Count Odd 3 Digit Numbers (Ignore None)

Count Odd 3 Digit Numbers (Ignore None)

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Write a function count_odd_three_digit_nums(nums) that takes a list of integers nums and returns the count of numbers that are: Odd. Three-digit numbers (ignoring the negative sign if present). Not None. NOTE: This is a function type question, you don’t have to take input or print the output, just have to complete the required function definition. Examples: count_odd_three_digit_nums([101, -203, None, 99, 300]) → 2 Explanation: 101 and -203 are odd three-digit numbers, ignoring None. count_odd_three_digit_nums([None, 120, 301, -401, 78]) → 2 Explanation: 301 and -401 qualify as odd three-digit numbers. count_odd_three_digit_nums([10, 305, 507, 99]) → 2 Explanation: 305 and 507 qualify as odd three-digit numbers.

Perfect ✅ You want a function that counts how many numbers in a list are:

• Odd
• Three-digit numbers (ignoring negative sign)
• Not None

Here’s the Python implementation:

def count_odd_three_digit_nums(nums: list) -> int:
    count = 0
    for n in nums:
        if n is not None:  # Ignore None values
            if 100 <= abs(n) <= 999 and n % 2 != 0:  # Check 3-digit and odd
                count += 1
    return count

✅ Explanation:

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1. We loop through each element in the list.
2. If the element is None, skip it.
3. abs(n) ensures we handle negative numbers when checking the three-digit condition.
4. Check that the number is odd (n % 2 != 0).
5. Increment the counter accordingly.

🔍 Examples:

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print(count_odd_three_digit_nums([101, -203, None, 99, 300]))   # 2
print(count_odd_three_digit_nums([None, 120, 301, -401, 78]))  # 2
print(count_odd_three_digit_nums([10, 305, 507, 99]))          # 2

Would you like me to also extend this so it works for both positive and negative multi-digit ranges (e.g., checking for 4-digit odd numbers too, if needed)?

This error happens because the code tries to do abs(n) before checking if n is not None. If n is None, abs(n) will raise a TypeError because you can’t take the absolute value of a NoneType. The fix is to make sure to check n is not None before calling abs(n).

Here is the corrected version:

def count_odd_three_digit_nums(nums: list) -> int:
    count = 0
    for n in nums:
        if n is not None:  # Check this first!
            if 100 <= abs(n) <= 999 and n % 2 != 0:
                count += 1
    return count

If you already have that check and still see this error, double-check the indentation and order of your conditions. You must never call abs(n) if n could be None.

Compact version using list comprehension:

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def count_odd_three_digit_nums(nums: list) -> int:
    return sum(
        1 for n in nums
        if n is not None and 100 <= abs(n) <= 999 and n % 2 != 0
    )

Key Point

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• Always check n is not None before using abs(n) or n % 2
• Do not reorder the conditions

This will prevent the TypeError and correctly count your odd three-digit numbers ignoring None.