L2 · Working engineerDictionaries & hashing~4 min · 4 tests#27

Word Frequency Count

Count how often each word appears in a sentence with split() and a dictionary in O(n). Practise Counter, defaultdict and dict.get in Python.

The problem

Return a dictionary mapping each word in sentence (split on whitespace) to how many times it appears. Words are case-sensitive.

Examples

Example 1

Input

word_frequency('python is easy and python is powerful')

Expected output

{'python': 2, 'is': 2, 'easy': 1, 'and': 1, 'powerful': 1}

Example 2
Input
```
word_frequency('a a a')
```
Expected output
```
{'a': 3}
```

+ 2 hidden tests on Submit.

Edge cases to ask about

Empty sentence
Repeated spaces
Case

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · word_frequency

def word_frequency(sentence):
    """Return {word: count}."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
list.count per word	O(w²)	O(w)	Recounts for each word.
bestOne pass with a dict	O(n)	O(w)	Or Counter(sentence.split()).

Walkthrough of the optimal approach (try it yourself first)

Split on whitespace, then count with a dict. In production you would also normalise case and strip punctuation — say that, then do what the question asks.

Complexity: O(n) time, O(w) space. split() and the counting loop are both linear in the sentence length; w distinct words are stored.

Reveal the reference solution

def word_frequency(sentence):
    freq = {}
    for word in sentence.split():
        freq[word] = freq.get(word, 0) + 1
    return freq

Follow-ups interviewers ask

Top 3 most frequent words.
Process a file too big for memory.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Word Frequency Count in Python?

The optimal solution runs in O(n) time and O(w) auxiliary space. split() and the counting loop are both linear in the sentence length; w distinct words are stored.

What is the brute-force approach, and how do you optimise it?

list.count per word: O(w²) time, O(w) space. Recounts for each word. One pass with a dict: O(n) time, O(w) space. Or Counter(sentence.split()).

What follow-up questions do interviewers ask about Word Frequency Count?

Top 3 most frequent words. Process a file too big for memory.