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Solve the following recurrences using master method T(n) = 9T(n/3) + n^3 T(n) = 8T(n/2) +n^3

Answer» TION:We compare the given recurrence relation with T(N) = aT(n/b) + θ (nklogpn).Then, we have-a = 3b = 2K = 2p = 0 Now, a = 3 and bk = 22 = 4.Clearly, a < bk.So, we FOLLOW case-03. Since p = 0, so we have-T(n) = θ (nklogpn)T(n) = θ (n2log0n) THUS,T(n) = θ (n2)


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