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2 Answers

Best answer
2 votes
2 votes

B) $O(n^2)$.

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1 votes
T(n) = T(n/4) + T(n/2)  +  c n^2

I think what U can do is, U can write subproblems of size (n/4) is nearly equivalent (n/2).

Now The Recurrence Relation looks something like this: T(n)  <= T(n/2) + T(n/2) + c n^2

T(n) <= 2 T(n/2) + cn^2

Now u can apply masters theorm

Ans will be O(n^2) ie; B

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