Omid - Challenge 10

data-challenges
advanced-exercises
🔰 The question table shows inventory levels for materials required to produce products, with specific combinations (1 A, 2 B, 3 C per product).
Published

March 24, 2026

Illustration for Omid - Challenge 10

Challenge Description

🔰 The question table shows inventory levels for materials required to produce products, with specific combinations (1 A, 2 B, 3 C per product).

Solutions

library(tidyverse)
library(readxl)

input = read_excel("files/CH-0010.xlsx", range = "B2:D17")
test  = read_excel("files/CH-0010.xlsx", range = "G2:H7")

result = input %>%
  mutate(requirements = case_when(
    Material == "A" ~ 1,
    Material == "B" ~ 2,
    Material == "C" ~ 3
  ), product_available = Inventory%/%requirements) %>%
  group_by(Date) %>%
  mutate(min_available = min(product_available), 
         usage = min_available*requirements) %>%
  summarise(usage = sum(usage),
            inventory = sum(Inventory),
            Efficiency_rate = usage/inventory)

identical(result$Efficiency_rate, test$`Efficeincy Rate`)
# [1] TRUE

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Aggregates or ranks values at the relevant grouping level

    • Builds the intermediate columns that drive the final result

  • 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

path = "CH-0010.xlsx"
input_data = pd.read_excel(path, usecols="B:D", skiprows=1, nrows=16)
test = pd.read_excel(path, usecols="G:H", skiprows=1, nrows=6)

requirements_map = {"A": 1, "B": 2, "C": 3}

result = input_data.assign(
    requirements=lambda df: df["Material"].map(requirements_map),
    product_available=lambda df: df["Inventory"] // df["requirements"],
)
result = (
    result.groupby("Date", as_index=False)
    .apply(
        lambda g: pd.Series(
            {
                "usage": (g["product_available"].min() * g["requirements"]).sum(),
                "inventory": g["Inventory"].sum(),
            }
        )
    )
    .reset_index(drop=True)
)
result["Efficiency_rate"] = result["usage"] / result["inventory"]

print(result["Efficiency_rate"].equals(test["Efficeincy Rate"]))

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Aggregates or ranks values at the relevant grouping level

    • Builds the intermediate columns that drive the final result

  • 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.