Excel BI - Excel Challenge 749

excel-challenges

excel-formulas

🔰 List Portmanteau words

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 749

Challenge Description

🔰 List Portmanteau words

Solutions

library(tidyverse)
library(readxl)

path = "Excel/700-799/749/749 Portmanteau Words v2.xlsx"
input1 = read_excel(path, range = "A1:A10")
input2 = read_excel(path, range = "B1:C10")
test  = read_excel(path, range = "D1:D6")

get_all_substrings =  function(x) {
  if (is.na(x) || x == "") return(character(0))
  len =  nchar(x)
  unique(c(substring(x, 1, 1:len), substring(x, len:1, len)))
}

result =  input2 %>%
  mutate(port1 = map2(Word1, Word2, ~ {
    s1 =  get_all_substrings(.x)
    s2 =  get_all_substrings(.y)
    ports =  as.vector(outer(s1, s2, paste0))
    intersect(ports, input1$Word)
  })) %>%
  select(port1) %>%
  unnest(port1) %>%
  arrange(port1)

all.equal(result$port1, test$`Answer Expected`, check.attributes = FALSE)
# [1] TRUE

Logic: Read the workbook ranges needed for the challenge; Derive the required intermediate columns.
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.

import pandas as pd

path = "700-799/749/749 Portmanteau Words v2.xlsx"
input1 = pd.read_excel(path, usecols="A", nrows=10)
input2 = pd.read_excel(path, usecols="B:C", nrows=10)
test = pd.read_excel(path, usecols="D", nrows=5)

def get_all_substrings(x):
    if not isinstance(x, str) or not x:
        return []
    return list({x[:i] for i in range(1, len(x)+1)} | {x[i-1:] for i in range(len(x), 0, -1)})

ports = [
    word
    for _, row in input2.iterrows()
    for word in set(
        a + b
        for a in get_all_substrings(row['Word1'])
        for b in get_all_substrings(row['Word2'])
    ) & set(input1.iloc[:, 0])
]

result = pd.DataFrame({'port1': sorted(ports)})

print(test['Answer Expected'].equals(result['port1']))
# True

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.