Challenge — Code Olympics · Mixed Skills

Eight problems. For each:

• Read the spec.
• Decide which lesson's skill applies — the tag tells you.
• Write the solution. Don't peek at the hint until you've given it a real try.
• Time yourself if you want. Target: 6 minutes per problem.

Solve at least 5 to feel ready for the capstone. All 8 = you're flying.

Given an integer, return the sum of its digits. Use a list comprehension.

assert digit_sum(1234) == 10
assert digit_sum(0) == 0
assert digit_sum(99999) == 45

def digit_sum(n):
    return sum(int(ch) for ch in str(abs(n)))

Given a paragraph of text, return a dict mapping each lower-cased word to how many times it appears. Ignore punctuation.

word_freq("Hello, hello, world.")
# → {"hello": 2, "world": 1}

import re

def word_freq(text):
    words = re.findall(r"[a-z]+", text.lower())
    counts = {}
    for w in words:
        counts[w] = counts.get(w, 0) + 1
    return counts

Given a list of names, group them by first letter into a dict-of-lists.

group_by_first(["Anna", "Bob", "Alice", "Carl", "Andy"])
# → {"A": ["Anna", "Alice", "Andy"], "B": ["Bob"], "C": ["Carl"]}

def group_by_first(names):
    out = {}
    for n in names:
        out.setdefault(n[0], []).append(n)
    return out

Two words are anagrams if they have the same letters. Ignore case and spaces.

assert anagram("Listen", "Silent")
assert anagram("Astronomer", "Moon starer")
assert not anagram("hello", "world")

def anagram(a, b):
    def normalise(s):
        return sorted(c for c in s.lower() if c.isalpha())
    return normalise(a) == normalise(b)

Given a path to a file containing one number per line (with blanks and non-number lines mixed in), return the sum of the valid numbers. Handle missing file by returning 0.

def file_sum(path):
    total = 0
    try:
        with open(path, encoding="utf-8") as f:
            for line in f:
                try:
                    total += float(line.strip())
                except ValueError:
                    pass
    except FileNotFoundError:
        pass
    return total

Given a text, return a list of every Malaysian phone number. Use regex.

find_phones("Call 012-3456789 or +60 12-1234567 not 123 or 12-3456")
# → ["012-3456789", "+60 12-1234567"]

import re

def find_phones(text):
    return re.findall(r"\+?60[\s\-]?\d{2}-?\d{7,8}|\d{2,3}-\d{7,8}", text)

Given students.json with this shape:

[
  {"name": "Aisyah",  "scores": [82, 91, 76]},
  {"name": "Wei Jie", "scores": [74, 68, 91]},
  {"name": "Priya",   "scores": [95, 97, 96]}
]

Write a function that loads this file and returns a list of dicts with name, average (1dp). Sort by average descending.

import json

def student_averages(path):
    with open(path, encoding="utf-8") as f:
        students = json.load(f)
    out = []
    for s in students:
        avg = round(sum(s["scores"]) / len(s["scores"]), 1)
        out.append({"name": s["name"], "average": avg})
    out.sort(key=lambda x: x["average"], reverse=True)
    return out

Write safe_average(numbers) that returns the average of a list — and the string "no data" if the list is empty. Don't use if. Use try/except.

def safe_average(numbers):
    try:
        return sum(numbers) / len(numbers)
    except ZeroDivisionError:
        return "no data"

Eight tiny problems, eight different L2 skills. Comprehensions, dicts, group-by, sets via sorted comparison, file safety, regex, JSON, try/except. Real software is sequences of small wins like these — five-minute functions composed into bigger features. Whichever problems felt slow are the ones to revisit; whichever felt fast are the ones to trust.