Time Complexity

Question

Comments

@MiNiPanda

1^2 + 2^2 +....+ n^2 = n(n+1)(2n+1)/6

O(n^3)

1st loop -> O(n)

=> O(n^4)

correct ?

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Can u tell me what is answer given?

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O(n* logn * loglogn ) though i don't know how :P

Answer 1

time complexity will be $O\left ( n^{n+1} \right )\cong O\left ( n! \right )$

Here is the calculation

So, for $j$ it will be $O(n^{n})$ and for $k$ it will be $O(n)$

So, T.C. will be  $O(n^{n+1})$

Comments

ohk

finally, is it right then?

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mam, till now i didn't check it.......... i will check and comment

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ohk..

Answer 2

nlognlog(lonn!)=O(nsquare logn loglogn)