T(n) = 10T(n/2) + 100n3
a=10, b= 2, k= 3 ,p=0Here a > bk
So According to Masters theorem (Rule 1)-
T(n) = $\theta$( nlogba )
= $\theta$( nlog210)
= $\theta$(n3). ( log210 >3)
Ans- A
so 2^10=.2^(2^3.2)(that is greater than 3
=n^3.2 > f(n)
so it will be ⦶(n^3)
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