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The given implementation can be improved by just changing the upperbound of the range to be int(math.sqrt(n))+2
Still probably not the most efficient approach, but there's no point in checking numbers above the sqrt(n)+1.
Edit:
Actually I think it does not matter since the return statement will exit the function before those higher values are tested.
Your code is much more efiicient
What you said is correct, if the number has a large factor, it will be time consuming.Please refer to my solution.