Thanks for @MartinR @MartinR for the hint, and @BorisTheSpider @BorisTheSpider to point out the problem with my original answer that didn't subtract.
Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.
Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.
Use theThe Math, @Heslacher ...
... well ok, in the case of one or two arguments, there's a simple math solution. Instead of iterating over the numbers in the range and performing modulo, you can you could:
The code becomesWith a helper function:
int sum =private 0;
foreachint SumOfMultiples(int multipletarget, inint multiplesmultiple) {
int count = (target - 1) / multiple;
sum +=return multiple * count * (count + 1) / 2;
}
return sum;
... unfortunately ...
As @BorisTheSpider pointed out in This gives a comment, andstraight \$O(1)\$ solution for one multiple.
@Edward demonstrated in his answerFor two multiples a and b,
this algorithm results in double-countingyou cannot simply sum them as that would include the common multiples.
So in case of 3 and 5their multiples twice. So after summing, you would have to subtractsubtract the multiples of 3 * 5a * b:
return SumOfMultiples(target, a)
+ SumOfMultiples(target, b)
- SumOfMultiples(target, a * b);
Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.
Use the Math, @Heslacher ...
Instead of iterating over the numbers in the range and performing modulo, you can:
The code becomes:
int sum = 0;
foreach (int multiple in multiples) {
int count = (target - 1) / multiple;
sum += multiple * count * (count + 1) / 2;
}
return sum;
... unfortunately ...
As @BorisTheSpider pointed out in a comment, and @Edward demonstrated in his answer, this algorithm results in double-counting the common multiples. So in case of 3 and 5, you would have to subtract multiples of 3 * 5.
Use The Math, @Heslacher ...
... well ok, in the case of one or two arguments, there's a simple math solution. Instead of iterating over the numbers in the range and performing modulo, you could:
With a helper function:
private int SumOfMultiples(int target, int multiple) {
int count = (target - 1) / multiple;
return multiple * count * (count + 1) / 2;
}
This gives a straight \$O(1)\$ solution for one multiple.
For two multiples a and b,
you cannot simply sum them as that would include the multiples of their multiples twice. So after summing, you would have to subtract the multiples of a * b:
return SumOfMultiples(target, a)
+ SumOfMultiples(target, b)
- SumOfMultiples(target, a * b);
Thanks for @MartinR for the hint, and @BorisTheSpider to point out the problem with my original answer that didn't subtract.
... unfortunately ...
As @BorisTheSpider pointed out in a comment, and @Edward demonstrated in his answer, this algorithm results in double-counting the common multiples. So in case of 3 and 5, you would have to subtract multiples of 3 * 5.
It's easy to handle two parameters like 3 and 5 in this example. But handling arbitrary number of parameters, the math gets a lot trickier.
In conclusion, when multiples.Length <= 2, a fairly simple math solution exists. When multiples.Length > 2 the implementation needs more work, and more math.
Your solution with iteration may be a bit inefficient, but it's simple and it works.
... unfortunately ...
As @BorisTheSpider pointed out in a comment, and @Edward demonstrated in his answer, this algorithm results in double-counting the common multiples. So in case of 3 and 5, you would have to subtract multiples of 3 * 5.
It's easy to handle two parameters like 3 and 5 in this example. But handling arbitrary number of parameters, the math gets a lot trickier.
In conclusion, when multiples.Length <= 2, a fairly simple math solution exists. When multiples.Length > 2 the implementation needs more work, and more math.
Your solution with iteration may be a bit inefficient, but it's simple and it works.