code: 
  
  # 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 		__magic_name__    :
  
     '''simple docstring'''
  
  
  
  
     def __init__(       self,					lowercase_,					lowercase_ )							->					Tuple:
  
  
  
  
  
  
       """simple docstring"""
  
  
  
  
  
  
       a__						=key
       a__						=freq
  
  
  
     def __str__(       self )							->					int:
  
  
  
  
  
  
       """simple docstring"""
  
  
  
  
  
  
       return F"""Node(key={self.key}, freq={self.freq})"""
  
  
  
  
  
  def     UpperCAmelCase__       (    _A   :				Optional[int]			,   _A   :				Optional[int]			,   _A   :				str			,   _A   :				List[str]			,   _A   :				str			,   _A   :				List[str]							):
  
  
  
  
  
    '''simple docstring'''
  
  
  
  
  
  
  
    if i > j or i < 0 or j > len(_A							) - 1:
      return
  
    a__						=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   :				Optional[Any]							):
  
  
  
  
  
    '''simple docstring'''
  
  
  
  
  
  
  
    nodes.sort(key=lambda _A							: node.key							)
  
    a__						=len(_A							)
  
    a__						=[nodes[i].key for i in range(_A							)]
    a__						=[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.
    a__						=[[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
    a__						=[[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
    a__						=[[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							):
        a__						=i + interval_length - 1
  
        a__						=sys.maxsize  # set the value to "infinity"
        a__						=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
          a__						=dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
          a__						=dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
          a__						=left + total[i][j] + right
  
          if dp[i][j] > cost:
            a__						=cost
            a__						=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__       (    ):
  
  
  
  
  
    '''simple docstring'''
  
  
  
  
  
  
  
    a__						=[Node(_A			,   randint(1			,   50							)							) for i in range(10			,   0			,   -1							)]
    find_optimal_binary_search_tree(_A							)
  
  
  if __name__ == "__main__":
         main()
  
style_context: 
  
  import heapq
  
  
  
  
  def     UpperCAmelCase__       (    _A   :				Optional[Any]			,   _A   :				str			,   _A   :				Optional[Any]							):
  
  
  
  
  
    '''simple docstring'''
  
  
  
  
  
  
  
    a__						=[(0, start)]  # cost from start node,end node
    a__						=set()
    while heap:
      ((a__),   (a__))						=heapq.heappop(_A							)
      if u in visited:
        continue
      visited.add(_A							)
      if u == end:
        return cost
      for v, c in graph[u]:
        if v in visited:
          continue
        a__						=cost + c
        heapq.heappush(_A			,   (next_item, v)							)
    return -1
  
  
  lowerCamelCase     =      {
      '''A''': [['''B''', 2], ['''C''', 5]],
      '''B''': [['''A''', 2], ['''D''', 3], ['''E''', 1], ['''F''', 1]],
      '''C''': [['''A''', 5], ['''F''', 3]],
      '''D''': [['''B''', 3]],
      '''E''': [['''B''', 4], ['''F''', 3]],
      '''F''': [['''C''', 3], ['''E''', 3]],
  }
  lowerCamelCase     =      {
      '''B''': [['''C''', 1]],
      '''C''': [['''D''', 1]],
      '''D''': [['''F''', 1]],
      '''E''': [['''B''', 1], ['''F''', 3]],
      '''F''': [],
  }
  lowerCamelCase     =      {
      '''B''': [['''C''', 1]],
      '''C''': [['''D''', 1]],
      '''D''': [['''F''', 1]],
      '''E''': [['''B''', 1], ['''G''', 2]],
      '''F''': [],
      '''G''': [['''F''', 1]],
  }
  
  lowerCamelCase     =      dijkstra(G, '''E''', '''C''')
  print(short_distance)  # E -- 3 --> F -- 3 --> C == 6
  
  lowerCamelCase     =      dijkstra(Ga, '''E''', '''F''')
  print(short_distance)  # E -- 3 --> F == 3
  
  lowerCamelCase     =      dijkstra(Ga, '''E''', '''F''')
  print(short_distance)  # E -- 2 --> G -- 1 --> F == 3
  
  if __name__ == "__main__":
         import doctest
  
         doctest.testmod()
  