The Letter Ledger

An easy Python interview practice problem on DataDriven. Write the Python and run it against test cases, with instant feedback.

Domain
Python
Difficulty
easy
Seniority
L4
Asked in
Nintendo

Problem

Return how often each character of text appears as [char, count] pairs, sorted by character code from lowest to highest, so 'A' comes before 'a' and punctuation lands wherever its code falls. We count every character exactly as written except spaces, through a CharFrequency class whose get_counts() returns those sorted pairs and which char_frequency(text) builds and returns.

Every character has a count to answer for.

Example 1

Input:

text = "hello world"

Output:

[["d",1],["e",1],["h",1],["l",3],["o",2],["r",1],["w",1]]

Example 2

Input:

text = "Hello, World!"

Output:

[["!",1],[",",1],["H",1],["W",1],["d",1],["e",1],["l",3],["o",2],["r",1]]

Worked solution and explanation

Why this problem exists in real interviews

This tests class design with frequency counting, combining OOP basics with the dict accumulation pattern. It probes constructor logic, method implementation, and sorted output.

Break down the requirements

Step 1: Count character frequencies in __init__

Build a frequency dict, ignoring spaces.

Step 2: get_counts returns sorted pairs

Return [character, count] two-element lists sorted alphabetically by character.

Step 3: Provide the driver entrypoint

Expose a top-level char_frequency(text) driver that constructs the class and returns get_counts(), so the harness has a single callable entrypoint.

The solution

Class-based character frequency tracker
def char_frequency(text):
    class CharFrequency:
        def __init__(self, text: str):
            self.freq = {}
            for ch in text:
                if ch == ' ':
                    continue
                if ch in self.freq:
                    self.freq[ch] += 1
                else:
                    self.freq[ch] = 1

        def get_counts(self) -> list:
            result = []
            for key in sorted(self.freq.keys()):
                result.append([key, self.freq[key]])
            return result

    return CharFrequency(text).get_counts()

Common follow-up questions