Project Euler: #8

Question

When trying to answer this problem:

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?

I get 2091059712 however Euler says the answer is incorrect is there anything I may be doing incorrectly?

public class LargestProductThirteen{
  public static void main( String[] args ) {
    final String num = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
    long greatestProduct = 0;
    for (int i = 0; i < num.length() - 12; i++) {
        long sum = Character.getNumericValue(num.charAt(i))* 
            Character.getNumericValue(num.charAt(i+1))* 
            Character.getNumericValue(num.charAt(i+2))* 
            Character.getNumericValue(num.charAt(i+3))* 
            Character.getNumericValue(num.charAt(i+4))* 
            Character.getNumericValue(num.charAt(i+5))* 
            Character.getNumericValue(num.charAt(i+6))* 
            Character.getNumericValue(num.charAt(i+7))* 
            Character.getNumericValue(num.charAt(i+8))* 
            Character.getNumericValue(num.charAt(i+9))* 
            Character.getNumericValue(num.charAt(i+10))* 
            Character.getNumericValue(num.charAt(i+11))* 
            Character.getNumericValue(num.charAt(i+12));
        if (sum > greatestProduct)
            greatestProduct = sum;
    }
    System.out.println(greatestProduct);
  }
}

Project Euler specifically stated not to post the answers on public forums — Marvin, Jul 18 '19 at 18:36
I did not see anyone talk about this. If we are to find the product of the consecutive 13 digits and one of the digit is a 0, then we can skip past as the product will be 0. In essence, you can actually skip past the digits until you find a non zero value. None of the algorithms below talks about this. — Joe Ferndz, Jan 03 '21 at 07:25

score 8 · Accepted Answer · answered May 29 '14 at 20:02

You are performing integer arithmetic when you are multiplying all those characters' numeric values together. With high digits, and 13 of them, it is likely that such a product would overflow an int, whose max value is about 2 billion (10 digits).

The Character.getNumericValue method returns an int. Cast the first return value as a long to force long math.

long sum = (long) Character.getNumericValue(num.charAt(i))* 
   Character.getNumericValue(num.charAt(i+1))* 
   ...

Incidentally, even though you have greatestProduct already, for some reason you defined this variable as sum. Just for semantics' sake, I would name it product.

score 3 · Answer 2 · answered May 29 '14 at 20:08

You are currently doing integer multiplication (e.g. multiplication of int(s)). Given your existing code - the easiest solution that I see is to change your sum calculation,

// This is a product, not a sum.
long sum = Long.valueOf(num.charAt(i))
        * Long.valueOf(num.charAt(i + 1))
        * Long.valueOf(num.charAt(i + 2))
        * Long.valueOf(num.charAt(i + 3))
        * Long.valueOf(num.charAt(i + 4))
        * Long.valueOf(num.charAt(i + 5))
        * Long.valueOf(num.charAt(i + 6))
        * Long.valueOf(num.charAt(i + 7))
        * Long.valueOf(num.charAt(i + 8))
        * Long.valueOf(num.charAt(i + 9))
        * Long.valueOf(num.charAt(i + 10))
        * Long.valueOf(num.charAt(i + 11))
        * Long.valueOf(num.charAt(i + 12));

score 2 · Answer 3 · answered May 29 '14 at 20:13

That is a good puzzle.

Your code is failing because you are using Character.getNumericValue. You could use BigDecimal instead.

Hints
* Don't think that the whole number is a long String.
* You are looking for the max number, not the result.
* Adjacent goes in all directions.
* Don't try to use longs (or any other number containers). Regardless of the result, 9*9*7 is greater than 7*7*7. You don't have to multiply them to know that, so build a structure that finds the greatest between the inputs.

score 1 · Answer 4 · answered Jun 02 '14 at 01:41

Building on Alexandre Santos comments, treat each 13 digit number as a string of 13 chars, you can sort the chars within the string, so 98031 becomes 01389, order of operands doesn't matter with multiplication (9*8*3 = 3*8*9). Now the problem is just finding the largest string ( compare strings with <). You only have to do the multiplication once, after you have found the largest string (by sort order not size.

score 1 · Answer 5 · answered Nov 11 '14 at 22:01

Another implementation (using for instead of 13 similar things):

public static void main(String[] args) throws Exception {
    String x = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
    int numberOfDigits = 13;
    int dif = x.length() - numberOfDigits;
    long max = 0;
    for ( int i = 0 ; i <= dif ; i++ ) {
        long p = 1;
        int sup = i + numberOfDigits;       
        for ( int j = i ; j < sup ; j++ ) {
            p*=Character.getNumericValue(x.charAt(j));
        }
        if ( p > max ) {
            max = p; 
        }
    }
    System.out.println(max);
}

It can be improved using the following observation:

0 * ? = 0

simple example:

if ( x.charAt(i) == '0' ) {
    continue;
}

Project Euler: #8

5 Answers5

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