Explain how nC2 is Θ(n^2)

Question

In the book CLRS, in page 69 it's said nC2 is Θ(n^2) in the unit divide and conquer(U - 4). Can anyone expalin how this result is true?

< CONSTn^2 for all n>0, CONST=1/2 - which means O(n^2) by definition — amit, Jun 20 '14 at 11:11
You can see it that way: you choose your first element (O(n) choices possible) then your second (O(n) again) so you have a O(n²) upper bound. — user189, Jun 20 '14 at 11:11
@shekharsuman It will be pretty much identical to show the same thing for big Omega, and from there it is big theta by definition. — amit, Jun 20 '14 at 11:17
Possible duplicate of [The complexity of n choose 2 is in Theta (n^2)?](https://stackoverflow.com/questions/19415988/the-complexity-of-n-choose-2-is-in-theta-n2) — smit shah, Oct 11 '19 at 05:09

score 6 · Accepted Answer · edited Jun 20 '20 at 09:12

6

nC2 = n*(n-1)/2 = (n^2-n)/2;

Hint :

(n^2-n)/2 will be greater than c1*(n^2) for some constants like c1 < 1/4 and for value of n = 2. Similarly it will be smaller than c2*n^2 for values of c2 = 1 and n = 2. So, here is it's a sandwich like situation. Similarly, it'll be sandwiched for other values of n and constants c. Hence, nC2 is Θ(n^2).

edited Jun 20 '20 at 09:12

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answered Jun 20 '14 at 11:12

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Explain how nC2 is Θ(n^2)

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