13time_complacity.py

#wrong approach 1


import time 
start = time.time()
for i in range(1,101):
    print(i)
print(time.time()-start)

#you can check with this method but it take different time in different system on acc to its configurations......
# not for extremely small input
# time varies have  time problem 
# time varies if implementation changes..... 

#wrong approach 2
#here we will take the code and count the how many operation taken by the code 
#order of growth , it is the graph time and growth 

#questions for time complexity 

##########################  Question no 1  #############################
"""time complicity of the following code ?"""
L=[1,2,3,4,5]
sum = 0
for i in L:
  sum = sum+i
print(sum)

product = 1
for i in L:
 product = product*i
print(product)
"""here the time complicity is O(n)which is linear because the 
   loop is running n times where n is the length of the array """
##########################  Question end  #############################


##########################  Question no 2  #############################
"""time complicity of the following code ?"""
L = [1, 2, 3, 4, 5]
for i in L:
    for j in L:
        print("({}, {})".format(i, j))
"""here the time complicity is O(n^2) which 
   is quadratic because the loop is nested"""
##########################  Question end  #############################


##########################  Question no 3  #############################
"""what will happen here"""
#Linear search
"""if the input size will increase the time will also increase """
##########################  Question end  #############################



##########################  Question no 4  #############################
"""time complicity of the following code ?"""
def intToStr(i):
   digits = "0123456789"
   if i == 0:
      return "0"
   result = ""
   while i > 0:
      result= digits[i%10]+result
      i = i//10
      return result
   """here the time complicity is O(log n) which is logarithmic"""
##########################  Question end  #############################



##########################  Question no 5 #############################
"""time complicity of the following code ?"""
n = 1000
i, j, k = 0, 0, 0
for i in range(n // 2, n + 1):
    j = 2
    while j <= n:
        k = k + n // 2
        j *= 2

"""here the time complicity is O(n_log_n) which is logarithmic """
##########################  Question end  #############################



##########################  Question no 6 #############################
"""what will happen here"""
#Binary search
"""the time complicity of the binary search is O(Log n) because in binary searching 
   we divide the array in half and then the half array in half till the outcome will shown"""
##########################  Question end  #############################


##########################  Question no 7 #############################
"""time complicity of the following code?"""
L=[1,2,3,4,5]
for i in range(0,len(L)):
   for j in range(i + 1,len(L)):
      print({} , {}.format(L[i],L[j]))
"""time complicity of this code is O(N^2) """
##########################  Question end  #############################