doubt

Question

Consider the following function
find (int n)
{
if (n < 2 ) then return;
else
{
sum= 0;
for (i= 1; i ≤ 4; i++)

find (n/2);
for (i=1; i≤ n*n; i++)
sum= sum + 1;
}
}
Assume that the division operation takes constant time and “sum” is global variable. What is the time complexity of “find (n)” ?

Comments

I think $\Theta (n^2 logn)$

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yes, an answer is given same, get ?how did you get?

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Recurrence Relation for above code is T(n) = 4T(n/2) + n^2

Solve this you will get :)

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i didnt got that please explain

Answer 1

T (n)= $\theta (n^2logn)$