Euler Problem 8 – Largest product in a series

Largest product in a series

Problem 8

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

The first glance at the problem is quite intimidating as there are 1000 digits in it. As soon as I went through the problem i knew I had to use one of the rollapply function from the zoo library in R. I think this is the only one problem where there is going to be little help using R because of rollapply function.

library(zoo)
largestProductInASeries = function(numberInCharacterForm, windowSize) {
numberVector = as.numeric(unlist(strsplit(numberInCharacterForm, "")))
productVector = rollapply(numberVector, windowSize, prod)
maxProduct = max(productVector)
baseIndex = which.max(productVector)
adjNumbers = numberVector[baseIndex:(baseIndex + windowSize - 1)]
cat(sprintf("The maximum product of %d adjacent numbers are : %f\n", windowSize, maxProduct))
cat(sprintf("And the numbers are : %s", paste0(adjNumbers, collapse = " , ")))
}

The first step is we require the entire number series in a string format because I do not think any language would support a 1000 digit number. So getting the input in string format we split every character and then convert it into numeric. So for example if we enter number as “23412”, it will break it down into numeric vector of size 5, as 2, 3, 4, 1 and 2.

Now, what rollapply does is it will take rolling product of every windowSize numbers from the numberVector. So continuing our example, if we give input as

rollapply(c(2, 3, 4, 1, 2), 2, prod)
It would give result as :
#[1]  6 12  4  2

First 2 and 3 are multiplied, then 3 and 4, then 4 and 1 and so on.

We then save the maximum product from the above vector and store it in maxProduct. With which.max we get which is the first element of the series. Finally we print the results.

Checked it on a sample list.

largestProductInASeries("982186", 2)
The maximum product of 2 adjacent numbers are : 72
And the numbers are : 9 & 8

Then used it on the problem data.

x =("7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450")

largestProductInASeries(x, 4)
The maximum product of 4 adjacent numbers are : 5832
And the numbers are : 9 , 9 , 8 , 9

This gave the correct answer for the example given. However, when I tried to use the same with window size as 13. It gave an error :

largestProductInASeries(x, 13)

Error in sprintf("The maximum product of %d adjacent numbers are : %d\n",  : 
  invalid format '%d'; use format %f, %e, %g or %a for numeric objects

The product 13 adjacent numbers was too big to fit in %d, so changed %d to %f and it worked.

cat(sprintf("The maximum product of %d adjacent numbers are : %f\n", windowSize, maxProduct))

> largestProductInASeries(x, 13)
The maximum product of 13 adjacent numbers are : 23514624000.000000
And the numbers are : 5 , 5 , 7 , 6 , 6 , 8 , 9 , 6 , 6 , 4 , 8 , 9 , 5
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