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Not one person in the comments has attempted to answer any of the questions either.
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Haha good try. Hope your interview goes well
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for(var i=0;i<=100;i++){ if((i%2)==1) console.log(i); }
btw % is the modulo operator, x%y returns the remainder of division of x by y
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Thank you holy shit I was beginning to think no one has ever seen a fizz buzz before
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Slightly simpler, start at 1 and increment by 2 so you don't have to check whether i is odd.
for (var i = 1; i < 100; i += 2) { console.log(i); }-
Strictly speaking this one does not find the odd numbers, it just prints them.
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for (i%1=0; i+2; int) odd++; cout(3)
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Will you give me the position if I answer the problems? 😀
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Sure! I'll hire you without even answering the questions. Of course I'm not the op, I dont work in the it field (any more) and none of my open positions involve programming... But you have a job with my company whenever you need one.
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(0..=100).filter(|n| n % 2 == 1).for_each(|n| println!("{n}"))
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