Omid - Challenge 4

data-challenges
advanced-exercises
🔰 Calculate the mission’s income, where the daily rate starts at $1 on the first day and increases by $1 for the next continous days
Published

March 24, 2026

Illustration for Omid - Challenge 4

Challenge Description

🔰 Calculate the mission’s income, where the daily rate starts at $1 on the first day and increases by $1 for the next continous days

Solutions

library(tidyverse)
library(readxl)

path = "files/CH-004.xlsx"
input = read_excel(path, range = "B2:C17")
test  = read_excel(path, range = "G2:H5")

result = input %>%
  group_by(Person) %>%
  complete(`Mission Date` = seq.Date(min(as.Date(`Mission Date`)), max(as.Date(`Mission Date`)), by = "day")) %>%
  ungroup() %>%
  mutate(wday = wday(`Mission Date`, label = TRUE, locale = "en"),
         is_weekend = wday %in% c("Sat", "Sun")) %>%
  full_join(input %>% mutate(n = 1), by = c("Person", "Mission Date")) %>%
  mutate(n = ifelse(lag(wday) == "Fri" & lag(n) == 1, 1, n), 
         n = ifelse(lag(wday) == "Sat" & lag(n) == 1, 1, n)) %>%
  mutate(cons = consecutive_id(n == 1), .by = Person) %>%
  replace_na(list(n = 0)) %>%
  mutate(cum1 = cumsum(n), .by = c(Person, cons)) %>%
  summarise(cum = sum(cum1), .by = Person)

  • 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-004.xlsx"
input_data = pd.read_excel(path, usecols="B:C", skiprows=1, nrows=16)
test = pd.read_excel(path, usecols="G:H", skiprows=1, nrows=4)

input_data["Mission Date"] = pd.to_datetime(input_data["Mission Date"])

frames = []
for person, g in input_data.groupby("Person"):
    full_dates = pd.DataFrame({"Mission Date": pd.date_range(g["Mission Date"].min(), g["Mission Date"].max(), freq="D")})
    full_dates["Person"] = person
    frames.append(full_dates)
result = pd.concat(frames, ignore_index=True)
result["wday"] = result["Mission Date"].dt.day_name().str[:3]
result = result.merge(input_data.assign(n=1), on=["Person", "Mission Date"], how="left")

for person, idx in result.groupby("Person").groups.items():
    sub = result.loc[idx].copy()
    sub["n"] = sub["n"].where(~((sub["wday"].shift() == "Fri") & (sub["n"].shift() == 1)), 1)
    sub["n"] = sub["n"].where(~((sub["wday"].shift() == "Sat") & (sub["n"].shift() == 1)), 1)
    result.loc[idx, "n"] = sub["n"].values

out = []
for person, g in result.groupby("Person"):
    g = g.copy()
    g["flag"] = g["n"].eq(1)
    g["cons"] = g["flag"].ne(g["flag"].shift(fill_value=False)).cumsum()
    g["n"] = g["n"].fillna(0)
    g["cum1"] = g.groupby("cons")["n"].cumsum()
    out.append({"Person": person, "cum": g["cum1"].sum()})

result2 = pd.DataFrame(out)
print(result2.equals(test))

  • 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

    • Applies the rule iteratively until the output stabilizes

  • 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.