CODEWITHSUNDEEP
javaarrays
Abhilash Muthuraj
Abhilash Muthuraj

Reputation: 2028

Why do I get an ArrayIndexOutOfBoundsException in this prime number check?

I was finding out highest prime factor which divides num, as shown in program, there's a issue with array and 

arr[j] = i;
j++;

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
    at primenum.main(primenum.java:13)

//to find highest prime factor
public class primenum {
    public static void main(String[] args) {
          double num = 600851475143.0;
           int j = 1;
          int arr[] = {j};

          for(int i=2; i<=num/2; i++)
          {
              if((num%i) == 0 )
              {
                  arr[j] = i;
                  j++;
              }

          }
          // take the last item from array, coz its last big prime
          System.out.println("largest prime is "+ arr[j-1]);

    }
}

What is best way to solve this problem??

I'm solving this problem by,

For prime I need to do more, but I'm stuck in initial stage.

Upvotes: 2

Views: 386

Answers (5)

Hank Gay
Hank Gay

Reputation: 71979

This line

int arr[] = {j};

Creates an array that only contains the value of j when it is executed. You probably want

int arr[] = new int[j];

UPDATE: Based on the answer you left below, trial division is taking too long. The Sieve of Eratosthenes is a classic algorithm that is pretty efficient, but the Sieve of Atkin is one of the most advanced algorithms for finding primes.

Upvotes: 3

President James K. Polk
President James K. Polk

Reputation: 42010

It looks like you are finding all divisors of num; one of these will be the largest prime factor. Two related facts alone should help make the problem tractable for smallish numbers:
1. If d is a divisor, then so is num/d.
2. you needn't check for any divisors greater than the sqrt(num).

To keep track of divisors, use a Set object.

Upvotes: 1

Daniel
Daniel

Reputation: 657

Looks like you start j = 1 and your array only has one element in it to begin, so on the first pass through your for loop you look for arr[1], but the first element in an array is at arr[0]. Java arrays are zero indexed meaning if you have 10 elements in the array they are located in arr[0] to arr[9].

Upvotes: 0

tonio
tonio

Reputation: 10541

Arrays in java are not lists: once allocated, your array won't grow magically.

You created the array with: int arr[] = {j}; thus the array has one cell only.

You should initialise your array with at least num/2 cells, with something like int arr[] = new int[num/2]; arr[0] = j;

Upvotes: 0

bkritzer
bkritzer

Reputation: 1416

By creating array arr[] = {j}, you have created an array which contains simply j, or 1. That means the length of the array is 1, because it contains 1 element. Thus, arr[1] is out of bounds. Java does not dynamically resize arrays, so you must create a sufficiently large array to contain all of the data you plan to hold. Either that or use something like an ArrayList, which is dynamically resizeable.

Upvotes: 2

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