'''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 ,						snake_case_ ,						snake_case_		):




       '''simple docstring'''

       UpperCAmelCase_				:  Union[str, Any]  = key
       UpperCAmelCase_				:  Tuple  = freq






  def __str__(					self		):




       '''simple docstring'''

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







def    _lowerCamelCase       ( lowerCamelCase_   :				str      ,	lowerCamelCase_   :				List[str]      ,	lowerCamelCase_   :				Dict      ,	lowerCamelCase_   :				List[Any]      ,	lowerCamelCase_   :				int      ,	lowerCamelCase_   :				Optional[int]  ):



     """simple docstring"""





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

     UpperCAmelCase_				:  Optional[int]  = 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_   :				Optional[int]  ):



     """simple docstring"""





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

     UpperCAmelCase_				:  int  = len(lowerCamelCase_  )

     UpperCAmelCase_				:  Dict  = [nodes[i].key for i in range(lowerCamelCase_  )]
     UpperCAmelCase_				:  Any  = [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.
     UpperCAmelCase_				:  Optional[Any]  = [[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
     UpperCAmelCase_				:  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
     UpperCAmelCase_				:  Optional[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  ):
               UpperCAmelCase_				:  Optional[Any]  = i + interval_length - 1

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

                    if dp[i][j] > cost:
                         UpperCAmelCase_				:  Dict  = cost
                         UpperCAmelCase_				:  Tuple  = 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       ( ):



     """simple docstring"""





     UpperCAmelCase_				:  Any  = [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





class        __magic_name__		:







			def __init__(      self							:    List[str]							,     lowercase_							:    int       ):
										lowercase_   :     List[str]									=			vertices
										lowercase_   :     Optional[Any]									=			[[0] * vertices for _ in range(lowercase_       )]







			def       SCREAMING_SNAKE_CASE_	(      self							:    int							,     lowercase_							:    list[int]       ):
										print("""Vertex \t Distance from Source"""       )
										for vertex in range(self.vertices       ):
																	print(lowercase_							,     """\t\t"""							,     distances_from_source[vertex]       )







			def       SCREAMING_SNAKE_CASE_	(      self							:    str							,     lowercase_							:    list[int]							,     lowercase_							:    list[bool]       ):
										lowercase_   :     Dict									=			1E7
										lowercase_   :     int									=			0

										# Search not nearest vertex not in the shortest path tree
										for vertex in range(self.vertices       ):
																	if distances_from_source[vertex] < minimum and visited[vertex] is False:
																								lowercase_   :     Tuple									=			distances_from_source[vertex]
																								lowercase_   :     str									=			vertex
										return min_index







			def       SCREAMING_SNAKE_CASE_	(      self							:    Union[str, Any]							,     lowercase_							:    int       ):
										lowercase_   :     Optional[Any]									=			[int(1E7       )] * self.vertices  # distances from the source
										lowercase_   :     List[Any]									=			0
										lowercase_   :     Any									=			[False] * self.vertices

										for _ in range(self.vertices       ):
																	lowercase_   :     Tuple									=			self.minimum_distance(lowercase_							,     lowercase_       )
																	lowercase_   :     Union[str, Any]									=			True

																	# Update dist value of the adjacent vertices
																	# of the picked vertex only if the current
																	# distance is greater than new distance and
																	# the vertex in not in the shortest path tree
																	for v in range(self.vertices       ):
																								if (
																								    self.graph[u][v] > 0
																								    and visited[v] is False
																								    and distances[v] > distances[u] + self.graph[u][v]
																								):
																															lowercase_   :     Any									=			distances[u] + self.graph[u][v]

										self.print_solution(lowercase_       )


if __name__ == "__main__":
						_lowercase		:					Any													=		Graph(9)
						_lowercase		:					Any													=		[
						    [0, 4, 0, 0, 0, 0, 0, 8, 0],
						    [4, 0, 8, 0, 0, 0, 0, 11, 0],
						    [0, 8, 0, 7, 0, 4, 0, 0, 2],
						    [0, 0, 7, 0, 9, 14, 0, 0, 0],
						    [0, 0, 0, 9, 0, 10, 0, 0, 0],
						    [0, 0, 4, 14, 10, 0, 2, 0, 0],
						    [0, 0, 0, 0, 0, 2, 0, 1, 6],
						    [8, 11, 0, 0, 0, 0, 1, 0, 7],
						    [0, 0, 2, 0, 0, 0, 6, 7, 0],
						]
						graph.dijkstra(0)





