"""simple docstring"""
def 		__a  (      _SCREAMING_SNAKE_CASE						,	_SCREAMING_SNAKE_CASE  )      ->List[Any]:
  a__:	Tuple     				=      len(_SCREAMING_SNAKE_CASE  )
  # Create a stack and push index of fist element to it
  a__:	List[str]     				=      []
  st.append(0  )

  # Span value of first element is always 1
  a__:	List[Any]     				=      1

  # Calculate span values for rest of the elements
  for i in range(1						,	_SCREAMING_SNAKE_CASE  ):
    # Pop elements from stack while stack is not
    # empty and top of stack is smaller than price[i]
    while len(_SCREAMING_SNAKE_CASE  ) > 0 and price[st[0]] <= price[i]:
      st.pop()

    # If stack becomes empty, then price[i] is greater
    # than all elements on left of it, i.e. price[0],
    # price[1], ..price[i-1]. Else the price[i]  is
    # greater than elements after top of stack
    a__:	Optional[int]     				=      i + 1 if len(_SCREAMING_SNAKE_CASE  ) <= 0 else (i - st[0])

    # Push this element to stack
    st.append(_SCREAMING_SNAKE_CASE  )







def 		__a  (      _SCREAMING_SNAKE_CASE						,	_SCREAMING_SNAKE_CASE  )      ->Any:
  for i in range(0						,	_SCREAMING_SNAKE_CASE  ):
    print(arr[i]						,	end=' '  )


# Driver program to test above function
lowercase__   			=      [10, 4, 5, 90, 120, 80]
lowercase__   			=      [0 for i in range(len(price) + 1)]

# Fill the span values in array S[]
calculation_span(price, S)


print_array(S, len(price))





# 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()


