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I'm a little late but your response made me curious. I inititially would have agreed with you, until I put it to the test. I think your statement is only valid for small numbers and/or small prime factors, though.
I tested both solutions with 1000 random, unsorted numbers. The results: This solution took about 29 seconds and your solution took about 22:30 minutes with the same numbers in the same order. This is mainly due to two cases with large numbers, where the prime factors where either only two factors or the number itself. These took 7:30 minutes and 14:42 minutes respectively. So I guess, this solution is not that bad after all, is it?
Though one could improve it a little with the usage of StringBuilder, a while-loop instead of the for-loop and by checking 2 and only the odd numbers afterwards.
No check for -ve?
okay i get it now. You used the isSquared() method which checks if a number is positiv or negativ
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wtf
Nice
Niceeeee !
Nice one!
The modulo operator (%) returns the remainder after dividing two numbers. If a number is divisible by another, the remainder will be 0, ergo the modulo operation will return 0. In this case, we are checking if a number is an integer. Math.sqrt(n) % 1 will only return a value other than 0 if the return value of Math.sqrt(n) is a decimal, since every integer is divisible by 1. Hope that helps. :)
One line solution, nice!
This really helped me simplify my code. Definitely did not think about it like this!
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Cool solution. Correct me if I'm wrong, but I think that it would be faster to compute the case fac = 2 separately, and then only cycle through odd #s
That's really smart!
Nice one!
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