# Grooving Project Euler # 8

I recently watched a video course on groovy and started working on some of the Project euler problems to flex my groovy* muscle.

(This is problem #8)

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

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

This is my code:

def number = """
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
""".replace('\n', '')

def digits = 13

def ans = 0
def ansStr = ""

for (i in 0..number.size()-digits) {
def str = number[i..<(i+digits)]
def prod = str.inject(1) {acc, it -> acc * (it as long)}
if (prod > ans) {
ans = prod
ansStr = str
}
}

def formattedStr = ansStr//.chars().collect {it as char}.join('*')
println "$ans,$formattedStr"


Is there any way to improve this code to be more like the groovy* way of doing this?

* sorry about that, such a bad name for a language :)

## 1 Answer

Your code is too imperative while groovy favours Functional Programming:

def largest_product =
(0 .. number.size() - digits)
.map {number[i..<(i + digits)].inject(1) {acc, it -> acc * (it as long)}}
.max

println largest_product


As you can see, the code is now without all those confusing variables, and the algorithm is more clearly expressed.

• @JAtkin I mean collect, the Groovy name of map. – Caridorc Nov 9 '15 at 14:26