L2 · Working engineerStrings~4 min · 4 tests#21

Count Character Frequency in a String

Count how many times each character appears in a Python string with a dict, then with collections.Counter. O(n) time, practised with hidden tests.

The problem

Return a dictionary mapping each character of s to the number of times it appears.

Examples

Example 1

Input

char_frequency('programming')

Expected output

{'p': 1, 'r': 2, 'o': 1, 'g': 2, 'a': 1, 'm': 2, 'i': 1, 'n': 1}

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

+ 2 hidden tests on Submit — spaces count too.

Edge cases to ask about

Empty string
Spaces and punctuation
Upper vs lower 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 · char_frequency

def char_frequency(s):
    """Return {character: 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
s.count() per character	O(n·k)	O(k)	One full scan per distinct character.
bestOne counting pass	O(n)	O(k)	dict.get or collections.Counter.

Walkthrough of the optimal approach (try it yourself first)

One pass, one dict: freq[ch] = freq.get(ch, 0) + 1. collections.defaultdict(int) and Counter are the idiomatic alternatives — know all three.

Complexity: O(n) time, O(k) space. One pass over the string; one dict entry per distinct character.

Reveal the reference solution

def char_frequency(s):
    freq = {}
    for ch in s:
        freq[ch] = freq.get(ch, 0) + 1
    return freq

The brute force, for comparison

def char_frequency(s):
    return {ch: s.count(ch) for ch in s}

Follow-ups interviewers ask

Return the most common character.
Ignore case and non-letters.

Frequently asked interview questions

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

What is the time complexity of Count Character Frequency in a String in Python?

The optimal solution runs in O(n) time and O(k) auxiliary space. One pass over the string; one dict entry per distinct character.

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

s.count() per character: O(n·k) time, O(k) space. One full scan per distinct character. One counting pass: O(n) time, O(k) space. dict.get or collections.Counter.

What follow-up questions do interviewers ask about Count Character Frequency in a String?

Return the most common character. Ignore case and non-letters.