__UpperCamelCase						)]
      A_	     =							[nodes[i].freq for i in range(__UpperCamelCase						)]

      # 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_	     =							[[freqs[i] if i == j else 0 for j in range(__UpperCamelCase						)] for i in range(__UpperCamelCase						)]
      # sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
      # array
      A_	     =							[[freqs[i] if i == j else 0 for j in range(__UpperCamelCase						)] for i in range(__UpperCamelCase						)]
      # stores tree roots that will be used later for constructing binary search tree
      A_	     =							[[i if i == j else 0 for j in range(__UpperCamelCase						)] for i in range(__UpperCamelCase						)]

      for interval_length in range(2							,n + 1						):
            for i in range(n - interval_length + 1						):
                  A_	     =							i + interval_length - 1

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

                        if dp[i][j] > cost:
                              A_	     =							cost
                              A_	     =							r

      print("Binary search tree nodes:"						)
      for node in nodes:
            print(__UpperCamelCase						)

      print(f'''\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}.'''						)
      print_binary_search_tree(__UpperCamelCase							,__UpperCamelCase							,0							,n - 1							,-1							,__UpperCamelCase						)

def 			__snake_case	(   ):



      """simple docstring"""

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


if __name__ == "__main__":
      main()


import warnings

from ...utils import logging
from .image_processing_mobilenet_va import MobileNetVaImageProcessor


__a						:Optional[int]            = logging.get_logger(__name__)

class      _a				( snake_case_							):




      """simple docstring"""

      def __init__(      self       :							List[Any]	, *UpperCAmelCase       :							Any	, **UpperCAmelCase       :							int  ):
            warnings.warn(
                "The class MobileNetV2FeatureExtractor is deprecated and will be removed in version 5 of Transformers."
                " Please use MobileNetV2ImageProcessor instead."	, UpperCAmelCase	, )
            super().__init__(*UpperCAmelCase	, **UpperCAmelCase  )