# 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 						snake_case_ :






   def __init__(   self       :  Tuple			,		lowercase_       :  Optional[int]			,		lowercase_       :  Optional[int]						)      ->		int:
       lowercase__				:     Optional[int]					      =       key
       lowercase__				:     Any					      =       freq







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



def     lowercase_       (    _lowerCamelCase							: List[str]	,			_lowerCamelCase							: Any	,			_lowerCamelCase							: Union[str, Any]	,			_lowerCamelCase							: Union[str, Any]	,			_lowerCamelCase							: int	,			_lowerCamelCase							: Tuple):
    if i > j or i < 0 or j > len(snake_case__) - 1:
        return

    lowercase__				:     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(snake_case__	,			snake_case__	,			snake_case__	,			node - 1	,			key[node]	,			snake_case__)
    print_binary_search_tree(snake_case__	,			snake_case__	,			node + 1	,			snake_case__	,			key[node]	,			snake_case__)

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

    lowercase__				:     Optional[Any]					      =       len(snake_case__)

    lowercase__				:     int					      =       [nodes[i].key for i in range(snake_case__)]
    lowercase__				:     Union[str, Any]					      =       [nodes[i].freq for i in range(snake_case__)]

    # 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.
    lowercase__				:     Tuple					      =       [[freqs[i] if i == j else 0 for j in range(snake_case__)] for i in range(snake_case__)]
    # sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
    # array
    lowercase__				:     List[str]					      =       [[freqs[i] if i == j else 0 for j in range(snake_case__)] for i in range(snake_case__)]
    # stores tree roots that will be used later for constructing binary search tree
    lowercase__				:     Optional[Any]					      =       [[i if i == j else 0 for j in range(snake_case__)] for i in range(snake_case__)]

    for interval_length in range(2	,			n + 1):
        for i in range(n - interval_length + 1):
            lowercase__				:     Union[str, Any]					      =       i + interval_length - 1

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

                if dp[i][j] > cost:
                    lowercase__				:     Tuple					      =       cost
                    lowercase__				:     Optional[int]					      =       r

    print("Binary search tree nodes:")
    for node in nodes:
        print(snake_case__)

    print(f'''\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}.''')
    print_binary_search_tree(snake_case__	,			snake_case__	,			0	,			n - 1	,			-1	,			snake_case__)

def     lowercase_       (    ):
    # A sample binary search tree
    lowercase__				:     Dict					      =       [Node(snake_case__	,			randint(1	,			50)) for i in range(10	,			0	,			-1)]
    find_optimal_binary_search_tree(snake_case__)


if __name__ == "__main__":
      main()




from typing import Callable

def     lowercase_       (    _lowerCamelCase							: Callable):

    def wrapper(_lowerCamelCase							: str):
        lowercase__				:     Tuple					      =       function_with_docstring.__doc__
        return target_function

    return wrapper



