Omid - Challenge 9

data-challenges
advanced-exercises
🔰 Find the length of Largest repetition of a specific pattern!
Published

March 24, 2026

Illustration for Omid - Challenge 9

Challenge Description

🔰 Find the length of Largest repetition of a specific pattern!

Solutions

library(tidyverse)
library(readxl)

input = read_excel("files/CH-009.xlsx", range = "B2:D32")
test1 = read_excel("files/CH-009.xlsx", range = "F2:G5") %>% janitor::clean_names()
test2 = read_excel("files/CH-009.xlsx", range = "I2:J5") %>% janitor::clean_names()


result = input %>%
  group_by(Product) %>% 
  summarise(seq = str_c(Result, collapse = "")) %>%
  ungroup()

# pattern +--

process_pattern1 = function(string) {
  string = string %>%
    str_replace_all("\\+\\-\\-", "3") %>%
    str_replace_all("\\-\\-3", "/5") %>%
    str_replace_all("3\\+\\-", "5/") %>%
    str_replace_all("\\-3", "/4") %>%
    str_replace_all("3\\+", "4/") %>%
    str_replace_all("/4\\+", "/5/") %>%
    str_replace_all("44", "8") %>%
    str_replace_all("35", "8") 
  
  numbers = as.numeric(str_extract_all(string, "")[[1]]) %>% max(., na.rm = TRUE) 
  return(numbers)
}

result1 = result %>%
  rowwise() %>%
  mutate(max_seq = map_dbl(seq, process_pattern1))
  
identical(result1$max_seq, test1$x)
#> [1] TRUE

# pattern +-

process_pattern2 = function(string) {
  string = string %>%
    str_replace_all("\\+\\-", "2") %>%
    str_replace_all("^\\-2", "3") %>%
    str_replace_all("2\\+$", "3") %>%
    str_replace_all("\\-3", "4") %>%
    str_replace_all("22", "4") %>%
    str_replace_all("32|23", "5") %>%
    str_replace_all("44", "8")

    numbers = as.numeric(str_extract_all(string, "")[[1]]) %>% max(., na.rm = TRUE)

  
  return(numbers)
}

result2 = result %>%
  rowwise() %>%
  mutate(max_seq = map_dbl(seq, process_pattern2))

identical(result2$max_seq, test2$x)
#> [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

    • 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

input_data = pd.read_excel("CH-009.xlsx", usecols="B:D", skiprows=1, nrows=31)
test1 = pd.read_excel("CH-009.xlsx", usecols="F:G", skiprows=1, nrows=4)
test2 = pd.read_excel("CH-009.xlsx", usecols="I:J", skiprows=1, nrows=4)

result = input_data.groupby("Product", as_index=False)["Result"].agg(seq="".join)

def process_pattern1(string):
    string = (string.replace("+--", "3")
                    .replace("--3", "/5")
                    .replace("3+-", "5/")
                    .replace("-3", "/4")
                    .replace("3+", "4/")
                    .replace("/4+", "/5/")
                    .replace("44", "8")
                    .replace("35", "8"))
    digits = [int(ch) for ch in string if ch.isdigit()]
    return max(digits) if digits else 0

def process_pattern2(string):
    string = (string.replace("+-", "2")
                    .replace("-2", "3", 1) if string.startswith("-2") else string)
    if string.endswith("2+"):
        string = string[:-2] + "3"
    string = (string.replace("-3", "4")
                    .replace("22", "4")
                    .replace("32", "5")
                    .replace("23", "5")
                    .replace("44", "8"))
    digits = [int(ch) for ch in string if ch.isdigit()]
    return max(digits) if digits else 0

result1 = result.assign(max_seq=result["seq"].map(process_pattern1))
result2 = result.assign(max_seq=result["seq"].map(process_pattern2))

print(result1["max_seq"].equals(test1.iloc[:, 1]))
print(result2["max_seq"].equals(test2.iloc[:, 1]))

  • 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.