Jessica - Week 2 assignments #3
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MeiTzy222
reviewed
Dec 1, 2017
| if i**3 < limit: | ||
| cubedList.append(i) | ||
| return cubedList[-1] #returns the last number (highest cube) on cubedList. Can add .sort() to ensure list sorted lowest to highest. | ||
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Like it! Clean and clear codes! Well done.
MeiTzy222
reviewed
Dec 1, 2017
| newLine+= (width*character) + "\n" | ||
| else: | ||
| newLine+= (character + " "*(width-2) + character) + "\n" #creates the hollow space in between two borders | ||
| return newLine | ||
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Overall, your solution looks a bit complicated to me. Wonder if it is possible to somewhat simplify it...
MeiTzy222
reviewed
Dec 1, 2017
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| numberList=[] #creates an empty list 'numberList' | ||
| for i in range(max_number+1): #Stop is max_number +1 to ensure given number is also tested. | ||
| if _is_prime(i)==True: #calls _is_prime(number) function |
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I see you have leveraged on the first function you built. Good!
MeiTzy222
reviewed
Dec 1, 2017
| for i in range (3,number,2): #checks if number is divisible by any number between 3 and number. 2 steps to skip even numbers. This can be improved with int(n**0.5) + 1 which I have not figured out yet. | ||
| if number%i == 0: | ||
| return False | ||
| return True #loop ends here if above conditions not met. We have a PRIME number! | ||
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Food for thought: Apart from tackling this by finding out what is a prime number, maybe it will be even more efficient if we can think about what is not a prime number...
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This PR will be reviewed by @MeiTzy222