




'''simple docstring'''

from itertools import permutations




def    snake_case__     (  a = 5	)			->       int:





			'''simple docstring'''






			if gon_side < 3 or gon_side > 5:
						raise ValueError("""gon_side must be in the range [3, 5]"""	)

			# Since it's 16, we know 10 is on the outer ring
			# Put the big numbers at the end so that they are never the first number
			snake_case__          =						list(range(gon_side + 1       ,      0       ,      -1	)	)
			snake_case__          =						list(range(gon_side + 2       ,      gon_side * 2 + 1	)	)

			for perm in permutations(small_numbers + big_numbers	):
						snake_case__          =						generate_gon_ring(a       ,      list(a	)	)
						if is_magic_gon(a	):
									return int("""""".join(str(a	) for n in numbers	)	)

			snake_case__          =						F"""Magic {gon_side}-gon ring is impossible"""
			raise ValueError(a	)




def    snake_case__     (  a       ,      a	)			->       list[int]:





			'''simple docstring'''






			snake_case__          =						[0] * (gon_side * 3)
			snake_case__          =						perm[0:3]
			perm.append(perm[1]	)

			snake_case__          =						1 if gon_side < 5 else 2

			for i in range(1       ,      len(a	) // 3 + magic_number	):
						snake_case__          =						perm[2 * i + 1]
						snake_case__          =						result[3 * i - 1]
						snake_case__          =						perm[2 * i + 2]

			return result




def    snake_case__     (  a	)			->       bool:





			'''simple docstring'''






			if len(a	) % 3 != 0:
						raise ValueError("""a gon ring should have a length that is a multiple of 3"""	)

			if min(numbers[::3]	) != numbers[0]:
						return False

			snake_case__          =						sum(numbers[:3]	)

			return all(sum(numbers[i : i + 3]	) == total for i in range(3       ,      len(a	)       ,      3	)	)


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


# 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  UpperCamelCase_      :


  """simple docstring"""



  def __init__(						self     :					Any  ,	A_     :					Union[str, Any]  ,	A_     :					Optional[Any]						)       ->			Union[str, Any]:
      SCREAMING_SNAKE_CASE__         =    key
      SCREAMING_SNAKE_CASE__         =    freq







  def __str__(						self     :					Union[str, Any]						)       ->			List[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

    SCREAMING_SNAKE_CASE__         =    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						)

    SCREAMING_SNAKE_CASE__         =    len(__A						)

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

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

                if dp[i][j] > cost:
                    SCREAMING_SNAKE_CASE__         =    cost
                    SCREAMING_SNAKE_CASE__         =    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
    SCREAMING_SNAKE_CASE__         =    [Node(__A			,							randint(1			,							50						)						) for i in range(10			,							0			,							-1						)]
    find_optimal_binary_search_tree(__A						)


if __name__ == "__main__":
  main()


