I'm not sure of the problem that you had in the first place, but I can point out that in your "bigger" method, if a number is greater than one then it is bigger, however the else statement says that the number *can* also be equal[==] to the second argument, so for example bigger(1,1) it would check if 1 > 1 and return false, so it will return that b[the second 1] is bigger! Hope this helps :D
Answer:
O(N!), O(2N), O(N2), O(N), O(logN)
Explanation:
N! grows faster than any exponential functions, leave alone polynomials and logarithm. so O( N! ) would be slowest.
2^N would be bigger than N². Any exponential functions are slower than polynomial. So O( 2^N ) is next slowest.
Rest of them should be easier.
N² is slower than N and N is slower than logN as you can check in a graphing calculator.
NOTE: It is just nitpick but big-Oh is not necessary about speed / running time ( many programmers treat it like that anyway ) but rather how the time taken for an algorithm increase as the size of the input increases. Subtle difference.
Answer:
scope of pet name is limited to pet class and color is accessible to the whole program
Explanation:
