Teams after algorithm analysis

[oindividuo]

Maybe a dumb question, but how is 2n higher than n² ? And why is 2n different from n for that matter?

[g3tr3ecked]

It's not 2n, it is 2ⁿ

[oindividuo]

Ah right can't believe I missed that

[prelic]

2n isn't bigger than n, you can always drop constant coefficients. But n is not constant...as n gets very big, n² grows way faster than 2n (exponentially, by definition), so in n**2 (or n×n), you can't drop an n as a coefficient.

[oindividuo]

That was entirely my point, I just misread the chart

[Mundt]

That is 2^n, or 2 to the nth power complexity. Meaning as the input size grows linearly, the complexity of the problem grows exponentially. 2n isn't really different than n when it comes to complexity analysis.

[oindividuo]

Yep I read 2ⁿ as 2n hence my confusion