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@abhinav dongre

 

Can we transform it into a recurrence with a more usual form :

T(12n)

=T(4n)+T(3n)+5(12n)

=T(4n/3)+T(n)+20n+T(3n/4)+T(n)+15n+60n

=2T(n)+105n+T(4n/3)+T(3n/4) ….

 

So, that’s pretty close to 4T(n)+105n ..

The recurrence T′(n)

=4T′(n/12)+105n/12 ….

 

can be approached with the Master Theorem,

which tells us that c =log124 ≈ 0.56, and 105n is Ω(nc),

 

so T′(n), if regular, is Θ(n)..

 

 

1. https://gateoverflow.in/191487/T-n-t-n-4-t-3n-4-n

 

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