Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):

        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    even_palindrome_count = 0
    odd_palindrome_count = 0

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    odd_palindrome_count = 0

    for i in range(1, n+1):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0

    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):

        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):

                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1

        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):

            even_palindrome_count += 1
    return (even_palindrome_count, odd_palindrome_count)

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def even_odd_palindrome(n):
    """
    Given a positive integer n, return a tuple that has the number of even and odd
    integer palindromes that fall within the range(1, n), inclusive.

    Example 1:

        Input: 3
        Output: (1, 2)
        Explanation:
        Integer palindrome are 1, 2, 3. one of them is even, and two of them are odd.

    Example 2:

        Input: 12
        Output: (4, 6)
        Explanation:
        Integer palindrome are 1, 2, 3, 4, 5, 6, 7, 8, 9, 11. four of them are even, and 6 of them are odd.

    Note:
        1. 1 <= n <= 10^3
        2. returned tuple has the number of even and odd integer palindromes respectively.
    """
    def is_palindrome(n):
        return str(n) == str(n)[::-1]

    even_palindrome_count = 0
    odd_palindrome_count = 0

    for i in range(1, n+1):
        if i%2 == 1 and is_palindrome(i):
                odd_palindrome_count += 1
        elif i%2 == 0 and is_palindrome(i):
            even_palindrome_count += 1

def check(candidate):

    # Check some simple cases
    assert candidate(123) == (8, 13)
    assert candidate(12) == (4, 6)
    assert candidate(3) == (1, 2)
    assert candidate(63) == (6, 8)
    assert candidate(25) == (5, 6)
    assert candidate(19) == (4, 6)
    assert candidate(9) == (4, 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) == (0, 1), "This prints if this assert fails 2 (also good for debugging!)"


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

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

    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

        n[0] = n[0] * neg
        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):

        n[0] = n[0] * neg
        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):

        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):

return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1

        n[0] = n[0] * neg
        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1

        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1

return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1

        n[0] = n[0] * neg
        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1

        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1

return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg
        return sum(n)

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1

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

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]

        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]

return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

        n[0] = n[0] * neg
        return sum(n)

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]

        n[0] = n[0] * neg
        return sum(n)
    return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

def check(candidate):

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

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


Prompt:
def count_nums(arr):
    """
    Write a function count_nums which takes an array of integers and returns
    the number of elements which has a sum of digits > 0.
    If a number is negative, then its first signed digit will be negative:
    e.g. -123 has signed digits -1, 2, and 3.
    >>> count_nums([]) == 0
    >>> count_nums([-1, 11, -11]) == 1
    >>> count_nums([1, 1, 2]) == 3
    """
    def digits_sum(n):
        neg = 1
        if n < 0: n, neg = -1 * n, -1
        n = [int(i) for i in str(n)]
        n[0] = n[0] * neg

return len(list(filter(lambda x: x > 0, [digits_sum(i) for i in arr])))

def check(candidate):

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

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