05-03: Exercises — Sets¶

Notes reference: 05-03: Sets

Q1: Create a set and remove duplicates¶

Create a set from a list with duplicates: [3, 1, 4, 1, 5, 9, 2, 6, 5, 3].

Solution

nums   = [3, 1, 4, 1, 5, 9, 2, 6, 5, 3]
unique = set(nums)
print(unique)     # {1, 2, 3, 4, 5, 6, 9}  (order may vary)
print(len(nums), "→", len(unique))   # 10 → 7

Q2: add and remove¶

Create languages = {"Python", "Java", "C"}. Add "Rust", remove "Java", and discard "Go" safely (without error).

Solution

languages = {"Python", "Java", "C"}
languages.add("Rust")
languages.remove("Java")
languages.discard("Go")    # no error even if not present
print(languages)   # {'Python', 'C', 'Rust'}

Q3: Union and intersection¶

Given students in Math = {"Alice", "Bob", "Carol"} and Physics = {"Bob", "David", "Carol", "Eve"}, find: - all students (union) - students in both (intersection) - students only in Math (difference)

Solution

math    = {"Alice", "Bob", "Carol"}
physics = {"Bob", "David", "Carol", "Eve"}

print(math | physics)     # all students
print(math & physics)     # {'Bob', 'Carol'}
print(math - physics)     # {'Alice'}
print(math ^ physics)     # symmetric difference: {'Alice', 'David', 'Eve'}

Q4: Subset and superset¶

Check whether {"Python", "C"} is a subset of {"Python", "C", "Java", "Rust"}.

Solution

small = {"Python", "C"}
big   = {"Python", "C", "Java", "Rust"}

print(small.issubset(big))     # True
print(big.issuperset(small))   # True
print(small <= big)            # True  (operator shorthand)

Q5: Unique characters¶

Count the number of unique characters in "Chittagong".

Solution

word = "Chittagong"
unique_chars = set(word)
print(len(unique_chars))   # 7  (C,h,i,t,a,g,o,n — 'g' repeated)
print(unique_chars)

Q6: Set from two lists¶

Find words that appear in sentence A but NOT in sentence B.

Solution

a = set("the quick brown fox".split())
b = set("the lazy brown dog".split())

only_in_a = a - b
print(only_in_a)   # {'quick', 'fox'}

Q7: Membership test performance¶

Show that in is O(1) for sets vs O(n) for lists — demonstrate by checking membership.

Solution

large_list = list(range(1_000_000))
large_set  = set(large_list)

# Both return True, but set lookup is instant
print(999_999 in large_list)   # True  (slow — scans the list)
print(999_999 in large_set)    # True  (fast — hash lookup)
print(2_000_000 in large_set)  # False

⬅️ Previous: 05-02: Exercises — Tuples ➡️ Next: 05-04: Exercises — Dictionaries