Omid - Challenge 369

data-challenges
advanced-exercises
🔰 Group 1 Group 2 Grouping Group the rows in the question table into two sets from midle, and allocate the middle row to the group with the lower total sales.
Published

March 24, 2026

Illustration for Omid - Challenge 369

Challenge Description

🔰 Group 1 Group 2 Grouping Group the rows in the question table into two sets from midle, and allocate the middle row to the group with the lower total sales.

Solutions

library(tidyverse)
library(readxl)

path <- "300-399/369/CH-369 Custom Grouping.xlsx"
input <- read_excel(path, range = "B3:E10")
test <- read_excel(path, range = "H3:I5")

result = input %>%
  mutate(rn = row_number()) %>%
  mutate(
    group = case_when(
      row_number() < ceiling(n() / 2) ~ "Group 1",
      row_number() > ceiling(n() / 2) ~ "Group 2",
      TRUE ~ "Middle"
    )
  ) %>%
  group_by(group) %>%
  summarise(sum_value = sum(`Total Sales`, na.rm = TRUE)) %>%
  pivot_wider(names_from = group, values_from = sum_value) %>%
  mutate(
    `Group 1` = `Group 1` + if_else(`Group 1` < `Group 2`, Middle, 0),
    `Group 2` = `Group 2` + if_else(`Group 2` <= `Group 1`, Middle, 0)
  ) %>%
  select(`Group 1`, `Group 2`) %>%
  pivot_longer(everything(), names_to = "IDs", values_to = "Sales")

all.equal(result, test)
## [1] TRUE

  • Logic:

    • Reads the workbook ranges needed for the challenge

    • Reshapes the data into the grain required by the task

    • 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
import numpy as np

path = "300-399/369/CH-369 Custom Grouping.xlsx"
input = pd.read_excel(path, usecols="B:E", skiprows=2, nrows=8)
test = pd.read_excel(path, usecols="H:I", skiprows=2, nrows=2)

n = len(input)
mid = int(np.ceil(n / 2))
g = (
    input
    .assign(rn=lambda d: np.arange(1, n + 1))
    .assign(group=lambda d: np.select(
        [d.rn < mid, d.rn > mid],
        ["Group 1", "Group 2"],
        default="Middle"
    ))
    .groupby("group")["Total Sales"]
    .sum()
)
g1, g2, m = g.get("Group 1", 0), g.get("Group 2", 0), g.get("Middle", 0)

result = pd.DataFrame({
    "IDs": ["Group 1", "Group 2"],
    "Sales": [g1 + (m if g1 < g2 else 0),
              g2 + (m if g2 <= g1 else 0)]
})

print(result.equals(test))
# Output: 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 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.