
"""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				,lowerCamelCase_				,lowerCamelCase_		):



          '''simple docstring'''

          _UpperCAmelCase 						=       key
          _UpperCAmelCase 						=       freq







       def __str__(       self		):



          '''simple docstring'''

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



def       __UpperCAmelCase		(		a_      :		Union[str, Any]     ,	a_      :		Tuple     ,	a_      :		List[Any]     ,	a_      :		Optional[Any]     ,	a_      :		Union[str, Any]     ,	a_      :		str					)  ->							Any:
   if i > j or i < 0 or j > len(a_					) - 1:
      return

   _UpperCAmelCase 						=       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_      :		Union[str, Any]					)  ->							int:
   nodes.sort(key=lambda a_					: node.key					)

   _UpperCAmelCase 						=       len(a_					)

   _UpperCAmelCase 						=       [nodes[i].key for i in range(a_					)]
   _UpperCAmelCase 						=       [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.
   _UpperCAmelCase 						=       [[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
   _UpperCAmelCase 						=       [[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
   _UpperCAmelCase 						=       [[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					):
         _UpperCAmelCase 						=       i + interval_length - 1

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

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


if __name__ == "__main__":
  main()



"""simple docstring"""







def       __UpperCAmelCase		(		a_      :		list[int]					)  ->							int:

   # validate:
   if not isinstance(a_     ,	a_					) or not all(
       isinstance(a_     ,	a_					) and item >= 0 for item in citations					):
      raise ValueError("The citations should be a list of non negative integers."					)

   citations.sort()
   _UpperCAmelCase 						=       len(a_					)

   for i in range(a_					):
      if citations[len_citations - 1 - i] <= i:
         return i

   return len_citations


if __name__ == "__main__":
  import doctest

  doctest.testmod()

