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You're right, if we were talking prime numbers, sqrt(n) would make sense, but not for divisors.
[Edit]: Let me correct myself, if you add i and n / i at the same step, you're adding the other number that is also a divisor in same step, take for example 8, if i is 2, you'll end up adding sum += 2 + 8/2 which traduces to sum += 2 + 4, both 2 & 4 are divisors of 8, keep doing that and you'll soon notice that this way the i variable only needs to go up to sqrt(n).
Day 19: Interfaces
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Another point to note, is that After n/2, the only factor of n will be n itself. This reduces the running time of the whole algorithm
sqrt(n) will reduce it more.
just add i and n/i(if a pure int)
how is this correct for a number like 20? 10 is larger than sqrt(20), am confused here
You're right, if we were talking prime numbers,
sqrt(n)would make sense, but not for divisors.[Edit]: Let me correct myself, if you add
iandn / iat the same step, you're adding the other number that is also a divisor in same step, take for example8, ifiis2, you'll end up addingsum += 2 + 8/2which traduces tosum += 2 + 4, both2&4are divisors of8, keep doing that and you'll soon notice that this way theivariable only needs to go up tosqrt(n).yes i got it thank you very much
If n=1, then this would return 2, no? even though it should return 1.