


# 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[str]							,     _a       :			Optional[Any]							,     _a       :			Tuple					)     ->      Any:
      UpperCAmelCase    :					Dict			=      key
      UpperCAmelCase    :					Union[str, Any]			=      freq

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







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

     UpperCAmelCase    :					List[str]			=      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 	A (		A__			):
     nodes.sort(key=lambda A__			: node.key			)

     UpperCAmelCase    :					str			=      len(A__			)

     UpperCAmelCase    :					Optional[Any]			=      [nodes[i].key for i in range(A__			)]
     UpperCAmelCase    :					List[str]			=      [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    :					str			=      [[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    :					List[str]			=      [[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    :					List[Any]			=      [[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    :					Any			=      i + interval_length - 1

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

                    if dp[i][j] > cost:
                         UpperCAmelCase    :					List[Any]			=      cost
                         UpperCAmelCase    :					int			=      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 	A (		):
     # A sample binary search tree
     UpperCAmelCase    :					str			=      [Node(A__		,		randint(1		,		5_0			)			) for i in range(1_0		,		0		,		-1			)]
     find_optimal_binary_search_tree(A__			)


if __name__ == "__main__":
      main()







import os







def 	A (		):
     UpperCAmelCase    :					Any			=      os.path.dirname(os.path.realpath(A__			)			)
     UpperCAmelCase    :					Optional[int]			=      os.path.join(A__		,		"triangle.txt"			)

     with open(A__			) as f:
          UpperCAmelCase    :					Union[str, Any]			=      f.readlines()

     UpperCAmelCase    :					List[str]			=      [[int(A__			) for y in x.rstrip("\r\n"			).split(" "			)] for x in triangle]

     for i in range(1		,		len(A__			)			):
          for j in range(len(a[i]			)			):
               UpperCAmelCase    :					List[Any]			=      a[i - 1][j] if j != len(a[i - 1]			) else 0
               UpperCAmelCase    :					Optional[int]			=      a[i - 1][j - 1] if j > 0 else 0
               a[i][j] += max(A__		,		A__			)
     return max(a[-1]			)


if __name__ == "__main__":
      print(solution())



