Omid - Challenge 201

data-challenges
advanced-exercises
🔰 Question Result ID M AB C D E
Published

March 24, 2026

Illustration for Omid - Challenge 201

Challenge Description

🔰 Question Result ID M AB C D E

Solutions

library(tidyverse)
library(readxl)

path = "files/CH-201Column Splitting.xlsx"
input = read_excel(path, range = "B2:B7")
test  = read_excel(path, range = "D2:F7")

is_palindrome = function(x) {{stringi::stri_reverse(x) == x}}

result = input %>%
  mutate(is_palindrome = map_lgl(ID, is_palindrome)) %>%
  mutate(vowelloc = str_locate(ID, "[AEIOU]")[,1]) %>%
  mutate(is_even = nchar(ID) %% 2 == 0) %>%
  mutate(`Part 1` = case_when(
    is_palindrome & is_even ~ str_sub(ID, 1, nchar(ID)/2),
    is_palindrome & !is_even ~ str_sub(ID, 1, nchar(ID)/2),
    !is_palindrome ~ str_sub(ID, 1, vowelloc-1)
  ),
  Middle = case_when(
    is_palindrome & is_even ~ NA,
    is_palindrome & !is_even ~ str_sub(ID, nchar(ID)/2+1, nchar(ID)/2+1),
    !is_palindrome ~ str_sub(ID, vowelloc, vowelloc)
  ),
  `Part 2`  = case_when(
    is_palindrome & is_even ~ str_sub(ID, nchar(ID)/2+1, nchar(ID)),
    is_palindrome & !is_even ~ str_sub(ID, nchar(ID)/2+2, nchar(ID)),
    !is_palindrome ~ str_sub(ID, vowelloc+1, nchar(ID))
  )) %>%
  select(`Part 1`, Middle, `Part 2`)

all.equal(test, result, check.attributes = FALSE)
#> [1] TRUE

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Builds the intermediate columns that drive the final result

    • Parses the text patterns directly instead of relying on manual cleanup

  • Strengths:

    • The R solution stays close to the workbook rule and keeps the transformation compact.

  • Areas for Improvement:

    • The code assumes the sheet structure and source ranges remain stable.

  • Gem:

    • The strongest part of the solution is choosing the right intermediate representation before shaping the final output.
import pandas as pd
import re
import numpy as np

path = "CH-201Column Splitting.xlsx"

input = pd.read_excel(path, usecols="B", skiprows=1, nrows=6)
test = pd.read_excel(path, usecols="D:F", skiprows=1, nrows=6)

vowels = set("AEIOU")

def process_id(s):
    is_palindrome = s == s[::-1]
    is_even = len(s) % 2 == 0
    match = re.search(r"[AEIOU]", s)
    vowelloc = match.start() if match else None

    if is_palindrome:
        half = len(s) // 2
        if is_even:
            part1 = s[:half]
            middle = np.NaN
            part2 = s[half:]
        else:
            part1 = s[:half]
            middle = s[half]
            part2 = s[half+1:]
    else:
        if vowelloc is None:
            part1, middle, part2 = s, None, ""
        else:
            part1 = s[:vowelloc]
            middle = s[vowelloc]
            part2 = s[vowelloc+1:]
    return pd.Series({"Part 1": part1, "Middle": middle, "Part 2": part2})

result = input["ID"].apply(process_id)
result = result.reset_index(drop=True)

print(result.equals(test)) # True

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Parses the text patterns directly instead of relying on manual cleanup

  • Strengths:

    • The Python version follows the same rule in a direct dataframe-oriented implementation.

  • Areas for Improvement:

    • The code assumes the workbook layout remains stable, so any sheet redesign would require small adjustments.

  • Gem:

    • The implementation stays close to the original workbook rule instead of adding unnecessary abstraction.

Difficulty Level

This task is moderate:

  • The core logic is clear, but the correct transformation pattern is not obvious from the raw input.

  • The challenge combines multiple reshaping, grouping, or parsing steps.