Question:
def get_max_triples(n):
    """
    You are given a positive integer n. You have to create an integer array a of length n.
        For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.
        Return the number of triples (a[i], a[j], a[k]) of a where i < j < k,
    and a[i] + a[j] + a[k] is a multiple of 3.

    Example :
        Input: n = 5
        Output: 1
        Explanation:
        a = [1, 3, 7, 13, 21]
        The only valid triple is (1, 7, 13).
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def get_max_triples(n):
    """
    You are given a positive integer n. You have to create an integer array a of length n.
        For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.
        Return the number of triples (a[i], a[j], a[k]) of a where i < j < k,
    and a[i] + a[j] + a[k] is a multiple of 3.

    Example :
        Input: n = 5
        Output: 1
        Explanation:
        a = [1, 3, 7, 13, 21]
        The only valid triple is (1, 7, 13).
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def get_max_triples(n):
    """
    You are given a positive integer n. You have to create an integer array a of length n.
        For each i (1 ≤ i ≤ n), the value of a[i] = i * i - i + 1.
        Return the number of triples (a[i], a[j], a[k]) of a where i < j < k,
    and a[i] + a[j] + a[k] is a multiple of 3.
    """
    # The formula is a[i] = i^2 - i + 1.
    # We need to find the number of triples (i, j, k) such that 1 <= i < j < k <= n
    # and (a[i] + a[j] + a[k]) % 3 == 0.

    # Let's analyze a[i] % 3:
    # i = 1: 1^2 - 1 + 1 = 1  => 1 % 3 = 1
    # i = 2: 2^2 - 2 + 1 = 3  => 3 % 3 = 0
    # i = 3: 3^2 - 3 + 1 = 7  => 7 % 3 = 1
    # i = 4: 4^2 - 4 + 1 = 13 => 13 % 3 = 1
    # i = 5: 5^2 - 5 + 1 = 21 => 21 % 3 = 0
    # i = 6: 6^2 - 6 + 1 = 31 => 31 % 3 = 1

    # Pattern of a[i] % 3 for i = 1, 2, 3, 4, 5, 6...
    # i % 3 == 1: a[i] = 1 - 1 + 1 = 1 (mod 3)
    # i % 3 == 2: a[i] = 4 - 2 + 1 = 3 = 0 (mod 3)
    # i % 3 == 0: a[i] = 0 - 0 + 1 = 1 (mod 3)

    # So, a[i] % 3 is 0 if i % 3 == 2, and 1 otherwise.
    # Let c0 be the count of elements where a[i] % 3 == 0.
    # Let c1 be the count of elements where a[i] % 3 == 1.
    # Let c2 be the count of elements where a[i] % 3 == 2.

    # From the pattern:
    # c0 = count of i in [1, n] such that i % 3 == 2.
    # c1 = count of i in [1, n] such that i % 3 == 1 or i % 3 == 0.
    # c2 = 0 (since a[i] % 3 is never 2).

    c0 = n // 3 if n % 3 != 2 else (n // 3) + 1
    # More simply:
    c0 = (n + 1) // 3 if (n + 1) % 3 == 0 else (n // 3) # This is wrong.
    # Correct way to count i % 3 == 2 in range [1, n]:
    c0 = (n - 2) // 3 + 1 if n >= 2 else 0
    c1 = n - c0
    c2 = 0

    # For (a[i] + a[j] + a[k]) % 3 == 0:
    # Possible combinations of (rem_i, rem_j, rem_k):
    # 1. (0, 0, 0) -> combinations = c0 * (c0 - 1) * (c0 - 2) // 6
    # 2. (1, 1, 1) -> combinations = c1 * (c1 - 1) * (c1 - 2) // 6
    # 3. (0, 1, 2) -> combinations = c0 * c1 * c2
    # 4. (0, 0, 3) - not possible
    # Since c2 = 0, only (0,0,0) and (1,1,1) are possible.

    ans = 0
    if c0 >= 3:
        ans += (c0 * (c0 - 1) * (c0 - 2)) // 6
    if c1 >= 3:
        ans += (c1 * (c1 - 1) * (c1 - 2)) // 6

    return ans


Question:
from typing import List

def remove_duplicates(numbers: List[int]) -> List[int]:
    """ From a list of integers, remove all elements that occur more than once.
    Keep order of elements left the same as in the input.
    >>> remove_duplicates([1, 2, 3, 2, 4])
    [1, 3, 4]
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

from typing import List

def remove_duplicates(numbers: List[int]) -> List[int]:
    """ From a list of integers, remove all elements that occur more than once.
    Keep order of elements left the same as in the input.
    >>> remove_duplicates([1, 2, 3, 2, 4])
    [1, 3, 4]
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

from typing import List
from collections import Counter

def remove_duplicates(numbers: List[int]) -> List[int]:
    """ From a list of integers, remove all elements that occur more than once.
    Keep order of elements left the same as in the input.
    >>> remove_duplicates([1, 2, 3, 2, 4])
    [1, 3, 4]
    """
    counts = Counter(numbers)
    return [x for x in numbers if counts[x] == 1]


Question:
def add(x: int, y: int):
    """Add two numbers x and y
    >>> add(2, 3)
    5
    >>> add(5, 7)
    12
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def add(x: int, y: int):
    """Add two numbers x and y
    >>> add(2, 3)
    5
    >>> add(5, 7)
    12
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def add(x: int, y: int):
    """Add two numbers x and y
    >>> add(2, 3)
    5
    >>> add(5, 7)
    12
    """
    return x + y


Question:
def encode_shift(s: str):
    """
    returns encoded string by shifting every character by 5 in the alphabet.
    """
    return "".join([chr(((ord(ch) + 5 - ord("a")) % 26) + ord("a")) for ch in s])

def decode_shift(s: str):
    """
    takes as input string encoded with encode_shift function. Returns decoded string.
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def encode_shift(s: str):
    """
    returns encoded string by shifting every character by 5 in the alphabet.
    """
    return "".join([chr(((ord(ch) + 5 - ord("a")) % 26) + ord("a")) for ch in s])

def decode_shift(s: str):
    """
    takes as input string encoded with encode_shift function. Returns decoded string.
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def encode_shift(s: str):
    """
    returns encoded string by shifting every character by 5 in the alphabet.
    """
    return "".join([chr(((ord(ch) + 5 - ord("a")) % 26) + ord("a")) for ch in s])

def decode_shift(s: str):
    """
    takes as input string encoded with encode_shift function. Returns decoded string.
    """
    return "".join([chr(((ord(ch) - 5 - ord("a")) % 26) + ord("a")) for ch in s])


Question:
def is_multiply_prime(a):
    """Write a function that returns true if the given number is the multiplication of 3 prime numbers
    and false otherwise.
    Knowing that (a) is less then 100.
    Example:
    is_multiply_prime(30) == True
    30 = 2 * 3 * 5
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def is_multiply_prime(a):
    """Write a function that returns true if the given number is the multiplication of 3 prime numbers
    and false otherwise.
    Knowing that (a) is less then 100.
    Example:
    is_multiply_prime(30) == True
    30 = 2 * 3 * 5
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def is_multiply_prime(a):
    """Write a function that returns true if the given number is the multiplication of 3 prime numbers
    and false otherwise.
    Knowing that (a) is less then 100.
    Example:
    is_multiply_prime(30) == True
    30 = 2 * 3 * 5
    """
    def is_prime(n):
        if n < 2:
            return False
        for i in range(2, int(n**0.5) + 1):
            if n % i == 0:
                return False
        return True

    primes = [i for i in range(2, 100) if is_prime(i)]

    for i in range(len(primes)):
        for j in range(i, len(primes)):
            for k in range(j, len(primes)):
                if primes[i] * primes[j] * primes[k] == a:
                    return True
    return False


Question:
def string_sequence(n: int) -> str:
    """ Return a string containing space-delimited numbers starting from 0 upto n inclusive.
    >>> string_sequence(0)
    '0'
    >>> string_sequence(5)
    '0 1 2 3 4 5'
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def string_sequence(n: int) -> str:
    """ Return a string containing space-delimited numbers starting from 0 upto n inclusive.
    >>> string_sequence(0)
    '0'
    >>> string_sequence(5)
    '0 1 2 3 4 5'
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def string_sequence(n: int) -> str:
    """ Return a string containing space-delimited numbers starting from 0 upto n inclusive.
    >>> string_sequence(0)
    '0'
    >>> string_sequence(5)
    '0 1 2 3 4 5'
    """
    return ' '.join(str(i) for i in range(n + 1))


Question:
def encode_cyclic(s: str):
    """
    returns encoded string by cycling groups of three characters.
    """
    # split string to groups. Each of length 3.
    groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) // 3)]
    # cycle elements in each group. Unless group has fewer elements than 3.
    groups = [(group[1:] + group[0]) if len(group) == 3 else group for group in groups]
    return "".join(groups)

def decode_cyclic(s: str):
    """
    takes as input string encoded with encode_cyclic function. Returns decoded string.
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def encode_cyclic(s: str):
    """
    returns encoded string by cycling groups of three characters.
    """
    # split string to groups. Each of length 3.
    groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) // 3)]
    # cycle elements in each group. Unless group has fewer elements than 3.
    groups = [(group[1:] + group[0]) if len(group) == 3 else group for group in groups]
    return "".join(groups)

def decode_cyclic(s: str):
    """
    takes as input string encoded with encode_cyclic function. Returns decoded string.
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def encode_cyclic(s: str):
    """
    returns encoded string by cycling groups of three characters.
    """
    # split string to groups. Each of length 3.
    groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) // 3)]
    # cycle elements in each group. Unless group has fewer elements than 3.
    groups = [(group[1:] + group[0]) if len(group) == 3 else group for group in groups]
    return "".join(groups)

def decode_cyclic(s: str):
    """
    takes as input string encoded with encode_cyclic function. Returns decoded string.
    """
    # split string to groups. Each of length 3.
    groups = [s[(3 * i):min((3 * i + 3), len(s))] for i in range((len(s) + 2) // 3)]
    # reverse cycle elements in each group. Unless group has fewer elements than 3.
    # If group is 'abc' (encoded from 'cab'), it should become 'cab'.
    # The encode_cyclic logic is: group[0] moves to the end.
    # To reverse: the last element moves to the front.
    groups = [(group[-1] + group[:-1]) if len(group) == 3 else group for group in groups]
    return "".join(groups)


Question:
def check_if_last_char_is_a_letter(txt):
    '''
    Create a function that returns True if the last character
    of a given string is an alphabetical character and is not
    a part of a word, and False otherwise.
    Note: "word" is a group of characters separated by space.

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

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def check_if_last_char_is_a_letter(txt):
    '''
    Create a function that returns True if the last character
    of a given string is an alphabetical character and is not
    a part of a word, and False otherwise.
    Note: "word" is a group of characters separated by space.

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

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def check_if_last_char_is_a_letter(txt):
    '''
    Create a function that returns True if the last character
    of a given string is an alphabetical character and is not
    a part of a word, and False otherwise.
    Note: "word" is a group of characters separated by space.

    Examples:
    check_if_last_char_is_a_letter("apple pie") ➞ False
    check_if_last_char_is_a_letter("apple pi e") ➞ True
    check_if_last_char_is_a_letter("apple pi e ") ➞ False
    check_if_last_char_is_a_letter("") ➞ False
    '''
    if not txt:
        return False

    # The last character must be a letter
    if not txt[-1].isalpha():
        return False

    # To not be part of a word, it must be preceded by a space
    # or be the only character in the string.
    if len(txt) == 1 or txt[-2] == ' ':
        return True

    return False


Question:
from typing import List, Optional

def longest(strings: List[str]) -> Optional[str]:
    """ Out of list of strings, return the longest one. Return the first one in case of multiple
    strings of the same length. Return None in case the input list is empty.
    >>> longest([])

    >>> longest(['a', 'b', 'c'])
    'a'
    >>> longest(['a', 'bb', 'ccc'])
    'ccc'
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

from typing import List, Optional

def longest(strings: List[str]) -> Optional[str]:
    """ Out of list of strings, return the longest one. Return the first one in case of multiple
    strings of the same length. Return None in case the input list is empty.
    >>> longest([])

    >>> longest(['a', 'b', 'c'])
    'a'
    >>> longest(['a', 'bb', 'ccc'])
    'ccc'
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

from typing import List, Optional

def longest(strings: List[str]) -> Optional[str]:
    """ Out of list of strings, return the longest one. Return the first one in case of multiple
    strings of the same length. Return None in case the input list is empty.
    >>> longest([])
    None

    >>> longest(['a', 'b', 'c'])
    'a'
    >>> longest(['a', 'bb', 'ccc'])
    'ccc'
    """
    if not strings:
        return None

    return max(strings, key=len)


Question:
def is_happy(s):
    """You are given a string s.
    Your task is to check if the string is happy or not.
    A string is happy if its length is at least 3 and every 3 consecutive letters are distinct
    For example:
    is_happy(a) => False
    is_happy(aa) => False
    is_happy(abcd) => True
    is_happy(aabb) => False
    is_happy(adb) => True
    is_happy(xyy) => False
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def is_happy(s):
    """You are given a string s.
    Your task is to check if the string is happy or not.
    A string is happy if its length is at least 3 and every 3 consecutive letters are distinct
    For example:
    is_happy(a) => False
    is_happy(aa) => False
    is_happy(abcd) => True
    is_happy(aabb) => False
    is_happy(adb) => True
    is_happy(xyy) => False
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def is_happy(s):
    """You are given a string s.
    Your task is to check if the string is happy or not.
    A string is happy if its length is at least 3 and every 3 consecutive letters are distinct
    For example:
    is_happy(a) => False
    is_happy(aa) => False
    is_happy(abcd) => True
    is_happy(aabb) => False
    is_happy(adb) => True
    is_happy(xyy) => False
    """
    if len(s) < 3:
        return False

    for i in range(len(s) - 2):
        window = s[i:i+3]
        if len(set(window)) != 3:
            return False

    return True


Question:
def multiply(a, b):
    """Complete the function that takes two integers and returns
    the product of their unit digits.
    Assume the input is always valid.
    Examples:
    multiply(148, 412) should return 16.
    multiply(19, 28) should return 72.
    multiply(2020, 1851) should return 0.
    multiply(14,-15) should return 20.
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def multiply(a, b):
    """Complete the function that takes two integers and returns
    the product of their unit digits.
    Assume the input is always valid.
    Examples:
    multiply(148, 412) should return 16.
    multiply(19, 28) should return 72.
    multiply(2020, 1851) should return 0.
    multiply(14,-15) should return 20.
    """

Output:
# Reasoning

sandbox: passed=false returncode=1 timed_out=false
Traceback (most recent call last):
  File "/tmp/tmp5s_a9s0_/solution.py", line 46, in <module>
    check(multiply)
  File "/tmp/tmp5s_a9s0_/solution.py", line 43, in check
    assertion(candidate(*inp), exp, 0)
  File "/tmp/tmp5s_a9s0_/solution.py", line 36, in assertion
    assert exact_match
           ^^^^^^^^^^^
AssertionError

# Answer

def multiply(a, b):
    """Complete the function that takes two integers and returns
    the product of their unit digits.
    Assume the input is always valid.
    Examples:
    multiply(148, 412) should return 16.
    multiply(19, 28) should return 72.
    multiply(2020, 1851) should return 0.
    multiply(14,-15) should return 20.
    """
    return abs(a % 10) * abs(b % 10)


Question:
def check_dict_case(dict):
    """
    Given a dictionary, return True if all keys are strings in lower
    case or all keys are strings in upper case, else return False.
    The function should return False is the given dictionary is empty.
    Examples:
    check_dict_case({"a":"apple", "b":"banana"}) should return True.
    check_dict_case({"a":"apple", "A":"banana", "B":"banana"}) should return False.
    check_dict_case({"a":"apple", 8:"banana", "a":"apple"}) should return False.
    check_dict_case({"Name":"John", "Age":"36", "City":"Houston"}) should return False.
    check_dict_case({"STATE":"NC", "ZIP":"12345" }) should return True.
    """

Prompt:
Complete the following Python function. Reply with the complete implementation in a single ```python code block and no prose.

def check_dict_case(dict):
    """
    Given a dictionary, return True if all keys are strings in lower
    case or all keys are strings in upper case, else return False.
    The function should return False is the given dictionary is empty.
    Examples:
    check_dict_case({"a":"apple", "b":"banana"}) should return True.
    check_dict_case({"a":"apple", "A":"banana", "B":"banana"}) should return False.
    check_dict_case({"a":"apple", 8:"banana", "a":"apple"}) should return False.
    check_dict_case({"Name":"John", "Age":"36", "City":"Houston"}) should return False.
    check_dict_case({"STATE":"NC", "ZIP":"12345" }) should return True.
    """

Output:
# Reasoning

sandbox: passed=true returncode=0 timed_out=false

# Answer

def check_dict_case(dict):
    """
    Given a dictionary, return True if all keys are strings in lower
    case or all keys are strings in upper case, else return False.
    The function should return False is the given dictionary is empty.
    Examples:
    check_dict_case({"a":"apple", "b":"banana"}) should return True.
    check_dict_case({"a":"apple", "A":"banana", "B":"banana"}) should return False.
    check_dict_case({"a":"apple", 8:"banana", "a":"apple"}) should return False.
    check_dict_case({"Name":"John", "Age":"36", "City":"Houston"}) should return False.
    check_dict_case({"STATE":"NC", "ZIP":"12345" }) should return True.
    """
    if not dict:
        return False

    keys = list(dict.keys())

    # Check if all keys are strings
    if not all(isinstance(k, str) for k in keys):
        return False

    # Check if all keys are lowercase
    all_lower = all(k.islower() for k in keys)
    # Check if all keys are uppercase
    all_upper = all(k.isupper() for k in keys)

    return all_lower or all_upper