Related chapter: Data Structures Deep Dive

Exercise 1: Remove Duplicates

Write unique(items) that returns a list of unique items preserving order.

  unique([1, 2, 2, 3, 1, 4, 3])  # [1, 2, 3, 4]
unique(["a", "b", "a", "c"])    # ["a", "b", "c"]

Exercise 2: Two Sum

Given a list of numbers and a target, return indices of two numbers that add up to the target.

  two_sum([2, 7, 11, 15], 9)  # [0, 1]
two_sum([3, 2, 4], 6)       # [1, 2]
Hint

Use a dict to store {value: index} as you iterate.

Exercise 3: Anagram Checker

Write is_anagram(s1, s2) — case-insensitive, ignore spaces.

  is_anagram("listen", "silent")     # True
is_anagram("Hello", "Olelh")        # True
is_anagram("Python", "Java")       # False

Exercise 4: Merge Dictionaries

Write merge_dicts(*dicts) that merges multiple dicts. Later dicts override earlier keys.

  merge_dicts({"a": 1}, {"b": 2}, {"a": 3})
# {"a": 3, "b": 2}

Exercise 5: Stack with Min

Implement a Stack class with push, pop, peek, and get_min — all in O(1) time.

  s = Stack()
s.push(3)
s.push(1)
s.push(5)
s.get_min()  # 1
s.pop()
s.get_min()  # 1

Exercise 6: Word Grouping

Group a list of words by their sorted characters (anagram groups):

  group_anagrams(["eat", "tea", "tan", "ate", "nat", "bat"])
# [["eat", "tea", "ate"], ["tan", "nat"], ["bat"]]

Challenge: LRU Cache

Implement an LRUCache(capacity) class with get(key) and put(key, value) that evicts the least recently used item when capacity is exceeded.

  cache = LRUCache(2)
cache.put(1, "a")
cache.put(2, "b")
cache.get(1)     # "a"
cache.put(3, "c") # evicts key 2
cache.get(2)     # None
Solution (Exercise 2) 
  def two_sum(nums, target):
    seen = {}
    for i, n in enumerate(nums):
        complement = target - n
        if complement in seen:
            return [seen[complement], i]
        seen[n] = i
    return []

Exercise 7: Rotate List

Write rotate(lst, k) that rotates a list right by k positions.

  rotate([1, 2, 3, 4, 5], 2)  # [4, 5, 1, 2, 3]

Exercise 8: Frequency Sort

Sort words by frequency (most common first). Break ties alphabetically.

  frequency_sort(["i", "love", "python", "i", "love", "coding"])
# ['i', 'love', 'coding', 'python']

Next: Algorithm Exercises