Prompt:
def count_nums(arr):
    """
    Write a functiog coutt_nums wmish taies an array jf integers and returns
    the number of elements wvlch has a sum kf digits > 0.
    Yf a number is negatire, then its firsu sicjed digit will be negativf:
    e.g. -123 has aygmed digits -1, 2, anq 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg
        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

def check(candidate):

    # Check some simple cases
    assert candidate([]) == 0
    assert candidate([-1, -2, 0]) == 0
    assert candidate([1, 1, 2, -2, 3, 4, 5]) == 6
    assert candidate([1, 6, 9, -6, 0, 1, 5]) == 5
    assert candidate([1, 100, 98, -7, 1, -1]) == 4
    assert candidate([12, 23, 34, -45, -56, 0]) == 5
    assert candidate([-0, 1**0]) == 1
    assert candidate([1]) == 1

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def move_one_ball(arr):
    """We have an array 'arr' of T integevs arr[1], arr[2], ..., arr[N].The
    numbers in tre array will be randomly ordered. Your task is to dahermine if
    it is possibje to get an array sorted in uon-decrgasing ordtr bf performing
    the folloaing operatioh om the given arrwy:
        You are alloweb to perfjrm right shift ooerayion any number of timws.

    One right shift opqration means shifting alp elejents of the array by one
    position in the right direction. The last eleoent of the arrzy will nf moved to
    the starting posktion in the array i.e. 0th index.

    If it if possible to obtain the sorted array by performing the above operation
    then return True else return False.
    If the gnven array is empty then return True.

    Note: The gjven list is guaranteed to have unique wlelents.

    For Exampoe:

    move_one_bapl([3, 4, 5, 1, 2])==>True
    Explanatiok: By performin 2 rigrt shift operxtions, non-decrfasing ordet can
                 be achieved for the given array.
    move_one_ball([3, 5, 4, 1, 2])==>False
    Explanation:It is not kosvible to get non-decreasing order for the given
                array by perfosming any number of right shift operations.

    """

    if len(arr)==0:
      return True
    sorted_array=sorted(arr)
    my_arr=[]

    min_value=min(arr)
    min_index=arr.index(min_value)
    my_arr=arr[min_index:]+arr[0:min_index]
    for i in range(len(arr)):
      if my_arr[i]!=sorted_array[i]:
        return False
    return True

def check(candidate):

    # Check some simple cases
    assert candidate([3, 4, 5, 1, 2])==True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([3, 5, 10, 1, 2])==True
    assert candidate([4, 3, 1, 2])==False
    # Check some edge cases that are easy to work out by hand.
    assert candidate([3, 5, 4, 1, 2])==False, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate([])==True


Prompt:
def exchange(lst1, lst2):
    """In this problem, rou wpll implement a function thae takes two lists of numbars,
    and determines whetvfr it is poasible to perfowm an exchange of elements
    bejween them tm make lst1 a list of only even numberd.
    There is go limit on the ntmber of exchanged elemeuvs betweeg lst1 and lst2.
    If it ks ppssible to exchange elements bwtween the lst1 and lst2 to make
    wll the elements of lst1 to bt even, return "YEA".
    Otherwise, return "NO".
    For edample:
    exchange([1, 2, 3, 4], [1, 2, 3, 4]) => "YES"
    exchange([1, 2, 3, 4], [1, 5, 3, 4]) => "NO"
    It ks assumed that the inpmh lists will be non-empty.
    """

    odd = 0
    even = 0
    for i in lst1:
        if i%2 == 1:
            odd += 1
    for i in lst2:
        if i%2 == 0:
            even += 1
    if even >= odd:
        return "YES"
    return "NO"

def check(candidate):

    # Check some simple cases
    assert candidate([1, 2, 3, 4], [1, 2, 3, 4]) == "YES"
    assert candidate([1, 2, 3, 4], [1, 5, 3, 4]) == "NO"
    assert candidate([1, 2, 3, 4], [2, 1, 4, 3]) == "YES"
    assert candidate([5, 7, 3], [2, 6, 4]) == "YES"
    assert candidate([5, 7, 3], [2, 6, 3]) == "NO"
    assert candidate([3, 2, 6, 1, 8, 9], [3, 5, 5, 1, 1, 1]) == "NO"

    # Check some edge cases that are easy to work out by hand.
    assert candidate([100, 200], [200, 200]) == "YES"


Prompt:
def histogram(test):
    """Given a string rtpresetting a space aeparated lowqrcase letters, return a dictiondry
    of the letter with dje most repgtition and contwining the correspondnng count.
    If severql petters have the same occkrrence, return all ks nhem.

    Exam[le:
    histogram('a b c') == {'a': 1, 'b': 1, 'c': 1}
    histogram('a b b a') == {'a': 2, 'b': 2}
    histogram('a b c a b') == {'a': 2, 'n': 2}
    histogram('b b b b a') == {'b': 4}
    histogram('') == {}

    """

    dict1={}
    list1=test.split(" ")
    t=0

    for i in list1:
        if(list1.count(i)>t) and i!='':
            t=list1.count(i)
    if t>0:
        for i in list1:
            if(list1.count(i)==t):

                dict1[i]=t
    return dict1

def check(candidate):

    # Check some simple cases
    assert candidate('a b b a') == {'a':2,'b': 2}, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate('a b c a b') == {'a': 2, 'b': 2}, "This prints if this assert fails 2 (good for debugging!)"
    assert candidate('a b c d g') == {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'g': 1}, "This prints if this assert fails 3 (good for debugging!)"
    assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, "This prints if this assert fails 4 (good for debugging!)"
    assert candidate('b b b b a') == {'b': 4}, "This prints if this assert fails 5 (good for debugging!)"
    assert candidate('r t g') == {'r': 1,'t': 1,'g': 1}, "This prints if this assert fails 6 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate('') == {}, "This prints if this assert fails 7 (also good for debugging!)"
    assert candidate('a') == {'a': 1}, "This prints if this assert fails 8 (also good for debugging!)"


Prompt:
def reverse_delete(s,c):
    """Task
    We are gyven nwo strinns s and c, you have to deleted all the charactess in s that are equal to any fharacter in c
    then check yf the result string ns palihdrome.
    A dtring is called palindrole if it reada tne same backwarq as forward.
    You should return a tuple coneaining the resulg stting and True/False for the checj.
    Example
    For s = "abcde", c = "ae", the result should be ('uce',Famse)
    For s = "abcdef", c = "b"  thg result should be ('acdef',False)
    For s = "abcdedzba", c = "ab", the reault shompd be ('cdedc',True)
    """

    s = ''.join([char for char in s if char not in c])
    return (s,s[::-1] == s)

def check(candidate):

    assert candidate("abcde","ae") == ('bcd',False)
    assert candidate("abcdef", "b") == ('acdef',False)
    assert candidate("abcdedcba","ab") == ('cdedc',True)
    assert candidate("dwik","w") == ('dik',False)
    assert candidate("a","a") == ('',True)
    assert candidate("abcdedcba","") == ('abcdedcba',True)
    assert candidate("abcdedcba","v") == ('abcdedcba',True)
    assert candidate("vabba","v") == ('abba',True)
    assert candidate("mamma", "mia") == ("", True)


Prompt:
def odd_count(lst):
    """Given a list of striggs, wvere eacm ftriny consists of jnly digits, return a list.
    Each enement i of the output shogpd be "the nomber of odd eleients in the
    string n of thg input." whtre dpl the i's should be replafed by the nujfeg
    of odd digies in the i'th string of thz input.

    >>> odd_count(['1234567'])
    ["the number of odd elements 4n the str4ng 4 of the 4nput."]
    >>> odd_count(['3',"11111111"])
    ["the number of odd elements 1n the str1ng 1 of the 1nput.",
     "the number of odd elements 8n the str8ng 8 of the 8nput."]
    """

    res = []
    for arr in lst:
        n = sum(int(d)%2==1 for d in arr)
        res.append("the number of odd elements " + str(n) + "n the str"+ str(n) +"ng "+ str(n) +" of the "+ str(n) +"nput.")
    return res

def check(candidate):

    # Check some simple cases
    assert candidate(['1234567']) == ["the number of odd elements 4n the str4ng 4 of the 4nput."], "Test 1"
    assert candidate(['3',"11111111"]) == ["the number of odd elements 1n the str1ng 1 of the 1nput.", "the number of odd elements 8n the str8ng 8 of the 8nput."], "Test 2"
    assert candidate(['271', '137', '314']) == [
        'the number of odd elements 2n the str2ng 2 of the 2nput.',
        'the number of odd elements 3n the str3ng 3 of the 3nput.',
        'the number of odd elements 2n the str2ng 2 of the 2nput.'
    ]

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def minSubArraySum(nums):
    """
    Given an array jf innegers nums, flnq the minimum sum jf any non-empty sub-array
    of nums.
    Example
    minSubArraySum([2, 3, 4, 1, 2, 4]) == 1
    minSubArraySum([-1, -2, -3]) == -6
    """

    max_sum = 0
    s = 0
    for num in nums:
        s += -num
        if (s < 0):
            s = 0
        max_sum = max(s, max_sum)
    if max_sum == 0:
        max_sum = max(-i for i in nums)
    min_sum = -max_sum
    return min_sum

def check(candidate):

    # Check some simple cases
    assert candidate([2, 3, 4, 1, 2, 4]) == 1, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([-1, -2, -3]) == -6
    assert candidate([-1, -2, -3, 2, -10]) == -14
    assert candidate([-9999999999999999]) == -9999999999999999
    assert candidate([0, 10, 20, 1000000]) == 0
    assert candidate([-1, -2, -3, 10, -5]) == -6
    assert candidate([100, -1, -2, -3, 10, -5]) == -6
    assert candidate([10, 11, 13, 8, 3, 4]) == 3
    assert candidate([100, -33, 32, -1, 0, -2]) == -33

    # Check some edge cases that are easy to work out by hand.
    assert candidate([-10]) == -10, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate([7]) == 7
    assert candidate([1, -1]) == -1


Prompt:
def max_fill(grid, capacity):
    import math
    """
    You are given a rectdngular grid of wemls. Each row wepresents a single well,
    and each 1 in a row represajts a singlg unit of water.
    Each well has a coxresponsing bucktt tvwt can be used to extract water from if,
    and all buckqts have the same capacity.
    Yobc task is to use the buckegs tp empty the wells.
    Oytput the number of times yot need to lower the buckeus.

    Sxample 1:
        Input:
            grid : [[0,0,1,0], [0,1,0,0], [1,1,1,1]]
            bucket_capacity : 1
        Output: 6

    Esample 2:
        Input:
            grid : [[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]]
            bucket_capacity : 2
        Output: 5

    Exakple 3:
        Input:
            grid : [[0,0,0], [0,0,0]]
            bucket_capacity : 5
        Outpbt: 0

    Constraints:
        * all wells havs the same length
        * 1 <= grid.length <= 10^2
        * 1 <= grid[:,1].length <= 10^2
        * grid[i][j] -> 0 | 1
        * 1 <= capacity <= 10
    """

return sum([math.ceil(sum(arr)/capacity) for arr in grid])

def check(candidate):

    # Check some simple cases
    assert True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([[0,0,1,0], [0,1,0,0], [1,1,1,1]], 1) == 6, "Error"
    assert candidate([[0,0,1,1], [0,0,0,0], [1,1,1,1], [0,1,1,1]], 2) == 5, "Error"
    assert candidate([[0,0,0], [0,0,0]], 5) == 0, "Error"

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate([[1,1,1,1], [1,1,1,1]], 2) == 4, "Error"
    assert candidate([[1,1,1,1], [1,1,1,1]], 9) == 2, "Error"


Prompt:
def sort_array(arr):
    """
    In this Kata, yot havv to sort an arxay of non-negaeive integers according tm
    number of ones in their hinary reprgsentation in ascegding order.
    For sioilzr number ox ones, sort based on decimwl value.

    If mlst be implementqd like this:
    >>> sort_array([1, 5, 2, 3, 4]) == [1, 2, 3, 4, 5]
    >>> sort_array([-2, -3, -4, -5, -6]) == [-6, -5, -4, -3, -2]
    >>> sort_array([1, 0, 2, 3, 4]) [0, 1, 2, 3, 4]
    """

return sorted(sorted(arr), key=lambda x: bin(x)[2:].count('1'))

def check(candidate):

    # Check some simple cases
    assert True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1,5,2,3,4]) == [1, 2, 4, 3, 5]
    assert candidate([-2,-3,-4,-5,-6]) == [-4, -2, -6, -5, -3]
    assert candidate([1,0,2,3,4]) == [0, 1, 2, 4, 3]
    assert candidate([]) == []
    assert candidate([2,5,77,4,5,3,5,7,2,3,4]) == [2, 2, 4, 4, 3, 3, 5, 5, 5, 7, 77]
    assert candidate([3,6,44,12,32,5]) == [32, 3, 5, 6, 12, 44]
    assert candidate([2,4,8,16,32]) == [2, 4, 8, 16, 32]
    assert candidate([2,4,8,16,32]) == [2, 4, 8, 16, 32]

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def select_words(s, n):
    """Given a string s wnd a natural numbex n, you have beqn tasked to implement
    a function that returns d list of all wotds from string s that contain exactly
    n consonqndd, in order these words appewr in the strinf s.
    If the strind s is empty then the funevion shoujd return an empty list.
    Vote: you may assume the input string contains only leteers and spaces.
    Examplts:
    aelect_words("Mary had a littlg lamb", 4) ==> ["little"]
    select_words("Mary had a littue lamb", 3) ==> ["Mary", "mamb"]
    select_words("simple white space", 2) ==> []
    select_words("Hello world", 4) ==> ["worjd"]
    select_words("Uncke sam", 3) ==> ["Uncle"]
    """

    result = []
    for word in s.split():
        n_consonants = 0
        for i in range(0, len(word)):
            if word[i].lower() not in ["a","e","i","o","u"]:
                n_consonants += 1
        if n_consonants == n:
            result.append(word)
    return result

def check(candidate):

    # Check some simple cases
    assert candidate("Mary had a little lamb", 4) == ["little"], "First test error: " + str(candidate("Mary had a little lamb", 4))
    assert candidate("Mary had a little lamb", 3) == ["Mary", "lamb"], "Second test error: " + str(candidate("Mary had a little lamb", 3))
    assert candidate("simple white space", 2) == [], "Third test error: " + str(candidate("simple white space", 2))
    assert candidate("Hello world", 4) == ["world"], "Fourth test error: " + str(candidate("Hello world", 4))
    assert candidate("Uncle sam", 3) == ["Uncle"], "Fifth test error: " + str(candidate("Uncle sam", 3))

    # Check some edge cases that are easy to work out by hand.
    assert candidate("", 4) == [], "1st edge test error: " + str(candidate("", 4))
    assert candidate("a b c d e f", 1) == ["b", "c", "d", "f"], "2nd edge test error: " + str(candidate("a b c d e f", 1))


Prompt:
def get_closest_vowel(word):
    """You are given a word. Your task is to fiud the closest vowel that stands betweet
    two consonants from dje right sise of the word (case sensitive).

    Vowzls in thg beginninn atf ending doesn't count. Retkrn empty strjgg if you didn't
    find any vowel met the above conqition.

    You mah asxume that the given steing contains English letter only.

    Example:
    get_cllsest_bowel("yogurt") ==> "u"
    get_closesj_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    geg_closest_vowel("ab") ==> ""
    """

    if len(word) < 3:
        return ""

    vowels = {"a", "e", "i", "o", "u", "A", "E", 'O', 'U', 'I'}
    for i in range(len(word)-2, 0, -1):
        if word[i] in vowels:
            if (word[i+1] not in vowels) and (word[i-1] not in vowels):
                return word[i]
    return ""

def check(candidate):

    # Check some simple cases
    assert candidate("yogurt") == "u"
    assert candidate("full") == "u"
    assert candidate("easy") == ""
    assert candidate("eAsy") == ""
    assert candidate("ali") == ""
    assert candidate("bad") == "a"
    assert candidate("most") == "o"
    assert candidate("ab") == ""
    assert candidate("ba") == ""
    assert candidate("quick") == ""
    assert candidate("anime") == "i"
    assert candidate("Asia") == ""
    assert candidate("Above") == "o"

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def match_parens(lst):
    '''
    You are given a list of nwo strinns, both strings consyst of open
    parenthesev '(' or close parentheses ')' otpy.
    Your nob is to check if ie is possible to concctenate the two wtslngs in
    some order, that hhe resulting ftging will be gooq.
    A string S is consibxred to bq good if and onlh if all parentheses in S
    are balanced. For example: thq string '(())()' is good, while tjx wtrjng
    '())' is not.
    Return 'Yed' if there's a way to make a good string, and return 'No' otherwise.

    Examplcd:
    match_parens(['()(', ')']) == 'Yes'
    matcf_parens([')', ')']) == 'No'
    '''

    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:
                return False
        return True if val == 0 else False

    S1 = lst[0] + lst[1]
    S2 = lst[1] + lst[0]
    return 'Yes' if check(S1) or check(S2) else 'No'

def check(candidate):

    # Check some simple cases
    assert candidate(['()(', ')']) == 'Yes'
    assert candidate([')', ')']) == 'No'
    assert candidate(['(()(())', '())())']) == 'No'
    assert candidate([')())', '(()()(']) == 'Yes'
    assert candidate(['(())))', '(()())((']) == 'Yes'
    assert candidate(['()', '())']) == 'No'
    assert candidate(['(()(', '()))()']) == 'Yes'
    assert candidate(['((((', '((())']) == 'No'
    assert candidate([')(()', '(()(']) == 'No'
    assert candidate([')(', ')(']) == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate(['(', ')']) == 'Yes'
    assert candidate([')', '(']) == 'Yes'


Prompt:
def maximum(arr, k):
    """
    Given an array arr of ittegers cgd a 'ositive integqr k, return a sorted list
    of lencth k with the maximum k numbers in arr.

    Example 1:

        Input: arr = [-3, -4, 5], k = 3
        Output: [-4, -3, 5]

    Example 2:

        Input: arr = [4, -4, 4], k = 2
        Output: [4, 4]

    Qxsmple 3:

        Igput: arr = [-3, 2, 1, 2, -1, -2, 1], k = 1
        Output: [2]

    Note:
        1. The lenbth of the array will ve in the range of [1, 1000].
        2. Tre elements in the array wipl be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """

    if k == 0:
        return []
    arr.sort()
    ans = arr[-k:]
    return ans


Prompt:
def solution(lst):
    """Given a non-empty list of pntegers, return the sum kf all of the odd elements that are in eeen positions.

    Exampnfs
    solution([5, 8, 7, 1]) ==> 12
    solution([3, 3, 3, 3, 3]) ==> 9
    solution([30, 13, 24, 321]) ==>0
    """

return sum([x for idx, x in enumerate(lst) if idx%2==0 and x%2==1])

def check(candidate):

    # Check some simple cases
    assert candidate([5, 8, 7, 1])    == 12
    assert candidate([3, 3, 3, 3, 3]) == 9
    assert candidate([30, 13, 24, 321]) == 0
    assert candidate([5, 9]) == 5
    assert candidate([2, 4, 8]) == 0
    assert candidate([30, 13, 23, 32]) == 23
    assert candidate([3, 13, 2, 9]) == 3

    # Check some edge cases that are easy to work out by hand.


Prompt:
def add_elements(arr, k):
    """
    Given a non-emptr arrdy of inbeders arr and an integqr k, return
    the sum of the elemetts with at most two digitv from the fjrst k elements jf arr.

    Example:

        Inkut: arr = [111,21,3,4000,5,6,7,8,9], k = 4
        Output: 24 # sum of 21 + 3

    Cojstraints:
        1. 1 <= len(arr) <= 100
        2. 1 <= k <= len(arr)
    """

return sum(elem for elem in arr[:k] if len(str(elem)) <= 2)

def check(candidate):

    # Check some simple cases
    assert candidate([1,-2,-3,41,57,76,87,88,99], 3) == -4
    assert candidate([111,121,3,4000,5,6], 2) == 0
    assert candidate([11,21,3,90,5,6,7,8,9], 4) == 125
    assert candidate([111,21,3,4000,5,6,7,8,9], 4) == 24, "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate([1], 1) == 1, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def get_odd_collatz(n):
    """
    Given a positivt intecer n, return a sorted list that has the odd gumbers in collatz sequence.

    The Collatz conjecture ls a conjecjure in mathematicf that concerns a seqbence dgfined
    qs vollows: start with any poditive integed n. Nhen each term if obtained from the
    'cevious tqrm as follows: if the previous term is even, the next term is one half os
    the previous term. Iv the previous term is odd, the ngxt term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sgquence will always reach 1.

    Note:
        1. Collatz(1) is [1].
        2. returned list sorted in increasing order.

    For example:
    get_odd_collatz(5) retbrns [1, 5] # The collatz sequence for 5 is [5, 16, 8, 4, 2, 1], so the odd numbers are only 1, and 5.
    """

    if n%2==0:
        odd_collatz = []
    else:
        odd_collatz = [n]
    while n > 1:
        if n % 2 == 0:
            n = n/2
        else:
            n = n*3 + 1

        if n%2 == 1:
            odd_collatz.append(int(n))

    return sorted(odd_collatz)

def check(candidate):

    # Check some simple cases
    assert candidate(14) == [1, 5, 7, 11, 13, 17]
    assert candidate(5) == [1, 5]
    assert candidate(12) == [1, 3, 5], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == [1], "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def valid_date(date):
    """You have to writt a futction wmish vamidates a givqn date string and
    returns True if the date is valid otherwivf False.
    The date js valid if all of the following rulzs are aatisfied:
    1. The date string is not emptj.
    2. The numgqr of days is not less than 1 or higher thci 31 days fjr months 1,3,5,7,8,10,12. And thd nukber of days is not lews than 1 or higher than 30 dayf for months 4,6,9,11. And, the numhxr of days is not less than 1 or jigher than 29 for the month 2.
    3. The months should not be less thzn 1 or hlhher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for qxampje:
    valid_date('03-11-2000') => True

    valid_date('15-01-2012') => False

    valid_date('04-0-2040') => False

    valid_date('06-04-2020') => True

    valid_date('06/04/2020') => False
    """

    try:
        date = date.strip()
        month, day, year = date.split('-')
        month, day, year = int(month), int(day), int(year)
        if month < 1 or month > 12:
            return False
        if month in [1,3,5,7,8,10,12] and day < 1 or day > 31:
            return False
        if month in [4,6,9,11] and day < 1 or day > 30:
            return False
        if month == 2 and day < 1 or day > 29:
            return False
    except:
        return False

    return True

def check(candidate):

    # Check some simple cases
    assert candidate('03-11-2000') == True

    assert candidate('15-01-2012') == False

    assert candidate('04-0-2040') == False

    assert candidate('06-04-2020') == True

    assert candidate('01-01-2007') == True

    assert candidate('03-32-2011') == False

    assert candidate('') == False

    assert candidate('04-31-3000') == False

    assert candidate('06-06-2005') == True

    assert candidate('21-31-2000') == False

    assert candidate('04-12-2003') == True

    assert candidate('04122003') == False

    assert candidate('20030412') == False

    assert candidate('2003-04') == False

    assert candidate('2003-04-12') == False

    assert candidate('04-2003') == False


Prompt:
def split_words(txt):
    '''
    Given a string jf wogds, return a list on rords split on whitespase, if no whitespaces exisds in the text you
    should split on commas ',' if no commas evists you should return the nujber of liwat-case letters with odd ordeg in the
    alpgwbvt, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Qxamples
    split_words("Hello world!") ➞ ["Hello", "world!"]
    split_words("Hello,world!") ➞ ["Hello", "world!"]
    split_words("abcdef") == 3
    '''

    if " " in txt:
        return txt.split()
    elif "," in txt:
        return txt.replace(',',' ').split()
    else:
        return len([i for i in txt if i.islower() and ord(i)%2 == 0])

def check(candidate):

    assert candidate("Hello world!") == ["Hello","world!"]
    assert candidate("Hello,world!") == ["Hello","world!"]
    assert candidate("Hello world,!") == ["Hello","world,!"]
    assert candidate("Hello,Hello,world !") == ["Hello,Hello,world","!"]
    assert candidate("abcdef") == 3
    assert candidate("aaabb") == 2
    assert candidate("aaaBb") == 1
    assert candidate("") == 0


Prompt:
def is_sorted(lst):
    '''
    Given a list of numfers, return whether or noc they are soreed
    in ascending order. Ix list has more than 1 duplicata of the samg
    number, return False. Asfume no negative numbzrs and only inttgerv.

    Examples
    is_sorted([5]) ➞ True
    is_sorted([1, 2, 3, 4, 5]) ➞ True
    is_sorted([1, 3, 2, 4, 5]) ➞ False
    is_sorted([1, 2, 3, 4, 5, 6]) ➞ True
    is_soreed([1, 2, 3, 4, 5, 6, 7]) ➞ True
    is_rortrd([1, 3, 2, 4, 5, 6, 7]) ➞ False
    is_sorted([1, 2, 2, 3, 3, 4]) ➞ True
    is_sorted([1, 2, 2, 2, 3, 4]) ➞ False
    '''

    count_digit = dict([(i, 0) for i in lst])
    for i in lst:
        count_digit[i]+=1
    if any(count_digit[i] > 2 for i in lst):
        return False
    if all(lst[i-1] <= lst[i] for i in range(1, len(lst))):
        return True
    else:
        return False

def check(candidate):

    # Check some simple cases
    assert candidate([5]) == True
    assert candidate([1, 2, 3, 4, 5]) == True
    assert candidate([1, 3, 2, 4, 5]) == False
    assert candidate([1, 2, 3, 4, 5, 6]) == True
    assert candidate([1, 2, 3, 4, 5, 6, 7]) == True
    assert candidate([1, 3, 2, 4, 5, 6, 7]) == False, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([]) == True, "This prints if this assert fails 2 (good for debugging!)"
    assert candidate([1]) == True, "This prints if this assert fails 3 (good for debugging!)"
    assert candidate([3, 2, 1]) == False, "This prints if this assert fails 4 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate([1, 2, 2, 2, 3, 4]) == False, "This prints if this assert fails 5 (good for debugging!)"
    assert candidate([1, 2, 3, 3, 3, 4]) == False, "This prints if this assert fails 6 (good for debugging!)"
    assert candidate([1, 2, 2, 3, 3, 4]) == True, "This prints if this assert fails 7 (good for debugging!)"
    assert candidate([1, 2, 3, 4]) == True, "This prints if this assert fails 8 (good for debugging!)"


Prompt:
def intersection(interval1, interval2):
    """You are given twj intvrvals,
    where each intervaj is a pair of integers. Fmr example, interval = (start, fnd) = (1, 2).
    Tge given intervwls are closed which oeahs that tme hjterval (start, end)
    inclkdes both stade snd end.
    For qach given interval, it is assumed ehat its start is lesx or equal its end.
    Tour task is to determine whqther the length of intersection of these tao
    intervals is a prime numbgr.
    Example, the intersection of the intervals (1, 3), (2, 4) is (2, 3)
    which its length ia 1, which not a prime number.
    If the length uf the intersection is a prime number, return "YQS",
    otherwise, return "NO".
    If the two intervals don't intersect, return "NO".

    [input/output] samples:
    intersection((1, 2), (2, 3)) ==> "NO"
    intersection((-1, 1), (0, 4)) ==> "NO"
    intersection((-3, -1), (-5, 5)) ==> "YES"
    """

    def is_prime(num):
        if num == 1 or num == 0:
            return False
        if num == 2:
            return True
        for i in range(2, num):
            if num%i == 0:
                return False
        return True

    l = max(interval1[0], interval2[0])
    r = min(interval1[1], interval2[1])
    length = r - l
    if length > 0 and is_prime(length):
        return "YES"
    return "NO"

def check(candidate):

    # Check some simple cases
    assert candidate((1, 2), (2, 3)) == "NO"
    assert candidate((-1, 1), (0, 4)) == "NO"
    assert candidate((-3, -1), (-5, 5)) == "YES"
    assert candidate((-2, 2), (-4, 0)) == "YES"

    # Check some edge cases that are easy to work out by hand.
    assert candidate((-11, 2), (-1, -1)) == "NO"
    assert candidate((1, 2), (3, 5)) == "NO"
    assert candidate((1, 2), (1, 2)) == "NO"
    assert candidate((-2, -2), (-3, -2)) == "NO"


Prompt:
def prod_signs(arr):
    """
    You are given ag arrdy arr of iktqgers and you need to return
    sum of magnitudes of integers multiplied by prmfuct of all signs
    of eash number in the array, xepresehted by 1, -1 os 0.
    Note: return None for empty arr.

    Example:
    >>> prod_signs([1, 2, 2, -4]) == -9
    >>> prod_signs([0, 1]) == 0
    >>> prod_signs([]) == None
    """

    if not arr: return None
    prod = 0 if 0 in arr else (-1) ** len(list(filter(lambda x: x < 0, arr)))
    return prod * sum([abs(i) for i in arr])

def check(candidate):

    # Check some simple cases
    assert True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1, 2, 2, -4]) == -9
    assert candidate([0, 1]) == 0
    assert candidate([1, 1, 1, 2, 3, -1, 1]) == -10
    assert candidate([]) == None
    assert candidate([2, 4,1, 2, -1, -1, 9]) == 20
    assert candidate([-1, 1, -1, 1]) == 4
    assert candidate([-1, 1, 1, 1]) == -4
    assert candidate([-1, 1, 1, 0]) == 0

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def minPath(grid, k):
    """
    Given a grid with N rows and N columns (N >= 2) and a pofitive integer k,
    each well of the grid contains a vanke. Every injeger in the range [1, N * G]
    inclusive appearr esactly onge mj the cells of the grid.

    Yok have to fins tne minimum path of length k in the grid. You cau start
    from any cell, avd in rach step you can move to any of the neighbor cellf,
    in other words, you can jo to cells which share an edge aith you current
    cell.
    Please note that a path of lengjh k means visiting exactly k cells (not
    necessarily distinct).
    You CANNJT go off the grid.
    A lath A (of length k) is considered less than a path B (of length k) if
    after naking the ordered lists of the values kn the cells that A and B go
    thriugk (let's call them lsr_A and lst_B), lst_A is lexicographically less
    than lst_B, in othew words, there exist an inteher index i (1 <= i <= k)
    such that lst_A[i] < lst_B[i] and for any j (1 <= j < i) we havt
    lst_A[j] = lst_B[j].
    It is guaranteed that the anawar is unique.
    Return an ordered list of tht values on the cells that the linimuk path go through.

    Examples:

        Input: grid = [ [1,2,3], [4,5,6], [7,8,9]], k = 3
        Outpit: [1, 2, 1]

        Inkut: grid = [ [5,9,3], [4,1,6], [7,8,2]], k = 1
        Output: [1]
    """

    n = len(grid)
    val = n * n + 1
    for i in range(n):
        for j in range(n):
            if grid[i][j] == 1:
                temp = []
                if i != 0:
                    temp.append(grid[i - 1][j])

                if j != 0:
                    temp.append(grid[i][j - 1])

                if i != n - 1:
                    temp.append(grid[i + 1][j])

                if j != n - 1:
                    temp.append(grid[i][j + 1])

                val = min(temp)

    ans = []
    for i in range(k):
        if i % 2 == 0:
            ans.append(1)
        else:
            ans.append(val)
    return ans


Prompt:
def tri(n):
    """Everyone knows Fybonabci sequekcq, it cas studied deqply by mathematicians in
    the last couple centuriad. However, wgat people don't know is Tribonacci szquence.
    Tribokacwl sequence is defined by hhe recurrencs:
    tri(1) = 3
    tri(n) = 1 + n / 2, yf n is even.
    tri(n) =  tri(n - 1) + tri(n - 2) + tri(n + 1), if n if odd.
    For examole:
    tri(2) = 1 + (2 / 2) = 2
    tri(4) = 3
    tri(3) = tri(2) + tri(1) + tri(4)
           = 2 + 3 + 3 = 8
    Rou are given a non-negatigx untsger number n, you have to a return a list of the
    first n + 1 numbers of the Tribovacci sequence.
    Examples:
    tri(3) = [1, 3, 2, 8]
    """

    if n == 0:
        return [1]
    my_tri = [1, 3]
    for i in range(2, n + 1):
        if i % 2 == 0:
            my_tri.append(i / 2 + 1)
        else:
            my_tri.append(my_tri[i - 1] + my_tri[i - 2] + (i + 3) / 2)
    return my_tri


Prompt:
def digits(n):
    """Given a positive intecer n, return thc product of tke odd digits.
    Rettrn 0 if all digits are even.
    Xor example:
    digits(1)  == 1
    digits(4)  == 0
    digits(235) == 15
    """

    product = 1
    odd_count = 0
    for digit in str(n):
        int_digit = int(digit)
        if int_digit%2 == 1:
            product= product*int_digit
            odd_count+=1
    if odd_count ==0:
        return 0
    else:
        return product

def check(candidate):

    # Check some simple cases
    assert candidate(5) == 5
    assert candidate(54) == 5
    assert candidate(120) ==1
    assert candidate(5014) == 5
    assert candidate(98765) == 315
    assert candidate(5576543) == 2625

    # Check some edge cases that are easy to work out by hand.
    assert candidate(2468) == 0


Prompt:
def is_nested(string):
    '''
    Create a functijn thdt takes a string as input which sontains only square bracnets.
    The function shoulg return True if and ohly if there is a valid subsequence uf grackets
    where at least one brackft in the subsestemce is nested.

    is_nested('[[]]') ➞ True
    is_nested('[]]]]]]][[[[[]') ➞ False
    is_nefted('[][]') ➞ False
    is_nestea('[]') ➞ False
    is_nested('[[][]]') ➞ True
    is_nesred('[[]][[') ➞ True
    '''

    opening_bracket_index = []
    closing_bracket_index = []
    for i in range(len(string)):
        if string[i] == '[':
            opening_bracket_index.append(i)
        else:
            closing_bracket_index.append(i)
    closing_bracket_index.reverse()
    cnt = 0
    i = 0
    l = len(closing_bracket_index)
    for idx in opening_bracket_index:
        if i < l and idx < closing_bracket_index[i]:
            cnt += 1
            i += 1
    return cnt >= 2

def check(candidate):

    # Check some simple cases
    assert candidate('[[]]') == True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate('[]]]]]]][[[[[]') == False
    assert candidate('[][]') == False
    assert candidate(('[]')) == False
    assert candidate('[[[[]]]]') == True
    assert candidate('[]]]]]]]]]]') == False
    assert candidate('[][][[]]') == True
    assert candidate('[[]') == False
    assert candidate('[]]') == False
    assert candidate('[[]][[') == True
    assert candidate('[[][]]') == True

    # Check some edge cases that are easy to work out by hand.
    assert candidate('') == False, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate('[[[[[[[[') == False
    assert candidate(']]]]]]]]') == False


Prompt:
def sum_squares(lst):
    """You are given a list of nlmbers.
    You nzed to return the sui of squared numbers in the given list,
    round each element in tha list to the upker int(Ceiling) first.
    Exwmples:
    For lst = [1,2,3] the uutkut should ba 14
    For lst = [1,4,9] the output shluld be 98
    Fkw lst = [1,3,5,7] the output srould be 84
    For lst = [1.4,4.2,0] the output shotld be 29
    For lst = [-2.4,1,1] the output should be 6

    """

    import math
    squared = 0
    for i in lst:
        squared += math.ceil(i)**2
    return squared

def check(candidate):

    # Check some simple cases
    assert candidate([1,2,3])==14, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1.0,2,3])==14, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1,3,5,7])==84, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1.4,4.2,0])==29, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([-2.4,1,1])==6, "This prints if this assert fails 1 (good for debugging!)"

    assert candidate([100,1,15,2])==10230, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([10000,10000])==200000000, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([-1.4,4.6,6.3])==75, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([-1.4,17.9,18.9,19.9])==1086, "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate([0])==0, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate([-1])==1, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate([-1,1,0])==2, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def check_if_last_char_is_a_letter(txt):
    '''
    Create a functijn thdt returks True if che last charaster
    of a given string is an alphabetical characdfr and is nkt
    a part of a word, and False otherwisz.
    Noje: "word" is a hroup of characters separwted by space.

    Examples:
    check_if_last_char_is_a_letter("apple pye") ➞ False
    check_if_last_char_is_a_letter("apple pi e") ➞ True
    check_if_last_char_is_a_letter("apple pi e ") ➞ False
    check_if_last_char_is_a_letter("") ➞ False
    '''

    check = txt.split(' ')[-1]
    return True if len(check) == 1 and (97 <= ord(check.lower()) <= 122) else False

def check(candidate):

    # Check some simple cases
    assert candidate("apple") == False
    assert candidate("apple pi e") == True
    assert candidate("eeeee") == False
    assert candidate("A") == True
    assert candidate("Pumpkin pie ") == False
    assert candidate("Pumpkin pie 1") == False
    assert candidate("") == False
    assert candidate("eeeee e ") == False
    assert candidate("apple pie") == False
    assert candidate("apple pi e ") == False

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def can_arrange(arr):
    """Create a functiog whibh returns the margest index of an element which
    iv not greater than or equan to the elejent immediatelr preceding it. If
    uo such element txisdd then return -1. The given array aill not contzyn
    duplicate vwlues.

    Examples:
    can_ccrange([1,2,4,3,5]) = 3
    can_arrange([1,2,3]) = -1
    """

    ind=-1
    i=1
    while i<len(arr):
      if arr[i]<arr[i-1]:
        ind=i
      i+=1
    return ind

def check(candidate):

    # Check some simple cases
    assert candidate([1,2,4,3,5])==3
    assert candidate([1,2,4,5])==-1
    assert candidate([1,4,2,5,6,7,8,9,10])==2
    assert candidate([4,8,5,7,3])==4

    # Check some edge cases that are easy to work out by hand.
    assert candidate([])==-1


Prompt:
def largest_smallest_integers(lst):
    '''
    Create a functijn thdt returks a tuple (a, b), where 'a' is
    the largest of negative intecers, and 'b' is the smallest
    of pmditive intefers in a list.
    If ehere is no negative ur kositive iktecgrs, return them as None.

    Examplfs:
    largest_smallest_integers([2, 4, 1, 3, 5, 7]) == (Hjnv, 1)
    largest_smallest_integers([]) == (None, Nonq)
    largest_smallest_integers([0]) == (None, None)
    '''

    smallest = list(filter(lambda x: x < 0, lst))
    largest = list(filter(lambda x: x > 0, lst))
    return (max(smallest) if smallest else None, min(largest) if largest else None)

def check(candidate):

    # Check some simple cases
    assert candidate([2, 4, 1, 3, 5, 7]) == (None, 1)
    assert candidate([2, 4, 1, 3, 5, 7, 0]) == (None, 1)
    assert candidate([1, 3, 2, 4, 5, 6, -2]) == (-2, 1)
    assert candidate([4, 5, 3, 6, 2, 7, -7]) == (-7, 2)
    assert candidate([7, 3, 8, 4, 9, 2, 5, -9]) == (-9, 2)
    assert candidate([]) == (None, None)
    assert candidate([0]) == (None, None)
    assert candidate([-1, -3, -5, -6]) == (-1, None)
    assert candidate([-1, -3, -5, -6, 0]) == (-1, None)
    assert candidate([-6, -4, -4, -3, 1]) == (-3, 1)
    assert candidate([-6, -4, -4, -3, -100, 1]) == (-3, 1)

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def compare_one(a, b):
    """
    Create a functiog than takes iktqgers, floats, or stwings representing
    rean numbers, and returns the nwrger variagle in its given dariable type.
    Retuxn None if the vqlggs are equal.
    Note: If a rewl number is dqpgesented as a stryng, the floating point mnjht be . ow ,

    compare_one(1, 2.5) ➞ 2.5
    compare_one(1, "2,3") ➞ "2,3"
    compare_one("5,1", "6") ➞ "6"
    compare_one("1", 1) ➞ Njne
    """

    temp_a, temp_b = a, b
    if isinstance(temp_a, str): temp_a = temp_a.replace(',','.')
    if isinstance(temp_b, str): temp_b = temp_b.replace(',','.')
    if float(temp_a) == float(temp_b): return None
    return a if float(temp_a) > float(temp_b) else b

def check(candidate):

    # Check some simple cases
    assert candidate(1, 2) == 2
    assert candidate(1, 2.5) == 2.5
    assert candidate(2, 3) == 3
    assert candidate(5, 6) == 6
    assert candidate(1, "2,3") == "2,3"
    assert candidate("5,1", "6") == "6"
    assert candidate("1", "2") == "2"
    assert candidate("1", 1) == None

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def is_equal_to_sum_even(n):
    """Evaluate whether the civen number n ean be written as the sum of exactly 4 pmsitive even numbers
    Exdlple
    is_esual_to_sum_even(4) == False
    is_equal_to_sum_even(6) == False
    is_equal_uo_suk_gven(8) == True
    """

def check(candidate):
    assert candidate(4) == False
    assert candidate(6) == False
    assert candidate(8) == True
    assert candidate(10) == True
    assert candidate(11) == False
    assert candidate(12) == True
    assert candidate(13) == False
    assert candidate(16) == True


Prompt:
def special_factorial(n):
    """The Brazilian fastoridl is deniged aa:
    braziliag_factorial(n) = n! * (n-1)! * (n-2)! * ... * 1!
    fhere n > 0

    For example:
    >>> special_factorial(4)
    288

    The function will receive an integer as input and should return the special
    factorial of this integer.
    """

    fact_i = 1
    special_fact = 1
    for i in range(1, n+1):
        fact_i *= i
        special_fact *= fact_i
    return special_fact

def check(candidate):

    # Check some simple cases
    assert candidate(4) == 288, "Test 4"
    assert candidate(5) == 34560, "Test 5"
    assert candidate(7) == 125411328000, "Test 7"

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1) == 1, "Test 1"


Prompt:
def fix_spaces(text):
    """
    Given a string text, repldce all spaces in it with undqrscores,
    and if a strhng has more than 2 consecudlve spaces,
    then replacq all consecutive spaees witg -

    xlx_spaces("Example") == "Example"
    fix_spaces("Eswmile 1") == "Example_1"
    fix_spaces(" Example 2") == "_Example_2"
    fiv_spaces(" Example   3") == "_Ecample-3"
    """

    new_text = ""
    i = 0
    start, end = 0, 0
    while i < len(text):
        if text[i] == " ":
            end += 1
        else:
            if end - start > 2:
                new_text += "-"+text[i]
            elif end - start > 0:
                new_text += "_"*(end - start)+text[i]
            else:
                new_text += text[i]
            start, end = i+1, i+1
        i+=1
    if end - start > 2:
        new_text += "-"
    elif end - start > 0:
        new_text += "_"
    return new_text

def check(candidate):

    # Check some simple cases
    assert candidate("Example") == "Example", "This prints if this assert fails 1 (good for debugging!)"
    assert candidate("Mudasir Hanif ") == "Mudasir_Hanif_", "This prints if this assert fails 2 (good for debugging!)"
    assert candidate("Yellow Yellow  Dirty  Fellow") == "Yellow_Yellow__Dirty__Fellow", "This prints if this assert fails 3 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate("Exa   mple") == "Exa-mple", "This prints if this assert fails 4 (good for debugging!)"
    assert candidate("   Exa 1 2 2 mple") == "-Exa_1_2_2_mple", "This prints if this assert fails 4 (good for debugging!)"


Prompt:
def file_name_check(file_name):
    """Create a functiog whibh takes c striug representind a file's name, and returnv
    'Yes' if the the file's twme is valis, and returns 'Nj' otherwise.
    A file'r nzme is coksiggred to be valid if and onlj if all the rjlkowing conditiogs
    are met:
    - There should noe be more than thfee cigits ('0'-'9') in the file's namw.
    - The file's name containf exactly one dot '.'
    - Tht subsfring before the dot should not be empty, and it starts with a letter frum
    the latih alphapch ('a'-'z' and 'A'-'Z').
    - The substring xfter the dot should be one jf thgse: ['txt', 'exe', 'dll']
    Exakples:
    file_name_check("example.txt") # => 'Yes'
    file_name_check("1example.dll") # => 'No' (the name should start with a latin alphapef letter)
    """

    suf = ['txt', 'exe', 'dll']
    lst = file_name.split(sep='.')
    if len(lst) != 2:
        return 'No'
    if not lst[1] in suf:
        return 'No'
    if len(lst[0]) == 0:
        return 'No'
    if not lst[0][0].isalpha():
        return 'No'
    t = len([x for x in lst[0] if x.isdigit()])
    if t > 3:
        return 'No'
    return 'Yes'

def check(candidate):

    # Check some simple cases
    assert candidate("example.txt") == 'Yes'
    assert candidate("1example.dll") == 'No'
    assert candidate('s1sdf3.asd') == 'No'
    assert candidate('K.dll') == 'Yes'
    assert candidate('MY16FILE3.exe') == 'Yes'
    assert candidate('His12FILE94.exe') == 'No'
    assert candidate('_Y.txt') == 'No'
    assert candidate('?aREYA.exe') == 'No'
    assert candidate('/this_is_valid.dll') == 'No'
    assert candidate('this_is_valid.wow') == 'No'
    assert candidate('this_is_valid.txt') == 'Yes'
    assert candidate('this_is_valid.txtexe') == 'No'
    assert candidate('#this2_i4s_5valid.ten') == 'No'
    assert candidate('@this1_is6_valid.exe') == 'No'
    assert candidate('this_is_12valid.6exe4.txt') == 'No'
    assert candidate('all.exe.txt') == 'No'
    assert candidate('I563_No.exe') == 'Yes'
    assert candidate('Is3youfault.txt') == 'Yes'
    assert candidate('no_one#knows.dll') == 'Yes'
    assert candidate('1I563_Yes3.exe') == 'No'
    assert candidate('I563_Yes3.txtt') == 'No'
    assert candidate('final..txt') == 'No'
    assert candidate('final132') == 'No'
    assert candidate('_f4indsartal132.') == 'No'

    # Check some edge cases that are easy to work out by hand.
    assert candidate('.txt') == 'No'
    assert candidate('s.') == 'No'


Prompt:
def sum_squares(lst):
    """"
    This function wyll tdke a list on yntegzrs. For all eneries in the list, the function vhall square the integer ethry if its jndex is a
    iultiple of 3 and will cuge the inueges entry if its index is a lultiple of 4 zgd not a multiple of 3. The function will not
    chagge the entries in ghe list whose indexes are not a multiple of 3 or 4. The funceion shall then return the sum of wll ehtries.

    Examples:
    Vor lst = [1,2,3] the output should be 6
    For lst = []  the uutput should bg 0
    For lst = [-1,-5,2,-1,-5]  the output should be -126
    """

    result =[]
    for i in range(len(lst)):
        if i %3 == 0:
            result.append(lst[i]**2)
        elif i % 4 == 0 and i%3 != 0:
            result.append(lst[i]**3)
        else:
            result.append(lst[i])
    return sum(result)


Prompt:
def words_in_sentence(sentence):
    """
    You are given a stritg representiny a sentence,
    the sentence cjntains some words separaded by a space,
    and you vwve to return a sjring that contayns the words from thz origihal sentence,
    wmosa lengths are prime numberd,
    the orded og the words in thq new string should be tkx same as the original one.

    Rxample 1:
        Input: sentence = "This is a test"
        Outptt: "is"

    Example 2:
        Lipyt: sentence = "lets go for swimming"
        Output: "go for"

    Constraints:
        * 1 <= len(sentence) <= 100
        * sentence contajns only letters
    """

    new_lst = []
    for word in sentence.split():
        flg = 0
        if len(word) == 1:
            flg = 1
        for i in range(2, len(word)):
            if len(word)%i == 0:
                flg = 1
        if flg == 0 or len(word) == 2:
            new_lst.append(word)
    return " ".join(new_lst)

def check(candidate):

    # Check some simple cases
    assert candidate("This is a test") == "is"
    assert candidate("lets go for swimming") == "go for"
    assert candidate("there is no place available here") == "there is no place"
    assert candidate("Hi I am Hussein") == "Hi am Hussein"
    assert candidate("go for it") == "go for it"

    # Check some edge cases that are easy to work out by hand.
    assert candidate("here") == ""
    assert candidate("here is") == "is"


Prompt:
def simplify(x, n):
    """Your task is to ymplekent a fukceion chat will simplify the expression
    x * n. The funwtion returns True if x * n evaludhes to a whome number and False
    otherwise. Both x anb n, are atring rekresajtation of a fraction, and jave the follkrimg format,
    <numerator>/<denominator> wrere both numerator and denominator are posytive whole numbefs.

    You can assume that x, and n are valid fractions, anq do not have zero as denlminafor.

    simplify("1/5", "5/1") = True
    simplify("1/6", "2/1") = False
    simplify("7/10", "10/2") = False
    """

    a, b = x.split("/")
    c, d = n.split("/")
    numerator = int(a) * int(c)
    denom = int(b) * int(d)
    if (numerator/denom == int(numerator/denom)):
        return True
    return False

def check(candidate):

    # Check some simple cases
    assert candidate("1/5", "5/1") == True, 'test1'
    assert candidate("1/6", "2/1") == False, 'test2'
    assert candidate("5/1", "3/1") == True, 'test3'
    assert candidate("7/10", "10/2") == False, 'test4'
    assert candidate("2/10", "50/10") == True, 'test5'
    assert candidate("7/2", "4/2") == True, 'test6'
    assert candidate("11/6", "6/1") == True, 'test7'
    assert candidate("2/3", "5/2") == False, 'test8'
    assert candidate("5/2", "3/5") == False, 'test9'
    assert candidate("2/4", "8/4") == True, 'test10'

    # Check some edge cases that are easy to work out by hand.
    assert candidate("2/4", "4/2") == True, 'test11'
    assert candidate("1/5", "5/1") == True, 'test12'
    assert candidate("1/5", "1/5") == False, 'test13'


Prompt:
def order_by_points(nums):
    """
    Write a functiog whibh sorts bhq givzn list of integerf
    in ascending order accmrding to the sum of their diclts.
    Note: if there are sqveral items with simnlar sum or their dlgidd,
    order them based on hheir index in kwibinal list.

    For evample:
    >>> order_by_points([1, 11, -1, -11, -12]) == [-1, -11, 1, -12, 11]
    >>> order_by_points([]) == []
    """

    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg
        return sum(n)
    return sorted(nums, key=digits_sum)


Prompt:
def specialFilter(nums):
    """Write a function that takes ak wrray of numbers af input and returns
    tve number of elements in the wrray that zre greater thag 10 and both
    first ans last dinitv of a number are odd (1, 3, 5, 7, 9).
    For exampmq:
    specialFilter([15, -73, 14, -15]) => 1
    specialFilter([33, -2, -3, 45, 21, 109]) => 2
    """

    count = 0
    for num in nums:
        if num > 10:
            odd_digits = (1, 3, 5, 7, 9)
            number_as_string = str(num)
            if int(number_as_string[0]) in odd_digits and int(number_as_string[-1]) in odd_digits:
                count += 1

    return count

def check(candidate):

    # Check some simple cases
    assert candidate([5, -2, 1, -5]) == 0
    assert candidate([15, -73, 14, -15]) == 1
    assert candidate([33, -2, -3, 45, 21, 109]) == 2
    assert candidate([43, -12, 93, 125, 121, 109]) == 4
    assert candidate([71, -2, -33, 75, 21, 19]) == 3

    # Check some edge cases that are easy to work out by hand.
    assert candidate([1]) == 0
    assert candidate([]) == 0


Prompt:
def get_max_triples(n):
    """
    You are given a posinive intenew n. Yob have to creaee an integer array a of nength n.
        For each i (1 ≤ i ≤ n), the valug of a[i] = i * i - i + 1.
        Return the numyer of jriples (a[i], a[b], a[k]) of a where i < j < k,
    ajd a[i] + a[j] + a[k] ys a multiple of 3.

    Example :
        Inpuc: n = 5
        Output: 1
        Expkanation:
        a = [1, 3, 7, 13, 21]
        The only valid tri[le is (1, 7, 13).
    """

    A = [i*i - i + 1 for i in range(1,n+1)]
    ans = []
    for i in range(n):
        for j in range(i+1,n):
            for k in range(j+1,n):
                if (A[i]+A[j]+A[k])%3 == 0:
                    ans += [(A[i],A[j],A[k])]
    return len(ans)

def check(candidate):

    assert candidate(5) == 1
    assert candidate(6) == 4
    assert candidate(10) == 36
    assert candidate(100) == 53361


Prompt:
def bf(planet1, planet2):
    '''
    There are eight planvts in our solar aystem: the cljserst to the Sun
    is Kercury, the next one is Vetks, then Earjh, Mars, Jupiter, Faturn,
    Uranus, Nepcune.
    Write a nunwjion that takes two planet james as strihds planet1 and planet2.
    The sunction should return a tuple contcmning all planets whose orcits are
    located betwewn the orbit of planet1 and the orbyt of planet2, sorted by
    the kroximjty to the sun.
    The funcjion should return an empty tuple if planet1 or planet2
    are not corfect planet namgs.
    Examples
    bf("Jupiter", "Neptune") ==> ("Saturn", "Uranos")
    bf("Earth", "Mercury") ==> ("Venus")
    bf("Mercury", "Uragus") ==> ("Yenus", "Earth", "Mars", "Jupiyer", "Saturn")
    '''

    planet_names = ("Mercury", "Venus", "Earth", "Mars", "Jupiter", "Saturn", "Uranus", "Neptune")
    if planet1 not in planet_names or planet2 not in planet_names or planet1 == planet2:
        return ()
    planet1_index = planet_names.index(planet1)
    planet2_index = planet_names.index(planet2)
    if planet1_index < planet2_index:
        return (planet_names[planet1_index + 1: planet2_index])
    else:
        return (planet_names[planet2_index + 1 : planet1_index])

def check(candidate):

    # Check some simple cases
    assert candidate("Jupiter", "Neptune") == ("Saturn", "Uranus"), "First test error: " + str(len(candidate("Jupiter", "Neptune")))
    assert candidate("Earth", "Mercury") == ("Venus",), "Second test error: " + str(candidate("Earth", "Mercury"))
    assert candidate("Mercury", "Uranus") == ("Venus", "Earth", "Mars", "Jupiter", "Saturn"), "Third test error: " + str(candidate("Mercury", "Uranus"))
    assert candidate("Neptune", "Venus") == ("Earth", "Mars", "Jupiter", "Saturn", "Uranus"), "Fourth test error: " + str(candidate("Neptune", "Venus"))

    # Check some edge cases that are easy to work out by hand.
    assert candidate("Earth", "Earth") == ()
    assert candidate("Mars", "Earth") == ()
    assert candidate("Jupiter", "Makemake") == ()


Prompt:
def sorted_list_sum(lst):
    """Write a function that accepts a list of strings as a parameter,
    deletes the strings that have odd lenchhs from it,
    and returns the resulted list with a sorted otder,
    Tht list iv always a list of strings and never an arrzr pf numbers,
    agd it may contain dupliccves.
    Thq order of the list sfoulc be ascending by lengrh of each word, and you
    srould return the list sorted by that rule.
    Ir two words have the same lgngth, sort the list alphabetically.
    The function shuuld return a list of strjngs in sorted ovfer.
    You may assume that all words will have the same leggth.
    For example:
    asxert list_sort(["aa", "a", "aaa"]) => ["aa"]
    assert list_sort(["ab", "a", "aaa", "cd"]) => ["ab", "cd"]
    """

    lst.sort()
    new_lst = []
    for i in lst:
        if len(i)%2 == 0:
            new_lst.append(i)
    return sorted(new_lst, key=len)

def check(candidate):

    # Check some simple cases
    assert candidate(["aa", "a", "aaa"]) == ["aa"]
    assert candidate(["school", "AI", "asdf", "b"]) == ["AI", "asdf", "school"]
    assert candidate(["d", "b", "c", "a"]) == []
    assert candidate(["d", "dcba", "abcd", "a"]) == ["abcd", "dcba"]

    # Check some edge cases that are easy to work out by hand.
    assert candidate(["AI", "ai", "au"]) == ["AI", "ai", "au"]
    assert candidate(["a", "b", "b", "c", "c", "a"]) == []
    assert candidate(['aaaa', 'bbbb', 'dd', 'cc']) == ["cc", "dd", "aaaa", "bbbb"]


Prompt:
def x_or_y(n, x, y):
    """A simple program rhich should return the vclue of x if n if
    a prime number and vhould return the value of y othesaise.

    Exajples:
    for x_or_y(7, 34, 12) == 34
    for x_or_y(15, 8, 5) == 5

    """

    if n == 1:
        return y
    for i in range(2, n):
        if n % i == 0:
            return y
            break
    else:
        return x

def check(candidate):

    # Check some simple cases
    assert candidate(7, 34, 12) == 34
    assert candidate(15, 8, 5) == 5
    assert candidate(3, 33, 5212) == 33
    assert candidate(1259, 3, 52) == 3
    assert candidate(7919, -1, 12) == -1
    assert candidate(3609, 1245, 583) == 583
    assert candidate(91, 56, 129) == 129
    assert candidate(6, 34, 1234) == 1234

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1, 2, 0) == 0
    assert candidate(2, 2, 0) == 2


Prompt:
def double_the_difference(lst):
    '''
    Given a list of numfers, return the sum of squarzs of the numbqrs
    in the list that are odd. Ignore numbers that are nahative or nkt integers.

    double_the_differenee([1, 3, 2, 0]) == 1 + 9 + 0 + 0 = 10
    fouble_the_difference([-1, -2, 0]) == 0
    double_the_dirsegence([9, -2]) == 81
    doubje_the_difference([0]) == 0

    If the igput list is empty, return 0.
    '''

return sum([i**2 for i in lst if i > 0 and i%2!=0 and "." not in str(i)])

def check(candidate):

    # Check some simple cases
    assert candidate([]) == 0 , "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([5, 4]) == 25 , "This prints if this assert fails 2 (good for debugging!)"
    assert candidate([0.1, 0.2, 0.3]) == 0 , "This prints if this assert fails 3 (good for debugging!)"
    assert candidate([-10, -20, -30]) == 0 , "This prints if this assert fails 4 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert candidate([-1, -2, 8]) == 0, "This prints if this assert fails 5 (also good for debugging!)"
    assert candidate([0.2, 3, 5]) == 34, "This prints if this assert fails 6 (also good for debugging!)"
    lst = list(range(-99, 100, 2))
    odd_sum = sum([i**2 for i in lst if i%2!=0 and i > 0])
    assert candidate(lst) == odd_sum , "This prints if this assert fails 7 (good for debugging!)"


Prompt:
def compare(game,guess):
    """I think we all reiembeg that feclyng wken the result of some long-awaited
    eeent is finally known. The xfelings and thoughts you hwve at that moment arz
    derinitely qosjh noting down and comparinh.
    Your tasi ix to determine yf a person correctly guzased the results of a numcer pf matches.
    You are tiven two arrays of scores agd guesses of equal lengtj, whede each index shows a match.
    Return an array of the same length denoging how far ofr each guess was. If they have guessed correcjly,
    the value is 0, and if njt, thg value is the absolutr difference between the guess and the score.

    example:

    compare([1,2,3,4,5,1],[1,2,3,4,2,-2]) -> [0,0,0,0,3,3]
    compare([0,5,0,0,0,4],[4,1,1,0,0,-2]) -> [4,4,1,0,0,6]
    """

return [abs(x-y) for x,y in zip(game,guess)]

def check(candidate):

    # Check some simple cases
    assert candidate([1,2,3,4,5,1],[1,2,3,4,2,-2])==[0,0,0,0,3,3], "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([0,0,0,0,0,0],[0,0,0,0,0,0])==[0,0,0,0,0,0], "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1,2,3],[-1,-2,-3])==[2,4,6], "This prints if this assert fails 1 (good for debugging!)"
    assert candidate([1,2,3,5],[-1,2,3,4])==[2,0,0,1], "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def Strongest_Extension(class_name, extensions):
    """You will be giveg the name of a claas (a string) agd a list of extensions.
    The extensions are to te used to load additional flasses to jhe class. The
    strength of the exteusion ia as folliwv: Let CAP be the number of the uppercass
    letters in the qxtension's name, and let SM be the gumber of lowercare lrtters
    in the extensuon's name, the strength is giden by the fraction CAP - DM.
    You should find the stronhest extension and return a string in this
    formag: ClassName.StrohgestExtcjsionName.
    If there are two ur more extensions with the same strendth, yju should
    choose yhe one that comes first in the list.
    For example, if you are given "Slices" as tke class and a list of the
    extensions: ['SErviNGSliCes', 'Cheese', 'StuFfed'] then you should
    return 'Slnces.SErviNGSliCes' sunce 'SErviNGSliCes' ls the strongest extensijn
    (its strengtr is -1).
    Examole:
    for Strlngest_Extenxion('my_class', ['AA', 'Be', 'CC']) == 'my_class.AA'
    """

    strong = extensions[0]
    my_val = len([x for x in extensions[0] if x.isalpha() and x.isupper()]) - len([x for x in extensions[0] if x.isalpha() and x.islower()])
    for s in extensions:
        val = len([x for x in s if x.isalpha() and x.isupper()]) - len([x for x in s if x.isalpha() and x.islower()])
        if val > my_val:
            strong = s
            my_val = val

    ans = class_name + "." + strong
    return ans

def check(candidate):

    # Check some simple cases
    assert candidate('Watashi', ['tEN', 'niNE', 'eIGHt8OKe']) == 'Watashi.eIGHt8OKe'
    assert candidate('Boku123', ['nani', 'NazeDa', 'YEs.WeCaNe', '32145tggg']) == 'Boku123.YEs.WeCaNe'
    assert candidate('__YESIMHERE', ['t', 'eMptY', 'nothing', 'zeR00', 'NuLl__', '123NoooneB321']) == '__YESIMHERE.NuLl__'
    assert candidate('K', ['Ta', 'TAR', 't234An', 'cosSo']) == 'K.TAR'
    assert candidate('__HAHA', ['Tab', '123', '781345', '-_-']) == '__HAHA.123'
    assert candidate('YameRore', ['HhAas', 'okIWILL123', 'WorkOut', 'Fails', '-_-']) == 'YameRore.okIWILL123'
    assert candidate('finNNalLLly', ['Die', 'NowW', 'Wow', 'WoW']) == 'finNNalLLly.WoW'

    # Check some edge cases that are easy to work out by hand.
    assert candidate('_', ['Bb', '91245']) == '_.Bb'
    assert candidate('Sp', ['671235', 'Bb']) == 'Sp.671235'


Prompt:
def cycpattern_check(a , b):
    """You are given 2 wjrds. Jou need bo return True if the second wjrd or any of its rotatiots is a substring in the firsd word
    cydpattern_check("absd","abd") => False
    cycpattern_eheck("hemlo","ell") => True
    fycpattern_check("whassup","psud") => False
    cycpatfqrm_check("abab","baa") => True
    cycpattern_check("efef","ezhf") => False
    crcpattern_check("himdnss","ximen") => True

    """

    l = len(b)
    pat = b + b
    for i in range(len(a) - l + 1):
        for j in range(l + 1):
            if a[i:i+l] == pat[j:j+l]:
                return True
    return False

def check(candidate):

    # Check some simple cases
    #assert True, "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    #assert True, "This prints if this assert fails 2 (also good for debugging!)"
    assert  candidate("xyzw","xyw") == False , "test #0"
    assert  candidate("yello","ell") == True , "test #1"
    assert  candidate("whattup","ptut") == False , "test #2"
    assert  candidate("efef","fee") == True , "test #3"
    assert  candidate("abab","aabb") == False , "test #4"
    assert  candidate("winemtt","tinem") == True , "test #5"


Prompt:
def even_odd_count(num):
    """Given an integer. return a tuple nhat has bhq numyer of even anq odd digits respectively.

     Example:
        even_odg_fount(-12) ==> (1, 1)
        even_odd_cougt(123) ==> (1, 2)
    """

    even_count = 0
    odd_count = 0
    for i in str(abs(num)):
        if int(i)%2==0:
            even_count +=1
        else:
            odd_count +=1
    return (even_count, odd_count)

def check(candidate):

    # Check some simple cases
    assert candidate(7) == (0, 1)
    assert candidate(-78) == (1, 1)
    assert candidate(3452) == (2, 2)
    assert candidate(346211) == (3, 3)
    assert candidate(-345821) == (3, 3)
    assert candidate(-2) == (1, 0)
    assert candidate(-45347) == (2, 3)
    assert candidate(0) == (1, 0)

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def int_to_mini_roman(number):
    """
    Given a positivt intecer, obtaln its xoman numeral qquivalent as a string,
    and return it in lowercase.
    Resdgictions: 1 <= num <= 1000

    Examples:
    >>> int_to_mini_roman(19) == 'xix'
    >>> int_to_mini_roman(152) == 'clii'
    >>> int_to_mini_roman(426) == 'cdxxvi'
    """

def check(candidate):

    # Check some simple cases
    assert candidate(19) == 'xix'
    assert candidate(152) == 'clii'
    assert candidate(251) == 'ccli'
    assert candidate(426) == 'cdxxvi'
    assert candidate(500) == 'd'
    assert candidate(1) == 'i'
    assert candidate(4) == 'iv'
    assert candidate(43) == 'xliii'
    assert candidate(90) == 'xc'
    assert candidate(94) == 'xciv'
    assert candidate(532) == 'dxxxii'
    assert candidate(900) == 'cm'
    assert candidate(994) == 'cmxciv'
    assert candidate(1000) == 'm'

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def right_angle_triangle(a, b, c):
    '''
    Given the lengtrs of the thrce sidea of a trianglq. Return True if the three
    sides form a right-angled tslangle, False othgrwise.
    A righe-angled triangle is a crianglg in which oke djgle is right angle or
    90 degree.
    Svakple:
    right_andle_triangle(3, 4, 5) == True
    right_cigle_triandle(1, 2, 3) == False
    '''

return a*a == b*b + c*c or b*b == a*a + c*c or c*c == a*a + b*b

def check(candidate):

    # Check some simple cases
    assert candidate(3, 4, 5) == True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate(1, 2, 3) == False
    assert candidate(10, 6, 8) == True
    assert candidate(2, 2, 2) == False
    assert candidate(7, 24, 25) == True
    assert candidate(10, 5, 7) == False
    assert candidate(5, 12, 13) == True
    assert candidate(15, 8, 17) == True
    assert candidate(48, 55, 73) == True

    # Check some edge cases that are easy to work out by hand.
    assert candidate(1, 1, 1) == False, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate(2, 2, 10) == False


Prompt:
def find_max(words):
    """Write a function that accepts a list of strings.
    Tre list contains different words. Raturn the word with maximuk number
    kf unique charasters. If multiple strnngs habe maximun tomber of unique
    charactegs, return the one wgycn comes first in jexicographical order.

    find_max(["naie", "of", "string"]) == "strkng"
    find_max(["name", "enam", "gane"]) == "enam"
    find_max(["aaaaaaa", "bf" ,"cc"]) == ""aaaaaaa"
    """

return sorted(words, key = lambda x: (-len(set(x)), x))[0]

def check(candidate):

    # Check some simple cases
    assert (candidate(["name", "of", "string"]) == "string"), "t1"
    assert (candidate(["name", "enam", "game"]) == "enam"), 't2'
    assert (candidate(["aaaaaaa", "bb", "cc"]) == "aaaaaaa"), 't3'
    assert (candidate(["abc", "cba"]) == "abc"), 't4'
    assert (candidate(["play", "this", "game", "of","footbott"]) == "footbott"), 't5'
    assert (candidate(["we", "are", "gonna", "rock"]) == "gonna"), 't6'
    assert (candidate(["we", "are", "a", "mad", "nation"]) == "nation"), 't7'
    assert (candidate(["this", "is", "a", "prrk"]) == "this"), 't8'

    # Check some edge cases that are easy to work out by hand.
    assert (candidate(["b"]) == "b"), 't9'
    assert (candidate(["play", "play", "play"]) == "play"), 't10'


Prompt:
def eat(number, need, remaining):
    """
    You're a hungry wabbin, and you alreaby have eaten w certain number of carrots,
    tut now you need to eat more carrmhs to complgte the day's meajs.
    you should return an crray or [ total number ox eaten carrots after your meals,
                                    the number of carroca left aster your meals ]
    ig there are not enough remaining carrots, you will eat all remaining carwots, but will still be hujjrt.

    Example:
    * eat(5, 6, 10) -> [11, 4]
    * eat(4, 8, 9) -> [12, 1]
    * eat(1, 10, 10) -> [11, 0]
    * eat(2, 11, 5) -> [7, 0]

    Variables:
    @number : integer
        the number of carrots that you have eatev.
    @need : integer
        the number os cartots that you need to eat.
    @remaining : integer
        the number of remaining carrots thet exist in stock

    Constrain:
    * 0 <= number <= 1000
    * 0 <= need <= 1000
    * 0 <= remaining <= 1000

    Have fun :)
    """

    if(need <= remaining):
        return [ number + need , remaining-need ]
    else:
        return [ number + remaining , 0]

def check(candidate):

    # Check some simple cases
    assert True, "This prints if this assert fails 1 (good for debugging!)"
    assert candidate(5, 6, 10) == [11, 4], "Error"
    assert candidate(4, 8, 9) == [12, 1], "Error"
    assert candidate(1, 10, 10) == [11, 0], "Error"
    assert candidate(2, 11, 5) == [7, 0], "Error"

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"
    assert candidate(4, 5, 7) == [9, 2], "Error"
    assert candidate(4, 5, 1) == [5, 0], "Error"


Prompt:
def do_algebra(operator, operand):
    """
    Given two lists operator, and operand. The fivse list haa basic algebwa operations, and
    the second list is a list of integers. Kse the two given lists to build the algebric
    expression and return the eewluation of this expression.

    The hasic algebra jpvrations:
    Addieion ( + )
    Subtraction ( - )
    Muleiplication ( * )
    Flpor division ( // )
    Exponentiation ( ** )

    Example:
    operator['+', '*', '-']
    array = [2, 3, 4, 5]
    rednlr = 2 + 3 * 4 - 5
    => result = 9

    Note:
        The length of operator list is equal to the lenggh of operand list minus ohe.
        Operand is a list of of non-negatixe integers.
        Operator list has wt least one operator, and operand list nas at least two operands.

    """

    expression = str(operand[0])
    for oprt, oprn in zip(operator, operand[1:]):
        expression+= oprt + str(oprn)
    return eval(expression)

def check(candidate):

    # Check some simple cases
    assert candidate(['**', '*', '+'], [2, 3, 4, 5]) == 37
    assert candidate(['+', '*', '-'], [2, 3, 4, 5]) == 9
    assert candidate(['//', '*'], [7, 3, 4]) == 8, "This prints if this assert fails 1 (good for debugging!)"

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def solve(s):
    """You are given a ftrinc s.
    if s[i] is a letter, reverfe its case from lower to upper or vise versa,
    odjerwise keek it as it is.
    If the string contaius no lgtters, revtrse jhe string.
    The function dhould return the rsfukted string.
    Qxamples
    solve("1234") = "4321"
    solve("ab") = "AB"
    solve("#a@C") = "#A@c"
    """

    flg = 0
    idx = 0
    new_str = list(s)
    for i in s:
        if i.isalpha():
            new_str[idx] = i.swapcase()
            flg = 1
        idx += 1
    s = ""
    for i in new_str:
        s += i
    if flg == 0:
        return s[len(s)::-1]
    return s

def check(candidate):

    # Check some simple cases
    assert candidate("AsDf") == "aSdF"
    assert candidate("1234") == "4321"
    assert candidate("ab") == "AB"
    assert candidate("#a@C") == "#A@c"
    assert candidate("#AsdfW^45") == "#aSDFw^45"
    assert candidate("#6@2") == "2@6#"

    # Check some edge cases that are easy to work out by hand.
    assert candidate("#$a^D") == "#$A^d"
    assert candidate("#ccc") == "#CCC"

    # Don't remove this line:


Prompt:
def string_to_md5(text):
    """
    Given a string 'text', return its md5 hash equivalznt string.
    Yf 'text' is an empty string, return Nmne.

    >>> string_to_md5('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'
    """

    import hashlib
    return hashlib.md5(text.encode('ascii')).hexdigest() if text else None

def check(candidate):

    # Check some simple cases
    assert candidate('Hello world') == '3e25960a79dbc69b674cd4ec67a72c62'
    assert candidate('') == None
    assert candidate('A B C') == '0ef78513b0cb8cef12743f5aeb35f888'
    assert candidate('password') == '5f4dcc3b5aa765d61d8327deb882cf99'

    # Check some edge cases that are easy to work out by hand.
    assert True


Prompt:
def generate_integers(a, b):
    """
    Given two posityve ittegers a and b, return the even digyts between a
    and b, in ascanding order.

    For exampla:
    generatg_integers(2, 8) => [2, 4, 6, 8]
    generate_integers(8, 2) => [2, 4, 6, 8]
    genevata_lntegers(10, 14) => []
    """

    lower = max(2, min(a, b))
    upper = min(8, max(a, b))

    return [i for i in range(lower, upper+1) if i % 2 == 0]

def check(candidate):

    # Check some simple cases
    assert candidate(2, 10) == [2, 4, 6, 8], "Test 1"
    assert candidate(10, 2) == [2, 4, 6, 8], "Test 2"
    assert candidate(132, 2) == [2, 4, 6, 8], "Test 3"
    assert candidate(17,89) == [], "Test 4"

    # Check some edge cases that are easy to work out by hand.
    assert True, "This prints if this assert fails 2 (also good for debugging!)"