text: """
      A Hamiltonian cycle (Hamiltonian circuit) is a graph cycle
      through a graph that visits each node exactly once.
      Determining whether such paths and cycles exist in graphs
      is the 'Hamiltonian path problem', which is NP-complete.
  
      Wikipedia: https://en.wikipedia.org/wiki/Hamiltonian_path
  """
  from typing import List
  
  
  def valid_connection(
      graph: List[List[int]], next_ver: int, curr_ind: int, path: List[int]
  ) -> bool:
      """
      Checks whether it is possible to add next into path by validating 2 statements
      1. There should be path between current and next vertex
      2. Next vertex should not be in path
      If both validations succeeds we return True saying that it is possible to connect
      this vertices either we return False
  
      Case 1:Use exact graph as in main function, with initialized values
      >>> graph = [[0, 1, 0, 1, 0],
      ...          [1, 0, 1, 1, 1],
      ...          [0, 1, 0, 0, 1],
      ...          [1, 1, 0, 0, 1],
      ...          [0, 1, 1, 1, 0]]
      >>> path = [0, -1, -1, -1, -1, 0]
      >>> curr_ind = 1
      >>> next_ver = 1
      >>> valid_connection(graph, next_ver, curr_ind, path)
      True
  
      Case 2: Same graph, but trying to connect to node that is already in path
      >>> path = [0, 1, 2, 4, -1, 0]
      >>> curr_ind = 4
      >>> next_ver = 1
      >>> valid_connection(graph, next_ver, curr_ind, path)
      False
      """
  
      # 1. Validate that path exists between current and next vertices
      if graph[path[curr_ind - 1]][next_ver] == 0:
          return False
  
      # 2. Validate that next vertex is not already in path
      return not any(vertex == next_ver for vertex in path)
  
  
  def util_hamilton_cycle(graph: List[List[int]], path: List[int], curr_ind: int) -> bool:
      """
      Pseudo-Code
      Base Case:
      1. Chceck if we visited all of vertices
          1.1 If last visited vertex has path to starting vertex return True either
              return False
      Recursive Step:
      2. Iterate over each vertex
          Check if next vertex is valid for transiting from current vertex
              2.1 Remember next vertex as next transition
              2.2 Do recursive call and check if going to this vertex solves problem
              2.3 if next vertex leads to solution return True
              2.4 else backtrack, delete remembered vertex
  
      Case 1: Use exact graph as in main function, with initialized values
      >>> graph = [[0, 1, 0, 1, 0],
      ...          [1, 0, 1, 1, 1],
      ...          [0, 1, 0, 0, 1],
      ...          [1, 1, 0, 0, 1],
      ...          [0, 1, 1, 1, 0]]
      >>> path = [0, -1, -1, -1, -1, 0]
      >>> curr_ind = 1
      >>> util_hamilton_cycle(graph, path, curr_ind)
      True
      >>> print(path)
      [0, 1, 2, 4, 3, 0]
  
      Case 2: Use exact graph as in previous case, but in the properties taken from
          middle of calculation
      >>> graph = [[0, 1, 0, 1, 0],
      ...          [1, 0, 1, 1, 1],
      ...          [0, 1, 0, 0, 1],
      ...          [1, 1, 0, 0, 1],
      ...          [0, 1, 1, 1, 0]]
      >>> path = [0, 1, 2, -1, -1, 0]
      >>> curr_ind = 3
      >>> util_hamilton_cycle(graph, path, curr_ind)
      True
      >>> print(path)
      [0, 1, 2, 4, 3, 0]
      """
  
      # Base Case
      if curr_ind == len(graph):
          # return whether path exists between current and starting vertices
          return graph[path[curr_ind - 1]][path[0]] == 1
  
      # Recursive Step
      for next in range(0, len(graph)):
          if valid_connection(graph, next, curr_ind, path):
              # Insert current vertex  into path as next transition
              path[curr_ind] = next
              # Validate created path
              if util_hamilton_cycle(graph, path, curr_ind + 1):
                  return True
              # Backtrack
              path[curr_ind] = -1
      return False
  
  
  def hamilton_cycle(graph: List[List[int]], start_index: int = 0) -> List[int]:
      r"""
      Wrapper function to call subroutine called util_hamilton_cycle,
      which will either return array of vertices indicating hamiltonian cycle
      or an empty list indicating that hamiltonian cycle was not found.
      Case 1:
      Following graph consists of 5 edges.
      If we look closely, we can see that there are multiple Hamiltonian cycles.
      For example one result is when we iterate like:
      (0)->(1)->(2)->(4)->(3)->(0)
  
      (0)---(1)---(2)
       |   /   \   |
       |  /     \  |
       | /       \ |
       |/         \|
      (3)---------(4)
      >>> graph = [[0, 1, 0, 1, 0],
      ...          [1, 0, 1, 1, 1],
      ...          [0, 1, 0, 0, 1],
      ...          [1, 1, 0, 0, 1],
      ...          [0, 1, 1, 1, 0]]
      >>> hamilton_cycle(graph)
      [0, 1, 2, 4, 3, 0]
  
      Case 2:
      Same Graph as it was in Case 1, changed starting index from default to 3
  
      (0)---(1)---(2)
       |   /   \   |
       |  /     \  |
       | /       \ |
       |/         \|
      (3)---------(4)
      >>> graph = [[0, 1, 0, 1, 0],
      ...          [1, 0, 1, 1, 1],
      ...          [0, 1, 0, 0, 1],
      ...          [1, 1, 0, 0, 1],
      ...          [0, 1, 1, 1, 0]]
      >>> hamilton_cycle(graph, 3)
      [3, 0, 1, 2, 4, 3]
  
      Case 3:
      Following Graph is exactly what it was before, but edge 3-4 is removed.
      Result is that there is no Hamiltonian Cycle anymore.
  
      (0)---(1)---(2)
       |   /   \   |
       |  /     \  |
       | /       \ |
       |/         \|
      (3)         (4)
      >>> graph = [[0, 1, 0, 1, 0],
      ...          [1, 0, 1, 1, 1],
      ...          [0, 1, 0, 0, 1],
      ...          [1, 1, 0, 0, 0],
      ...          [0, 1, 1, 0, 0]]
      >>> hamilton_cycle(graph,4)
      []
      """
  
      # Initialize path with -1, indicating that we have not visited them yet
      path = [-1] * (len(graph) + 1)
      # initialize start and end of path with starting index
      path[0] = path[-1] = start_index
      # evaluate and if we find answer return path either return empty array
      return path if util_hamilton_cycle(graph, path, 1) else []
  