Given that m <= n, is O(n^2) in O(m^2n)? I know that O(mn) is in O(n^2) but I do not know what happens when you add the extra term
CodePudding user response：
Even as m grows with n, and no matter the difference between m and n, n^2 will always be larger so (m^2)n would always be contained within n^2.
CodePudding user response：
Given that m <= n, is O(n^2) in O((m^2)n)
No, even if m grows with n.
Let m = sqrt(sqrt(n)).
O(n^2) is not contained in O((m^2)n) = O(((sqrt(sqrt(n)))^2)n) = O(sqrt(n)n)