Excel BI - Excel Challenge 719

excel-challenges

excel-formulas

🔰 List those entries where opening parenthesis is matched by a closing parenthesis in right order.

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 719

Challenge Description

🔰 List those entries where opening parenthesis is matched by a closing parenthesis in right order.

Solutions

library(tidyverse)
library(readxl)

path = "Excel/700-799/719/719 Parentheses Matching.xlsx"
input = read_excel(path, range = "A1:A10")
test = read_excel(path, range = "B1:B6")

balance = function(s) {
  stack = character()
  open = c("(" = ")", "{" = "}", "[" = "]")
  for (ch in strsplit(s, "")[[1]]) {
    if (ch %in% names(open)) stack = c(stack, open[ch])
    else if (ch %in% open && (length(stack) == 0 || tail(stack, 1) != ch)) return(FALSE)
    else if (ch %in% open) stack = stack[-length(stack)]
  }
  length(stack) == 0
}

result = input %>%
  mutate(result = map_lgl(Brackets, balance)) %>%
  filter(result) %>%
  select(-result)

all.equal(result$Brackets, test$`Answer Expected`)

Logic: Read the workbook ranges needed for the challenge; Derive the required intermediate columns; Iterate through the sequence until the rule is satisfied.
Strengths: The algorithm is explicit about the sequence rule, so the control flow is easy to validate against the prompt.
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 non-obvious part is the local rule inside the loop, because that rule determines the whole output.

import pandas as pd

path = "700-799/719/719 Parentheses Matching.xlsx"
input_df = pd.read_excel(path, usecols="A", nrows=10)
test_df = pd.read_excel(path, usecols="B", nrows=6)

def balance(s):
    stack = []
    openers = {'(': ')', '{': '}', '[': ']'}
    for ch in s:
        if ch in openers:
            stack.append(openers[ch])
        elif ch in openers.values():
            if not stack or stack[-1] != ch:
                return False
            stack.pop()
    return not stack

result = input_df[input_df['Brackets'].apply(balance)]

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.