No, you're 100% correct. O(n^[number]) = polynomial, with O(n^2) being quadratic...

		mabbo on May 27, 2020 \| parent \| context \| favorite \| on: Snowpack 2.0 No, you're 100% correct. O(n^[number]) = polynomial, with O(n^2) being quadratic. O([a number]^n) = exponential. I guess the authors aren't big on CS fundamentals?