Excel BI - Excel Challenge 821

🔰 Pascal’s triangle is a triangular array of numbers that starts with 1 at the top, with each subsequent row beginning and ending with 1, and every other number being the sum of the…

Illustration for Excel BI - Excel Challenge 821

Challenge Description

🔰 Pascal’s triangle is a triangular array of numbers that starts with 1 at the top, with each subsequent row beginning and ending with 1, and every other number being the sum of the two numbers directly above it. But current challenge is a variation. Here, every other number is either sum or product of the two numbers directly above it. Sum or Product is controlled by column B.

Solutions

R Solution

library(tidyverse)
library(readxl)

path = "Excel/800-899/821/821 Pascal Triangle Variation.xlsx"
input = read_excel(path, range = "B2:B10", col_names = FALSE) %>% pull(1)
test  = read_excel(path, range = "E2:U10", col_names = FALSE) %>% as.matrix()

generate_custom_pascal <- function(signs) {
  n <- length(signs) 
  triangle <- matrix(NA, n, 2 * n - 1)
  triangle[1, n] <- 1

  for (i in 2:n) {
    for (j in (n - (i - 1)):(n + (i - 1))) {
      left <- ifelse(j - 1 < 1, 0, triangle[i - 1, j - 1])
      right <- ifelse(j + 1 > (2 * n - 1), 0, triangle[i - 1, j + 1])
      if (is.na(left) & is.na(right)) {
        triangle[i, j] <- NA
      } else {
        left <- ifelse(is.na(left), 0, left)
        right <- ifelse(is.na(right), 0, right)
        if (signs[i] == "+") {
          triangle[i, j] <- left + right
        } else if (signs[i] == "*") {
          left <- ifelse(left == 0, 1, left)
          right <- ifelse(right == 0, 1, right)
          triangle[i, j] <- left * right
        }
      }
    }
  } 

  triangle
}

result <- generate_custom_pascal(input)

all(result == test, na.rm = TRUE)
# [1] TRUE

R Analysis

• Logic: Read the workbook ranges needed for the challenge; Apply the business rule conditions explicitly; 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.

Python Solution

import pandas as pd
import numpy as np

path = "800-899/821/821 Pascal Triangle Variation.xlsx"
input_signs = pd.read_excel(path, usecols="B", skiprows=1, nrows=9, header=None).values.flatten()
test = pd.read_excel(path, usecols="E:U", skiprows=1, nrows=9, header=None).values

def generate_custom_pascal(signs):
    n = len(signs)
    triangle = np.full((n, 2 * n - 1), np.nan)
    triangle[0, n - 1] = 1

    for i in range(1, n):
        for j in range(n - i - 1, n + i):
            left = 0 if j - 1 < 0 else triangle[i - 1, j - 1]
            right = 0 if j + 1 > (2 * n - 2) else triangle[i - 1, j + 1]
            
            if np.isnan(left) and np.isnan(right):
                triangle[i, j] = np.nan
            else:
                left = 0 if np.isnan(left) else left
                right = 0 if np.isnan(right) else right
                if signs[i] == "+":
                    triangle[i, j] = left + right
                elif signs[i] == "*":
                    left = 1 if left == 0 else left
                    right = 1 if right == 0 else right
                    triangle[i, j] = left * right
    return triangle


result = generate_custom_pascal(input_signs)
print(np.array_equal(result, test, equal_nan=True)) # Truessssssssssssss

Python

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.

Related Challenges

• Excel Challenge 822
• Excel Challenge 823
• Excel Challenge 824