Excel BI - Excel Challenge 699

excel-challenges

excel-formulas

🔰 Only one alphabet can be removed from Words list and it should become a palindrome.

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 699

Challenge Description

🔰 Only one alphabet can be removed from Words list and it should become a palindrome. List the palindrome word so generated. If no answer is possible, answer will be blank.

Solutions

library(tidyverse)
library(readxl)
library(stringi)

path = "Excel/699 Palindrome after one character removal.xlsx"
input = read_excel(path, range = "A1:A10")
test = read_excel(path, range = "B1:B10") %>%
  replace_na(list(`Answer Expected` = ""))

remove_each_char <- function(word) {
  map_chr(
    seq_len(nchar(word)),
    \(i) paste0(substr(word, 1, i - 1), substr(word, i + 1, nchar(word)))
  )
}
is_palindrome = function(word) {
  word == stri_reverse(word)
}

result = input %>%
  mutate(words = map(Words, remove_each_char)) %>%
  unnest_longer(words) %>%
  mutate(is_palindrome = map_lgl(words, is_palindrome)) %>%
  filter(is_palindrome) %>%
  select(-is_palindrome)

result2 = input %>%
  left_join(result, by = c("Words" = "Words")) %>%
  replace_na(list(words = ""))

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 = "699 Palindrome after one character removal.xlsx"
input = pd.read_excel(path, usecols="A", nrows=10, names=["Words"])
test = pd.read_excel(path, usecols="B", nrows=10, names=["Answer Expected"]).fillna("")

def is_palindrome(word):
    return word == word[::-1]

result = input["Words"].apply(
    lambda word: next((word[:i] + word[i+1:] for i in range(len(word)) if is_palindrome(word[:i] + word[i+1:])), "")
)
print(result)

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.