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 = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450.replace('\n', '')def digits = 13def ans = 0def 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, $formattedStrIs 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 :)
Grooving Project Euler # 8
programming challenge;groovy
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)}}    .maxprintln largest_productAs you can see, the code is now without all those confusing variables, and the algorithm is more clearly expressed.