library(tidyverse)
library(readxl)
path <- "Excel/800-899/850/850 First Missing Letter.xlsx"
input <- read_excel(path, range = "A1:A11")
test <- read_excel(path, range = "B1:B11") %>% replace_na(list(`Answer Expected` = ""))
words <- input$Data
state <- reduce(
words,
.init = list(
alphabet = letters,
results = character()
),
.f = function(acc, word) {
chars_now <- str_split_1(str_to_lower(word), "") %>% unique()
alphabet_new <- setdiff(acc$alphabet, chars_now)
chosen <- if (length(alphabet_new) == 0) "" else alphabet_new[1]
alphabet_new <- setdiff(alphabet_new, chosen)
list(
alphabet = alphabet_new,
results = c(acc$results, chosen)
)
}
)
result <- tibble(final_missing = state$results)
all.equal(result$final_missing, test$`Answer Expected`)
# [1] TRUEExcel BI - Excel Challenge 850
excel-challenges
excel-formulas
🔰 Reading from top to bottom, list the first missing letter from a to z considering all previous strings encountered when going from top to bottom.

Challenge Description
🔰 Reading from top to bottom, list the first missing letter from a to z considering all previous strings encountered when going from top to bottom. Once a missing letter is identified, it should not be identified for any succeeding string.
Solutions
- Logic: Read the workbook ranges needed for the challenge; Parse the packed text or string structure.
- 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
import string
path = "Excel/800-899/850/850 First Missing Letter.xlsx"
input = pd.read_excel(path, usecols="A", nrows=11)
test = pd.read_excel(path, usecols="B", nrows=11).fillna({"Answer Expected": ""})
words = input["Data"].tolist()
alphabet = list(string.ascii_lowercase)
results = []
for word in words:
chars_now = set(word.lower())
alphabet_new = [ch for ch in alphabet if ch not in chars_now]
chosen = "" if not alphabet_new else alphabet_new[0]
alphabet = [ch for ch in alphabet_new if ch != chosen]
results.append(chosen)
result = pd.DataFrame({"final_missing": results})
print(result["final_missing"].equals(test["Answer Expected"]))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.