Answer:
Answer: T(m) =3T (m/2) + 0 (m)
Explanation:
Algo A: Dividing into 3 sub problems and recurring for each with a combining step of linear time
T(m) =3T (m/2) + 0 (m)
Algo B: Dividing into 4sub problems with 1/4 size and solving each quadratic time
T(m) = 4{(m/4)²} + 0(m)
= 0(m)² + 0(m)
T(m) = 0(m²)
Algo C:
T(m) = 27(m/2) + 0(m)
Applying master theorem;
T(m) = a(T(m/b) + f(m)
log₂² = 1
mlogbᵃ = f(m)
Time complexity = 0(m x logm)
Also applying master theorem on Algo A;
m logbᵃ = m log₂³
f(m) = m
mlogbᵃ is greater than f(m)
Time complexity = mlog³₂
It's called a reverse dns lookup. Actually it's a somewhat regular lookup in the in-addr.arpa domain.
Answer:
See Explaination
Explanation:
def listmax(lst):
largest = None
for num in lst:
if largest is None or num > largest:
largest = num
return largest
mylist = [10, 20, 30]
x = listmax(mylist)
print(x)
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Explanation: