

"""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							:





      def __init__(	self     ,    SCREAMING_SNAKE_CASE__     ,    SCREAMING_SNAKE_CASE__		) ->       Dict:
         lowerCamelCase__						       =      key
         lowerCamelCase__						       =      freq





      def __str__(	self		) ->       List[str]:
         return f'Node(key={self.key}, freq={self.freq})'






def 						UpperCAmelCase__ (					A__						,     A__						,     A__						,     A__						,     A__						,     A__ )   ->   Optional[int]:




   """simple docstring"""
   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 						UpperCAmelCase__ (					A__ )   ->   Optional[Any]:




   """simple docstring"""
   nodes.sort(key=lambda A__ : 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 						UpperCAmelCase__ (					)   ->   Optional[int]:




   """simple docstring"""
   # 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()







"""simple docstring"""




SCREAMING_SNAKE_CASE_  :      Optional[int]							=      {0: [1, 2], 1: [0, 3], 2: [0], 3: [1], 4: [5, 6], 5: [4, 6], 6: [4, 5]}

SCREAMING_SNAKE_CASE_  :      str							=      {0: [1, 2, 3], 1: [0, 3], 2: [0], 3: [0, 1], 4: [], 5: []}



def 						UpperCAmelCase__ (					A__						,     A__						,     A__ )   ->   list:




   """simple docstring"""
   lowerCamelCase__						       =      True
   lowerCamelCase__						       =      []

   for neighbour in graph[vert]:
      if not visited[neighbour]:
         connected_verts += dfs(A__						,     A__						,     A__ )

   return [vert, *connected_verts]



def 						UpperCAmelCase__ (					A__ )   ->   list:




   """simple docstring"""
   lowerCamelCase__						       =      len(A__ )
   lowerCamelCase__						       =      graph_size * [False]
   lowerCamelCase__						       =      []

   for i in range(A__ ):
      if not visited[i]:
         lowerCamelCase__						       =      dfs(A__						,     A__						,     A__ )
         components_list.append(A__ )

   return components_list


if __name__ == "__main__":
     import doctest

     doctest.testmod()




