most efficient code for printing odd numbers from 1-99 [closed]

Answer 1

In the first code you are iterating over two generators first range(1, 100) and then over liszt whereas in the second case the iteration is only over liszt. Other than that the operation is same in both the cases, so the second method is more efficient.

Since every second number after 1 would be odd, a better solution could be:

for i in range(1, 100, 2):
    print(i)

Answer 2

print("\n".join(str(i) for i in range(1,100,2)))