Excel BI - Excel Challenge 798

excel-challenges

excel-formulas

🔰 Traversing from back, extract

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 798

Challenge Description

🔰 Traversing from back, extract

Solutions

library(tidyverse)
library(readxl)

path = "Excel/700-799/798/798 From back - Extract String Before a Repeated Character.xlsx"
input = read_excel(path, range = "A1:A11")
test  = read_excel(path, range = "B1:B11")

cut_at_first_global_dup <- function(x) {
  map_chr(x, ~{
    ch <- str_split(.x, "", simplify = TRUE)
    dups <- names(which(table(ch) > 1))
    if (length(dups) == 0) return(.x)
    start <- max(map_int(dups, ~ which(ch == .x)[1])) + 1
    str_sub(.x, start)
  })
}

result = input %>%
  mutate(`Answer Expected` = cut_at_first_global_dup(String))

all.equal(result$`Answer Expected`, test$`Answer Expected`)
result$`Answer Expected` == test$`Answer Expected`
# one solution different from expected

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

path = "700-799/798/798 From back - Extract String Before a Repeated Character.xlsx"
input = pd.read_excel(path, usecols="A", nrows=11)
test = pd.read_excel(path, usecols="B", nrows=11)

def cut_at_first_global_dup(x):
    result = []
    for s in x:
        chars = list(str(s))
        counts = pd.Series(chars).value_counts()
        dups = counts[counts > 1].index.tolist()
        if not dups:
            result.append(s)
            continue
        starts = [chars.index(d) for d in dups]
        start = max(starts) + 1
        result.append(''.join(chars[start:]))
    return result

input['Answer Expected'] = cut_at_first_global_dup(input.iloc[:, 0])
print(input['Answer Expected'] == test.iloc[:, 0])
# one solution different from 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.