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2 Answers

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1 votes

T(n) = 7T(n/2) + n^2

comparing with the equation (MASTER THEOREM)

T(n) = aT(n/b) + ⊖(n^k $\log_{n}p$)

we get ,a=7,b=2,k=2,p=0

now it satisfies a>b^k,

so case first of master theorem

T(n) = ⊖(n^$\log_{b}a$)

T(n)=⊖(n^3)

0 votes
0 votes

Compare with master theorem ,a=7,b=2 and f(n)=n2

T(n)=nlog27 =n2.81 i.e f(n) is polynomially smaller than t(n), therefore it is first case

 therefore, ans is (nlog7) or (n2.81)

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