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

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

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

        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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_closest_vowel("ab") ==> ""
    """
    if len(word) < 3:

    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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_closest_vowel("ab") ==> ""
    """
    if len(word) < 3:
        return ""

    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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_closest_vowel("ab") ==> ""
    """
    if len(word) < 3:
        return ""

    vowels = {"a", "e", "i", "o", "u", "A", "E", 'O', 'U', 'I'}

        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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_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+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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_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:

                return word[i]
    return ""

if (word[i+1] not in vowels) and (word[i-1] not in vowels):

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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_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):

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 get_closest_vowel(word):
    """You are given a word. Your task is to find the closest vowel that stands between
    two consonants from the right side of the word (case sensitive).

    Vowels in the beginning and ending doesn't count. Return empty string if you didn't
    find any vowel met the above condition.

    You may assume that the given string contains English letter only.

    Example:
    get_closest_vowel("yogurt") ==> "u"
    get_closest_vowel("FULL") ==> "U"
    get_closest_vowel("quick") ==> ""
    get_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]

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 two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''

        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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):

        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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0

            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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:

                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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':

            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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1

                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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:

            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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1

                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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_parens([')', ')']) == 'No'
    '''
    def check(s):
        val = 0
        for i in s:
            if i == '(':
                val = val + 1
            else:
                val = val - 1
            if val < 0:

        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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_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

    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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_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

    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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_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]

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 match_parens(lst):
    '''
    You are given a list of two strings, both strings consist of open
    parentheses '(' or close parentheses ')' only.
    Your job is to check if it is possible to concatenate the two strings in
    some order, that the resulting string will be good.
    A string S is considered to be good if and only if all parentheses in S
    are balanced. For example: the string '(())()' is good, while the string
    '())' is not.
    Return 'Yes' if there's a way to make a good string, and return 'No' otherwise.

    Examples:
    match_parens(['()(', ')']) == 'Yes'
    match_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 integers and a positive integer k, return a sorted list
    of length 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]

    Example 3:

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

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

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


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length 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]

    Example 3:

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

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

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


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length 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]

    Example 3:

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

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


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length 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]

    Example 3:

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

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []
    arr.sort()


Prompt:
def maximum(arr, k):
    """
    Given an array arr of integers and a positive integer k, return a sorted list
    of length 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]

    Example 3:

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

    Note:
        1. The length of the array will be in the range of [1, 1000].
        2. The elements in the array will be in the range of [-1000, 1000].
        3. 0 <= k <= len(arr)
    """
    if k == 0:
        return []
    arr.sort()
    ans = arr[-k:]


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

    Examples
    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-empty array of integers arr and an integer k, return
    the sum of the elements with at most two digits from the first k elements of arr.

    Example:

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

    Constraints:
        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 positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

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

        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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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:

    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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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 = []

        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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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:

    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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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]

        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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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:

            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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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:

        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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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

            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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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:

        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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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

            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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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:

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 get_odd_collatz(n):
    """
    Given a positive integer n, return a sorted list that has the odd numbers in collatz sequence.

    The Collatz conjecture is a conjecture in mathematics that concerns a sequence defined
    as follows: start with any positive integer n. Then each term is obtained from the
    previous term as follows: if the previous term is even, the next term is one half of
    the previous term. If the previous term is odd, the next term is 3 times the previous
    term plus 1. The conjecture is that no matter what value of n, the sequence will always reach 1.

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

    For example:
    get_odd_collatz(5) returns [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))

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 write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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
    """

        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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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:

        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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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 = 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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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('-')

        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

month, day, year = int(month), int(day), int(year)

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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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)

            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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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:

        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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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

            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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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:

        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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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

            return False
        if month == 2 and day < 1 or day > 29:
            return False
    except:
        return False

    return True

if month in [4,6,9,11] and day < 1 or day > 30:

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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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:

        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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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

            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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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:

    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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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

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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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:

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 valid_date(date):
    """You have to write a function which validates a given date string and
    returns True if the date is valid otherwise False.
    The date is valid if all of the following rules are satisfied:
    1. The date string is not empty.
    2. The number of days is not less than 1 or higher than 31 days for months 1,3,5,7,8,10,12. And the number of days is not less than 1 or higher than 30 days for months 4,6,9,11. And, the number of days is not less than 1 or higher than 29 for the month 2.
    3. The months should not be less than 1 or higher than 12.
    4. The date should be in the format: mm-dd-yyyy

    for example:
    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

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 of words, return a list of words split on whitespace, if no whitespaces exists in the text you
    should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
    alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Examples
    split_words("Hello world!") ➞ ["Hello", "world!"]
    split_words("Hello,world!") ➞ ["Hello", "world!"]
    split_words("abcdef") == 3
    '''

        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 split_words(txt):
    '''
    Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
    should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
    alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Examples
    split_words("Hello world!") ➞ ["Hello", "world!"]
    split_words("Hello,world!") ➞ ["Hello", "world!"]
    split_words("abcdef") == 3
    '''
    if " " in txt:

    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 split_words(txt):
    '''
    Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
    should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
    alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Examples
    split_words("Hello world!") ➞ ["Hello", "world!"]
    split_words("Hello,world!") ➞ ["Hello", "world!"]
    split_words("abcdef") == 3
    '''
    if " " in txt:
        return txt.split()

        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 split_words(txt):
    '''
    Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
    should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
    alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Examples
    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:

    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 split_words(txt):
    '''
    Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
    should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
    alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Examples
    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()

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 split_words(txt):
    '''
    Given a string of words, return a list of words split on whitespace, if no whitespaces exists in the text you
    should split on commas ',' if no commas exists you should return the number of lower-case letters with odd order in the
    alphabet, ord('a') = 0, ord('b') = 1, ... ord('z') = 25
    Examples
    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 numbers, return whether or not they are sorted
    in ascending order. If list has more than 1 duplicate of the same
    number, return False. Assume no negative numbers and only integers.

    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_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True
    is_sorted([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
    '''

    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

count_digit = dict([(i, 0) for i in lst])

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 is_sorted(lst):
    '''
    Given a list of numbers, return whether or not they are sorted
    in ascending order. If list has more than 1 duplicate of the same
    number, return False. Assume no negative numbers and only integers.

    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_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True
    is_sorted([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])

        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 is_sorted(lst):
    '''
    Given a list of numbers, return whether or not they are sorted
    in ascending order. If list has more than 1 duplicate of the same
    number, return False. Assume no negative numbers and only integers.

    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_sorted([1, 2, 3, 4, 5, 6, 7]) ➞ True
    is_sorted([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:

    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!)"