Project Euler: Problem 23 – Non-abundant sums

Problem 23:

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
 
A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.
 
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.
 
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Idea:

The problem gives us an upper limit on computations which helped tremendously. I generated a list of abundant numbers, and added the numbers 1 – 11 (all smaller than the smallest abundant number, and in hindsight, I could’ve added the numbers 1 – 23).
 
Then I looped from 12 to the upper limit defined in the problem above, and subtracted each of those numbers from each number in the list of abundant numbers. If the resulting difference was found in the list of abundant numbers, then the current number in the loop could be written as the sum of two abundant numbers and I could continue on to the next number; if not, then I added that number to the overall sum.

int answer = 0;
for(int i = 1; i < 12; answer+=i,i++);

HashSet<Integer> abundant = abundantNumbers(28123);

for(int i = 12; i < 28123; i++) {
   boolean abundantSum = false;
   //for each abundant number
   for(Integer num:abundant) {
      //if i - that number is in the list then it can
      //be written as sum of 2 abundant numbers, break
      if(abundant.contains(i - num)) {
         abundantSum = true;
         break;
      }
   }
   if(!abundantSum) {
      answer += i;
   }
}

System.out.println("Answer: "+answer);

Note: “abundantNumbers()” is a helper method I wrote that generates all the abundant numbers up to (and including) the given number.