






"""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 		a_      :







      '''simple docstring'''

      def __init__(self,     lowerCamelCase_,     lowerCamelCase_					):

       '''simple docstring'''



       lowerCamelCase__     :			List[Any]     				=						key
       lowerCamelCase__     :			str     				=						freq





      def __str__(self					):

       '''simple docstring'''



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






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

 lowerCamelCase__     :			List[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(_lowerCamelCase			,   _lowerCamelCase			,   _lowerCamelCase			,   node - 1			,   key[node]			,   _lowerCamelCase				)
 print_binary_search_tree(_lowerCamelCase			,   _lowerCamelCase			,   node + 1			,   _lowerCamelCase			,   key[node]			,   _lowerCamelCase				)






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

 lowerCamelCase__     :			Optional[Any]     				=						len(_lowerCamelCase				)

 lowerCamelCase__     :			List[Any]     				=						[nodes[i].key for i in range(_lowerCamelCase				)]
 lowerCamelCase__     :			int     				=						[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.
 lowerCamelCase__     :			Dict     				=						[[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
 lowerCamelCase__     :			Union[str, 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
 lowerCamelCase__     :			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				):
   lowerCamelCase__     :			List[Any]     				=						i + interval_length - 1

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

    if dp[i][j] > cost:
     lowerCamelCase__     :			Any     				=						cost
     lowerCamelCase__     :			str     				=						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_  (						):
 # A sample binary search tree
 lowerCamelCase__     :			Dict     				=						[Node(_lowerCamelCase			,   randint(1			,   50				)				) for i in range(10			,   0			,   -1				)]
 find_optimal_binary_search_tree(_lowerCamelCase				)


if __name__ == "__main__":
      main()





'''simple docstring'''







from __future__ import annotations






def       lowerCAmelCase__       (				lowerCamelCase     :			list   ):
       _A     :	Union[str, Any]      =					sorted(lowerCamelCase   )
       _A     :	Tuple      =					len(lowerCamelCase   )
       _A     :	int      =					length >> 1
       return (
           (sorted_list[mid_index] + sorted_list[mid_index - 1]) / 2
           if length % 2 == 0
           else sorted_list[mid_index]
       )






def       lowerCAmelCase__       (				):
       import doctest

       doctest.testmod()


if __name__ == "__main__":
   main()






