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Not really. sqrt is obtained by integer casting; what you suggested might end up not accounting for some of the divisors.
Say n = 6.
Then sqrt = (int) Math.sqrt(6) = 2
So your loop would run only once with this statement:
sum += 1+6
After that, i < sqrt so we exit the loop.
Since 2*2 != 6, you don't add 2 to the sum.
So your suggested modification would result in sum = 7 for n = 6, which is the wrong sum.
You have probably caught this but for anyone who have not, always pay close attention to the difference between (< vs <= ) when you do integer casting!
Day 19: Interfaces
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can you please tell me why you negation of the sum.
If sqrt is a divisor, we end up counting it twice. We do
sum -=sqrtto basically remove the double-count we did earlier.HackerRank solutions.
it would be slightly more efficient, though, to do
for(... i < sqrt ...)andif (sqrt*sqrt==n) sum += sqrt;in the end.Not really. sqrt is obtained by integer casting; what you suggested might end up not accounting for some of the divisors.
Say n = 6. Then sqrt = (int) Math.sqrt(6) = 2 So your loop would run only once with this statement: sum += 1+6
After that, i < sqrt so we exit the loop. Since 2*2 != 6, you don't add 2 to the sum.
So your suggested modification would result in sum = 7 for n = 6, which is the wrong sum.
You have probably caught this but for anyone who have not, always pay close attention to the difference between (< vs <= ) when you do integer casting!