Chapter 14 Intermediate 62 Questions

Practice Questions — Dictionaries in Python

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9 Easy

12 Medium

8 Hard

Topic-Specific Questions

Question 1

Easy

What is the output of the following code?

d = {"name": "Aarav", "age": 16}
print(d["name"])
print(d["age"])

Square brackets access the value associated with a key.

Aarav
16

Question 2

Easy

What is the output?

d = {"a": 1, "b": 2}
d["c"] = 3
print(d)
print(len(d))

Assigning to a new key adds it to the dictionary.

{'a': 1, 'b': 2, 'c': 3}
3

Question 3

Easy

What is the output?

d = {"x": 10, "y": 20}
print(d.get("x"))
print(d.get("z"))
print(d.get("z", 0))

.get() returns None for missing keys unless a default is specified.

10
None
0

Question 4

Easy

What is the output?

d = {"a": 1, "b": 2, "c": 3}
print("a" in d)
print(1 in d)
print("d" in d)

The 'in' operator checks keys, not values.

True
False
False

Question 5

Easy

What is the output?

d = {"a": 1, "b": 2, "c": 3}
print(list(d.keys()))
print(list(d.values()))

keys() returns all keys. values() returns all values.

['a', 'b', 'c']
[1, 2, 3]

Question 6

Easy

What is the output?

d = {"a": 1, "b": 2}
d["a"] = 10
print(d)

Assigning to an existing key updates its value.

{'a': 10, 'b': 2}

Question 7

Medium

What is the output?

d = {"a": 1, "b": 2, "c": 3}
val = d.pop("b")
print(val)
print(d)

pop() removes the key and returns its value.

2
{'a': 1, 'c': 3}

Question 8

Medium

What is the output?

d = {"a": 1, "b": 2}
for key, value in d.items():
    print(f"{key} -> {value}")

items() returns key-value pairs as tuples, which can be unpacked.

a -> 1
b -> 2

Question 9

Medium

What is the output?

squares = {x: x**2 for x in range(1, 5)}
print(squares)

Dictionary comprehension: {key_expr: value_expr for item in iterable}.

{1: 1, 2: 4, 3: 9, 4: 16}

Question 10

Medium

What is the output?

d1 = {"a": 1, "b": 2}
d2 = {"b": 3, "c": 4}
merged = {**d1, **d2}
print(merged)

** unpacking merges dictionaries. When keys overlap, the last one wins.

{'a': 1, 'b': 3, 'c': 4}

Question 11

Medium

What is the output?

d = {}
d.setdefault("count", 0)
d["count"] += 1
d.setdefault("count", 0)
d["count"] += 1
print(d)

setdefault() only sets the value if the key does not already exist.

{'count': 2}

Question 12

Medium

What is the output?

school = {
    "Aarav": {"marks": 90},
    "Priya": {"marks": 95}
}
print(school["Priya"]["marks"])

Chain keys to access nested dictionary values.

95

Question 13

Hard

What is the output?

text = "abba"
count = {}
for ch in text:
    count[ch] = count.get(ch, 0) + 1
print(count)

.get(ch, 0) returns 0 if the key does not exist, then adds 1.

{'a': 2, 'b': 2}

Question 14

Hard

What is the output?

d = {"a": 1, "b": 2, "a": 3}
print(d)
print(len(d))

Dictionary keys must be unique. What happens with duplicate keys in a literal?

{'a': 3, 'b': 2}
2

Question 15

Hard

What is the output?

d = dict.fromkeys(["a", "b", "c"], [])
d["a"].append(1)
print(d)

fromkeys() with a mutable default shares the same object for all keys.

{'a': [1], 'b': [1], 'c': [1]}

Question 16

Hard

What is the output?

d = {"a": 1, "b": 2, "c": 3}
filtered = {k: v for k, v in d.items() if v > 1}
print(filtered)

Dictionary comprehension with a filter condition.

{'b': 2, 'c': 3}

Question 17

Hard

What is the output?

d = {"x": 10, "y": 20, "z": 30}
result = max(d, key=d.get)
print(result)
print(d[result])

max(dict, key=dict.get) finds the key with the highest value.

z
30

Mixed & Application Questions

Question 1

Easy

What is the output?

d = {"name": "Rohan"}
d["name"] = "Vikram"
print(d)

Assigning to an existing key updates its value.

{'name': 'Vikram'}

Question 2

Easy

What is the output?

d = {"a": 1, "b": 2}
d.clear()
print(d)
print(type(d))

clear() removes all key-value pairs but the dictionary object remains.

{}
<class 'dict'>

Question 3

Easy

What is the output?

d = {1: "one", 2: "two", 3: "three"}
print(d[2])
print(list(d.keys()))

Dictionary keys can be integers, not just strings.

two
[1, 2, 3]

Question 4

Medium

What is the output?

d = {"a": 1, "b": 2}
d.update({"b": 20, "c": 30})
print(d)

update() adds new keys and overwrites existing ones.

{'a': 1, 'b': 20, 'c': 30}

Question 5

Medium

What is the output?

d = {"a": 1, "b": 2, "c": 3}
result = d.pop("b", "not found")
result2 = d.pop("z", "not found")
print(result)
print(result2)
print(d)

pop(key, default) removes and returns the value, or returns default if key is missing.

2
not found
{'a': 1, 'c': 3}

Question 6

Medium

What is the output?

original = {"a": 1, "b": 2}
inverted = {v: k for k, v in original.items()}
print(inverted)

Dictionary comprehension can swap keys and values.

{1: 'a', 2: 'b'}

Question 7

Medium

What is the output?

d1 = {"a": 1}
d2 = d1
d2["b"] = 2
print(d1)
print(d1 is d2)

d2 = d1 creates an alias, not a copy. Same as with lists.

{'a': 1, 'b': 2}
True

Question 8

Hard

What is the output?

data = [{"name": "Aarav", "marks": 90},
        {"name": "Priya", "marks": 95},
        {"name": "Rohan", "marks": 78}]
top = max(data, key=lambda x: x["marks"])
print(top["name"])

max() with a key function can find the dict with the highest value for a specific key.

Priya

Question 9

Medium

What is the difference between dict[key] and dict.get(key) when the key does not exist?

One raises an error, the other returns a default value.

dict[key] raises a KeyError when the key does not exist. dict.get(key) returns None (or a custom default if specified: dict.get(key, default)) without raising an error.

Question 10

Medium

Why does in check keys and not values in a dictionary? How do you check if a value exists?

Think about the purpose of dictionaries and what operation is most common.

Dictionaries are optimized for key-based access. Checking if a key exists is O(1) (constant time) because of hashing. Checking values would require O(n) linear search. Since key lookup is the primary use case, in defaults to checking keys. To check values, use value in dict.values().

Question 11

Hard

What is the output?

d = {}
for ch in "hello world":
    if ch != " ":
        d[ch] = d.get(ch, 0) + 1
print(d)
print(d["l"])

This counts character frequency, skipping spaces.

{'h': 1, 'e': 1, 'l': 3, 'o': 2, 'w': 1, 'r': 1, 'd': 1}
3

Question 12

Hard

What is the output?

d = {"a": [1, 2], "b": [3, 4]}
copy = d.copy()
copy["a"].append(99)
print(d)
print(copy)

dict.copy() is a shallow copy. Nested mutable objects are still shared.

{'a': [1, 2, 99], 'b': [3, 4]}
{'a': [1, 2, 99], 'b': [3, 4]}

Multiple Choice Questions

MCQ 1

Which of the following creates an empty dictionary?

A. d = []
B. d = ()
C. d = {}
D. d = set()

Answer: C
C is correct. Curly braces {} create an empty dictionary. [] (A) creates an empty list. () (B) creates an empty tuple. set() (D) creates an empty set (note: {} is a dict, not a set).

MCQ 2

What does dict.get('key') return if 'key' does not exist?

A. 0
B. False
C. None
D. KeyError

Answer: C
C is correct. .get(key) returns None when the key is not found (no error). To return a custom default, use .get(key, default). Option D describes the behavior of dict[key], not .get().

MCQ 3

What does the 'in' operator check for dictionaries?

A. Values
B. Keys
C. Both keys and values
D. Key-value pairs

Answer: B
B is correct. The in operator checks keys by default. "name" in d checks if "name" is a key. To check values, use value in d.values().

MCQ 4

Which types can be used as dictionary keys?

A. Only strings
B. Any immutable type (str, int, tuple)
C. Any type including lists
D. Only strings and integers

Answer: B
B is correct. Dictionary keys must be hashable (immutable). Strings, integers, floats, and tuples of immutables all work. Lists (C) and dictionaries cannot be keys because they are mutable.

MCQ 5

What does d.pop('key') do?

A. Returns the value without removing
B. Removes the key and returns the value
C. Removes the key and returns None
D. Removes the last item

Answer: B
B is correct. pop(key) removes the key-value pair and returns the value. It raises KeyError if the key does not exist (unless a default is provided). Option D describes popitem().

MCQ 6

What is the output of {x: x**2 for x in range(3)}?

A. {0, 1, 4}
B. {0: 0, 1: 1, 2: 4}
C. [(0, 0), (1, 1), (2, 4)]
D. {1: 1, 2: 4}

Answer: B
B is correct. Dictionary comprehension creates key-value pairs. For x = 0, 1, 2: the key is x and the value is x**2. Result: {0: 0, 1: 1, 2: 4}. Option A would be a set comprehension ({x**2 for x in range(3)}).

MCQ 7

What happens when you add a key that already exists in a dictionary?

A. A KeyError is raised
B. The new value replaces the old one
C. Both values are kept
D. The operation is ignored

Answer: B
B is correct. Dictionary keys are unique. Assigning to an existing key overwrites the previous value. No error is raised, and the old value is lost.

MCQ 8

What does d.items() return?

A. A list of keys
B. A list of values
C. A view of key-value tuples
D. A dictionary

Answer: C
C is correct. .items() returns a view object containing (key, value) tuples. It is not a regular list, but it can be iterated over and converted with list().

MCQ 9

What does setdefault('key', value) do if 'key' already exists?

A. Updates the key with the new value
B. Returns the existing value without changing it
C. Raises a KeyError
D. Deletes the existing key

Answer: B
B is correct. setdefault() only sets the value if the key does NOT exist. If the key already exists, it simply returns the current value without modifying anything.

MCQ 10

How do you safely copy a dictionary?

A. b = a
B. b = a.copy()
C. b = a.values()
D. b = dict(a.items())

Answer: B
B is correct. .copy() creates a shallow copy. Option A creates an alias (not a copy). Option C gives only values (not a dict). Option D also works but is less common and readable. Note: for nested dicts, use copy.deepcopy().

MCQ 11

Since which Python version are dictionaries guaranteed to maintain insertion order?

A. Python 2.7
B. Python 3.5
C. Python 3.7
D. Python 3.9

Answer: C
C is correct. In Python 3.6, CPython (the standard implementation) made dicts ordered as an implementation detail. In Python 3.7, it became part of the language specification. Before that, dictionaries were unordered.

MCQ 12

What is the output of {**{'a': 1}, **{'a': 2}}?

A. {'a': 1}
B. {'a': 2}
C. {'a': [1, 2]}
D. Error: duplicate key

Answer: B
B is correct. When merging with ** unpacking, the last value for a duplicate key wins. The second dict has 'a': 2, which overwrites the first dict's 'a': 1.

MCQ 13

What does dict.copy() create?

A. A deep copy
B. A shallow copy
C. An alias (same object)
D. A frozen dictionary

Answer: B
B is correct. .copy() creates a shallow copy: a new dict object, but nested mutable values (like lists) are still shared references. For a fully independent copy of nested structures, use copy.deepcopy().

MCQ 14

What does popitem() do in Python 3.7+?

A. Removes a random key-value pair
B. Removes the first inserted pair
C. Removes the last inserted pair
D. Removes all pairs

Answer: C
C is correct. In Python 3.7+, popitem() removes and returns the last inserted key-value pair (LIFO order). In older Python versions, it removed an arbitrary pair. Option D describes clear().

MCQ 15

Which method merges one dictionary into another in place?

A. merge()
B. update()
C. extend()
D. join()

Answer: B
B is correct. update() merges the given dictionary into the current one, modifying it in place. merge() (A) does not exist. extend() (C) is a list method. join() (D) is a string method.

MCQ 16

What is the time complexity of checking 'key in dict'?

A. O(n)
B. O(log n)
C. O(1) average
D. O(n log n)

Answer: C
C is correct. Dictionary key lookup uses hashing, which provides O(1) average-case time complexity. This is why dictionaries are preferred over lists for lookups. List membership checking (item in list) is O(n).

MCQ 17

What is the output of len({'a': 1, 'b': 2, 'c': 3})?

A. 6
B. 3
C. 2
D. 9

Answer: B
B is correct. len() on a dictionary returns the number of key-value pairs. This dictionary has 3 pairs, so the length is 3.

MCQ 18

What does d = {}; d[(1, 2)] = 'point' do?

A. Creates a dictionary with two keys: 1 and 2
B. Creates a dictionary with one key: the tuple (1, 2)
C. Raises TypeError
D. Creates a nested dictionary

Use sum(), len(), and dictionary comprehension.

marks = {'Aarav': 85, 'Priya': 92, 'Rohan': 78, 'Meera': 95, 'Vikram': 88}
avg = sum(marks.values()) / len(marks)
print(f"Average: {avg}")
above = {name: score for name, score in marks.items() if score > avg}
print(f"Above average: {above}")

Challenge 6: Nested Dictionary Access

Medium

Given a nested dictionary representing a school, write a program that prints each student's name and their total marks across all subjects.

school = {
    "Aarav": {"Maths": 90, "Science": 85, "English": 88},
    "Priya": {"Maths": 95, "Science": 92, "English": 91}
}

Sample Input

(No input required)

Sample Output

Aarav: 263 Priya: 278

Use nested loops or sum() on values.

school = {
    "Aarav": {"Maths": 90, "Science": 85, "English": 88},
    "Priya": {"Maths": 95, "Science": 92, "English": 91}
}
for name, subjects in school.items():
    total = sum(subjects.values())
    print(f"{name}: {total}")

Challenge 7: Two Lists to Dictionary

Medium

Given keys = ['name', 'age', 'city'] and values = ['Rohan', 16, 'Delhi'], write a program that creates a dictionary from these two lists using zip().

Sample Input

(No input required)

Sample Output

{'name': 'Rohan', 'age': 16, 'city': 'Delhi'}

Use zip() and dict().

keys = ['name', 'age', 'city']
values = ['Rohan', 16, 'Delhi']
result = dict(zip(keys, values))
print(result)

Challenge 8: Character Frequency with Sorting

Hard

Write a program that counts the frequency of each character in "programming" and prints them sorted by frequency (highest first). For ties, sort alphabetically.

Sample Input

(No input required)

Sample Output

g: 2 m: 2 r: 2 a: 1 i: 1 n: 1 o: 1 p: 1

Use .get() for counting. Use sorted() with a key.

text = "programming"
freq = {}
for ch in text:
    freq[ch] = freq.get(ch, 0) + 1
sorted_chars = sorted(freq.items(), key=lambda x: (-x[1], x[0]))
for ch, count in sorted_chars:
    print(f"{ch}: {count}")

Challenge 9: Find Duplicate Values

Hard

Given the dictionary {'a': 1, 'b': 2, 'c': 1, 'd': 3, 'e': 2}, write a program that finds and prints which values appear more than once, along with the keys that share them.

Sample Input

(No input required)

Sample Output

Value 1 is shared by: ['a', 'c'] Value 2 is shared by: ['b', 'e']

Use a dictionary to group keys by their values.

d = {'a': 1, 'b': 2, 'c': 1, 'd': 3, 'e': 2}
value_to_keys = {}
for key, value in d.items():
    value_to_keys.setdefault(value, []).append(key)
for value, keys in value_to_keys.items():
    if len(keys) > 1:
        print(f"Value {value} is shared by: {keys}")

Challenge 10: Flatten a Nested Dictionary

Hard

Write a program that flattens the nested dictionary {'a': 1, 'b': {'c': 2, 'd': {'e': 3}}} into a flat dictionary with dotted keys: {'a': 1, 'b.c': 2, 'b.d.e': 3}.

Sample Input

(No input required)

Sample Output

{'a': 1, 'b.c': 2, 'b.d.e': 3}

Use recursion or a stack-based approach.

def flatten(d, parent_key='', sep='.'):
    items = {}
    for k, v in d.items():
        new_key = parent_key + sep + k if parent_key else k
        if isinstance(v, dict):
            items.update(flatten(v, new_key, sep))
        else:
            items[new_key] = v
    return items

nested = {'a': 1, 'b': {'c': 2, 'd': {'e': 3}}}
print(flatten(nested))

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