x3c-solver

import re

X =[2,1,3,5,9,0]
C = (2,1,3),(5),(9,0)

# This algorithim belongs to Hope1995x on Reddit.


if len(X) % 3 > 0:
  print('no')
  quit()

# This will check for no's

convert_to_string = str(C)
new_list = (re.sub("[^0-9,]", "", convert_to_string))
new_list_one = new_list.split(',')
new_list_one = list(map(int, new_list_one))

if len(new_list_one) < len(X):
 print('no')
 quit()

using_this_for_second_loop=[];
what_if_there_is_no_repeating_elements = 0

# The first part of trying to solve the X3C problem
# is to loop through each indidvual index of C.

# I will then use a nested loop so that I can loop
# thorugh indexes as if they were lists.

# Indexes that contain elements that don't exist in X
# will not be appended to the new list.

# Indexes with repeating elements
# will not be appended to the new list.


# I used a function from https://thispointer.com/python-3-ways-to-check-if-there-are-duplicates-in-a-list/



def checkIfDuplicates_3(listOfElems):
    for j in range(0, len(C)):
        convert_index_into_list = str(C[j])
        new_list = (re.sub("[^0-9,]", "", convert_index_into_list))
        new_list = new_list.split(',')
        new_list = list(map(int, new_list))
        make_sure_all_elements_exist_in_X = all(elem in X for elem in new_list)
        print(new_list)
        if make_sure_all_elements_exist_in_X:
            for elem in new_list:
                if listOfElems.count(elem) < 2:
                    using_this_for_second_loop.append(C[j])
                    what_if_there_is_no_repeating_elements = 1

if what_if_there_is_no_repeating_elements == 0:
    using_this_for_second_loop = C


# I will then modify the list that will be used for the 2nd loop.
# This is to prevent semantic bugs.

# Here the only elements that do exist are the one's THAT DON'T REPEAT
# AND THE ONE'S THAT EXIST IN X.


preparing_the_answer = str(using_this_for_second_loop)
new_list_two = (re.sub("[^0-9,]", "", preparing_the_answer))
new_list_thr = new_list_two.split(',')

new_list_thr = list(map(int, new_list_thr))
making_sure_all_elements_in_X = all(elem in X for elem in new_list_thr)

if making_sure_all_elements_in_X:
    print('yes')
else:
    print('no')