I love Python
That’s 2*(n-1) comparisons, constant,
The problem can be solved with a worst(n).
Can be more faster
Python use TimSort which is the fastest sorting algorithm in current time
and max() and min() function also uses sorting btw
TimSort worst case complexity is in O(nlogn). We can definitely do it in less than that. How about O(N)?
def solve(self, A): minn, maxx = sys.maxsize, -sys.maxsize for i in range(len(A)): if A[i] < minn: minn = A[i] if A[i] > maxx: maxx = A[i] return maxx+minn