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B4cuJIuuvs.Teban-Sanchez

Return 100 100% efficient

Code
Diff
  • return_hundred=lambda n:(n**0)*100
    • def return_hundred(n):
    • num = n
    • n = n/num
    • while(n!=100):
    • if(n<100):
    • n *= 10
    • if(n>100):
    • n = n /10
    • return n
    • #this is goat.
    • return_hundred=lambda n:(n**0)*100