Excel BI - Excel Challenge 726

excel-challenges

excel-formulas

🔰 Find

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 726

Challenge Description

🔰 Find

Solutions

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

path = "Excel/700-799/726/726 Palindromes Odd and Even Times.xlsx"
test = read_excel(path, range = "A1:A1001", sheet = "Sheet2")

get_valid_palindromes <- function(limit = 1000, max_n = 10000000) {
  10:max_n %>%
    as.character() %>%
    keep(~ .x == stri_reverse(.x)) %>%
    keep(
      ~ {
        d <- as.integer(str_split(.x, "")[[1]])
        all(table(d[d %% 2 == 0]) %% 2 == 0) &&
          all(table(d[d %% 2 == 1]) %% 2 == 1)
      }
    ) %>%
    head(limit) %>%
    as.integer()
}

result = get_valid_palindromes(1000, 10000000)

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 numpy as np

def get_valid_palindromes(limit=1000, max_n=10_000_000):
    res = []
    for n in range(10, max_n + 1):
        s = str(n)
        if s != s[::-1]:
            continue
        d = np.array(list(map(int, s)))
        ev = np.bincount(d[d % 2 == 0], minlength=10)
        od = np.bincount(d[d % 2 == 1], minlength=10)
        if np.all(ev % 2 == 0) and np.all((od == 0) | (od % 2 == 1)):
            res.append(n)
            if len(res) == limit:
                break
    return res

# Example usage:
result = get_valid_palindromes(1000, 10_000_000)11111
print(result)e

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.