Excel BI - Excel Challenge 657

🔰 Find the largest substring number which has alternating odd/even or even/odd digits.

Illustration for Excel BI - Excel Challenge 657

Challenge Description

🔰 Find the largest substring number which has alternating odd/even or even/odd digits.

Solutions

R Solution

library(tidyverse)
library(readxl)

path = "Excel/657 Largest Alternating Even Odd or Odd Even Substring.xlsx"
input = read_excel(path, range = "A1:A10")
test  = read_excel(path, range = "B1:B10")

generate_substrings <- function(num_str) {
  nchar_num_str <- nchar(num_str)
  unique(flatten_chr(map(seq_len(nchar_num_str), function(start) {
    map_chr(start:nchar_num_str, function(end) substr(num_str, start, end))
  })))
}

is_alternating <- function(num_str) {
  all(diff(as.numeric(strsplit(num_str, "")[[1]]) %% 2) != 0)
}

largest_alternating_substring <- function(number) {
  num_str <- as.character(number)
  valid_substrings <- generate_substrings(num_str) %>%
    map_chr(~ gsub("^0+", "", .x) %>% ifelse(. == "", "0", .)) %>%
    keep(is_alternating)
  if (length(valid_substrings) == 0) return(NA)
  max(valid_substrings[nchar(valid_substrings) == max(nchar(valid_substrings))])
}

result = input %>%
  mutate(LAS = map_chr(Numbers, largest_alternating_substring))

all.equal(result$LAS, test$`Answer Expected`, check.attributes = FALSE)
# [1] TRUE

R Analysis

• Logic: Read the workbook ranges needed for the challenge; Derive the required intermediate columns; Apply the business rule conditions explicitly.
• Strengths: The code maps the workbook rule into a compact, reproducible pipeline.
• Areas for Improvement: The solution assumes the workbook layout and selected ranges remain stable, so any structural change in the sheet would require small adjustments.
• Gem: The elegant part is how little code is needed once the correct intermediate representation is chosen.

Python Solution

import pandas as pd

path = "657 Largest Alternating Even Odd or Odd Even Substring.xlsx"
input = pd.read_excel(path, usecols="A", nrows=10)
test = pd.read_excel(path, usecols="B", nrows=10).astype(str)

def generate_substrings(num_str):
    return list(set(num_str[i:j] for i in range(len(num_str)) for j in range(i + 1, len(num_str) + 1)))

def is_alternating(num_str):
    return all((int(num_str[i]) % 2) != (int(num_str[i + 1]) % 2) for i in range(len(num_str) - 1))

def largest_alternating_substring(number):
    num_str = str(number)
    valid_substrings = [s.lstrip('0') or '0' for s in generate_substrings(num_str) if is_alternating(s)]
    if not valid_substrings:
        return None
    max_length = max(len(s) for s in valid_substrings)
    return max(s for s in valid_substrings if len(s) == max_length)

input['LAS'] = input.iloc[:, 0].apply(largest_alternating_substring)

print(input['LAS'].equals(test['Answer Expected'])) # True

Python

The Python version keeps the algorithm explicit, which helps when the challenge depends on a greedy or iterative rule.

Difficulty Level

Easy / Medium

The business rule is clear, though the workbook still needs a few transformation steps to reach the expected output.

Related Challenges

• Excel Challenge 658
• Excel Challenge 659
• Excel Challenge 660