Question

For large problems sizes, an algorithm with time cost O(2^n) is faster than an algorithm
that has time cost O(N^2)

Is this true or false?

What I think is that if C^n, C = constant and C > 1, then it will grow faster than
O(N^2). Is this correct?

Answer 1

For large problems sizes, an algorithm with time cost O(2ⁿ) is faster than an algorithm that has time cost O(n²).

FALSE, because 2ⁿ > n² for n > 4, and greater means slower.

For C = constant and C > 1, Cⁿ grows faster than O(n²).

TRUE.

Here is a Wolfram|Alpha reference.