DCP40 - No-duplicated number

'''
Given an array of integers where every integer occurs three times except for one integer, which only occurs once, find and return the non-duplicated integer.
For example, given [6, 1, 3, 3, 3, 6, 6], return 1. Given [13, 19, 13, 13], return 19.
Do this in O(N) time and O(1) space.
'''
from timeit import default_timer as timer
from random import randrange, shuffle, _sha512

# Function 1
def solve1(myList):
    uniques = list()
    for i in range(len(myList)):
        for unique in uniques:
            if myList[i] == unique:
                break
        uniques.append(myList[i])

    counters = list()
    for i in range(len(uniques)):
        counters.append(0)
    for i in range(len(myList)):
        for unique in uniques:
            if myList[i] == unique:
                counters[i] += 1

    # So, now my counters list is ready
    for i in range(len(counters)):
        if counters[i] == 1:
            return uniques[i]

# Function 2
def solve2(myList):
    myList = sorted(myList)
    # Now, that "myList" is sorted, I can compare its element with the previous and next one. If my current element is equal with the previous one and
    # the next one, I am sure that this element appears exactly 3 times, so I walk on. Otherwise, I have to return this element.
    answer = -1000
    for i in range(1, len(myList)-1):
        if myList[i] != myList[i-1] and myList[i] != myList[i+1]:
            answer = myList[i]
    # The above code works when this element is only in the MEDIUM positions of the array, otherwise the value of "answer" remains -1000
    # This means that maybe the first or last element is the answer
    if answer == -1000:
        if myList[0] == myList[1]:
            # The answer is not in the list's beginning, but in the end
            answer = myList[len(myList)-1]
        else:
            answer = myList[0]
    return answer


# Function 3
def prettyPrintWithTime(function, myList):
    start = timer()
    result = function(myList)
    end = timer()
    elapsed = 10**6 * (end - start)
    print("Using function " + str(function.__name__) + ": " + str(myList) + " ----> " + str(result) + " (" + str(round(elapsed, 2)) + " μs).")


# Function 4
def generateRandomList(n1, n2):

    N = randrange(n1, n2)    # 7 <= N <= 11
    # My final list will contain form 7 * 3 + 1 = 22 elements to 11 * 3 + 1 = 34 elements
    myList = list()
    myList.append(randrange(20))
    for i in range(1, N):
        r = randrange(myList[len(myList)-1] + 1, myList[len(myList)-1] + 21)
        myList.append(r)

    # These will be the different elements except the last one, that I will keep it twice
    newList = list()
    for i in range(len(myList)):
        newList.append(myList[i])
        newList.append(myList[i])
        newList.append(myList[i])
    # Last step is to add a brand-new different value
    exists = False
    new = randrange(200)
    for element in myList:
        if element == new:
            exists = True
            break
    while exists == True:
        new = randrange(200)
        for element in myList:
            if element == new:
                exists = True
                break
    newList.append(new)
    shuffle(newList)
    return newList

# MAIN FUNCTION
list1 = [6, 1, 3, 3, 3, 6, 6]
list2 = [13, 19, 13, 13]
list3 = generateRandomList(7, 12)
prettyPrintWithTime(solve1, list1)
prettyPrintWithTime(solve2, list1)
prettyPrintWithTime(solve1, list2)
prettyPrintWithTime(solve2, list2)
prettyPrintWithTime(solve1, list3)
prettyPrintWithTime(solve2, list3)