S2Q1 · Count Odd 3-Digit Numbers (Ignore None)

📁 GitHub

⚡ Quick Reference

Function: count_odd_three_digit_nums(nums: list) -> int

Core idea: filter out None, then count elements that are three-digit (using abs) and odd.

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

Key rules: - Skip None values - check n is not None first - Three-digit: 100 ≤ abs(n) ≤ 999 (ignores negative sign) - Odd: n % 2 != 0 - All three conditions must hold

---

Problem Statement

Problem

Write a function count_odd_three_digit_nums(nums) that counts elements that are not None, are three-digit numbers (ignoring sign), and are odd.

Examples:

Input

[101, -203, None, 99, 300]

Output

2

Input

[None, 120, 301, -401, 78]

Output

2

Input

[10, 305, 507, 99]

Output

2

---

Tracing all examples

Example 1: [101, -203, None, 99, 300]

n	Not None?	abs(n)	3-digit?	Odd?	Count?
101	✅	101	✅	✅	✅
-203	✅	203	✅	✅	✅
None	❌	-	-	-	❌
99	✅	99	❌ (2 digits)	-	❌
300	✅	300	✅	❌ (even)	❌

Count = 2 ✓

Example 2: [None, 120, 301, -401, 78]

n	Not None?	3-digit?	Odd?	Count?
None	❌	-	-	❌
120	✅	✅	❌ (even)	❌
301	✅	✅	✅	✅
-401	✅	✅	✅	✅
78	✅	❌	-	❌

Count = 2 ✓

---

Solution approaches

Pythonic (generator expression)Explanatory (loop with counter)Using filter + lambdaUsing list comprehension + len

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

def count_odd_three_digit_nums(nums):
    count = 0
    for n in nums:
        if n is None:
            continue
        if 100 <= abs(n) <= 999 and n % 2 != 0:
            count += 1
    return count

def count_odd_three_digit_nums(nums):
    valid = filter(
        lambda n: n is not None and 100 <= abs(n) <= 999 and n % 2 != 0,
        nums
    )
    return len(list(valid))

def count_odd_three_digit_nums(nums):
    return len([
        n for n in nums
        if n is not None
        and 100 <= abs(n) <= 999
        and n % 2 != 0
    ])

---

Key takeaways

01

Check None first with "is not None"

Always check n is not None before calling abs(n) or n % 2. Calling abs(None) raises a TypeError. Python's short-circuit and ensures the later conditions only run when n is not None.

02

abs() handles negative three-digit numbers

abs(-203) = 203 which is in range [100, 999]. Without abs(), negative numbers would always fail the range check since -203 < 100.

03

n % 2 != 0 for odd - use original n, not abs(n)

In Python, -203 % 2 == 1 (odd) and -204 % 2 == 0 (even), so the odd check works correctly on negative numbers directly. Using abs(n) % 2 would also work but is redundant.