Question

Suppose you are inside of a sentineled, doubly-linked list class as specified in project 2. The method below correctly finds the first index of a value in that list, raising a ValueError exception if the item is not found. Some parts have been removed. Using the minimum possible syntactically valid spacing in your answers, fill in the missing code. Assume that the Node's public attributes are next, previous, and value. Note that none of your responses should contain colons. As a reminder, the linked list implementation provides the following methods:

insert_element_at(value, index) #cannot be used to append
append_element(value)
get_element_at(index)
remove_element_at(index)
__len__ #support for the len method to obtain the list's size

  def index_of(self, val):
     = self.__header.next
    index =
    while current :
      if :
        return
      index =
      
    raise ValueError

and what is the perforamnce of this code?

Answer 1

Here is the completed code for this problem. Comments are included, go through it, learn how things work and let me know if you have any doubts or if you need anything to change. If you are satisfied with the solution, please rate the answer. If not, PLEASE let me know before you rate, I’ll help you fix whatever issues. Thanks

Note: Please maintain proper code spacing (indentation), just copy the code part and paste it in your compiler/IDE directly, no modifications required.

#code

#the completed implementation of this method
def index_of(self, val):    #line1
    current=self.__header.next #setting self.__header.next (first node) as current. #line2
    index = 0 #setting 0 as index. #line3
    while current: #looping until current is None (no change needed). #line4
        if current.value==val: #checking if current.value is same as val. #line5
            return index #returning index. #line6
        index = index+1 #otherwise incrementing index. #line7
    raise ValueError #line8


'''
So the missing code in the above function are:
'current' on line 2, before = sign
'0' on line 3, after = sign
no error on line 4, although while current!=None is the right syntax
'current.value==val' after if before : on line 5
'index' on line 6, after return
'index+1' on line7 after = sign

The function loops through each and every node on the linked list to find val, so the performance is linear
(time complexity: O(n))
'''