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it will be $O(n^{2})$
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2 Answers

+1 vote

..............

by Boss (35.7k points)
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@abhishekmehta4u

We can't say for sure that complexity is O(n^2)

As g(n) is omega (n), not O(n)

When we don't know the upper bound of g(n), we can't find out the upper bound of g(n).h(n)

However, we can say that it can be omega(n) for sure, at least.
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can u explain the steps i am not able to understand  how does u get ???

if any links available then also it will be nice !!

@abhishekmehta4u
+1
@vijju

what is the max and min value for this term [g(n)*h(n)] , min would be n as g(n) cant go below n and h(n) can become constant that case would give minimum value,

now think what can be its max value,you cant predict it

so for term [g(n)*h(n)] what you can surely say is its minimum value would be n, i.e  omega(n)

now total exprsn is f(n)+[g(n)*h(n)], f(n) is theta(n) , it cant go beyond n or less than n ,so what you can say from total exprsn that value of this exprsn will always be greater or equal to n i.e omega(n)
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check this solution.

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