mínimo número de escalón de paso a-reducir a 1
#minimum-number-of-steps-to-reduce-number-to-1
def stepCount(n):
count = 0
while n > 1:
if n % 2 == 0: # bitmask: *0
n = n // 2
elif n == 3 or n % 4 == 1: # bitmask: 01
n = n - 1
else: # bitmask: 11
n = n + 1
count += 1
return count
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