Answer:
Answer explained
Explanation:
From the previous question we know that while searching for n^(1/r) we don't have to look for guesses less than 0 and greater than n. Because for less than 0 it will be an imaginary number and for rth root of a non negative number can never be greater than itself. Hence lowEnough = 0 and tooHigh = n.
we need to find 5th root of 47226. The computation of root is costlier than computing power of a number. Therefore, we will look for a number whose 5th power is 47226. lowEnough = 0 and tooHigh = 47226 + 1. Question that should be asked on each step would be "Is 5th power of number < 47227?" we will stop when we find a number whose 5th power is 47226.
Answer:
(a) What is the best case time complexity of the algorithm (assuming n > 1)?
Answer: O(1)
(b) What is the worst case time complexity of the algorithm?
Answer: O(n^4)
Explanation:
(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1)
.
(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).
It’s b just really got to make since an read it right that’s all
