Question

Days ago I had a job interview were they ask me how I would calculate the sum of all the numbers multiples of 2 or 5 from 1 to 10000 using a the c language. I did this:

int mult2_5()
{
     int i, sum=0;
     for(i = 1; i <= 10000; i++)
        if(i % 2 == 0 || i % 5 == 0)
          sum += i;
     return sum;  
}

I as wonder if it was any faster implementation that this one?

Answer 1

The modulus operator is inefficient. A more faster implementation would be something like this:

int multiply2_5(int max)
{
  int i, x2 = 0,x5 = 0,x10 = 0;

  for(i = 2; i < max; i+=2)    x2 += i;    // Store all multiples of 2   O(max/2)
  for(i = 5; i < max; i+=5)    x5 += i;    //  Store all multiples of 3  O(max/5)
  for(i = 10; i < max; i+=10)  x10 += i;   //   Store all multiples 10;  O(max/10)

  return x2+x5-x10;
}

In this solution I had to take out multiples of 10 because, 2 and 5 have 10 as multiple so on the second loop it will add multiples of 10 that already been added in the first loop; The three loops combine have O(8/10 max).

Another even better solution is if you take a mathematical approach.

You are trying to sum all numbers like this 2 + 4 + 6 + 8 … 10000 and 5 + 10 + 15 +20 + … 10000 this is the same of having 2 * (1 + 2 + 3 + 4 + … + 5000) and 5 * ( 1 + 2 + 3 + 4 + … + 2000), the sum of ‘n’ natural number is (n * (n + 1)) (source) so you can calculate in a constant time, as it follows:

int multiply2_5(int max)
{
    // x = 2 + 4 + 6 + ... = 2 * (1 + 2 + 3 +...)
    // y = 5 + 10 + 15 + ... = 5 * (1 + 2 + 3 +...)
    // The sun of n natural numbers is sn = (n (n + 1)) / 2
   int x2 = max/ 2;                      // 2 * ( 1 +2 + 3 … max/2)
   int x5 = max /5;                      // 5 * ( 1 +2 + 3 … max/5)
   int x10 = max/ 10;                  
   int sn2 = 0.5 * (x2 * (x2+1));        // (n * (n + 1)) / 2
   int sn5 = 0.5 * (x5 * (x5+1)); 
   int sn10 = 0.5 * (x10 * (x10+1));
   return (2*sn2) + (5 *sn5) - (10*sn10);
}