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:
li=list(map(str,input().strip().split()))#taking input of the string.
#swapping first and last element.
temp=li[0]
li[0]=li[-1]
li[-1]=temp
print(li)#printing the list.
Explanation:
I have taken the list li for taking the input of strings.Then after that swapping first and last element of the list.Then printing the list.
num1 = float(input("Enter the first number: "))
num2 = float(input("Enter the second number: "))
operation = input("Which operation are you performing? (a/s/m/d) ")
if operation == "a":
print("{} + {} = {}".format(num1, num2, num1+num2))
elif operation == "s":
print("{} - {} = {}".format(num1, num2, num1-num2))
elif operation == "m":
print("{} * {} = {}".format(num1, num2, num1*num2))
elif operation == "d":
print("{} / {} = {}".format(num1, num2, num1/num2))
I hope this helps!
Answer:
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