Assignment 1:find the range of numbers that are both squares and cubes

Author: PTHARRISH

300_Exercises/Simple/1.numbers_and_squares.py

@@ -1,3 +1,4 @@
+# Python to find the range of numbers that are squares within a specified limit
 # method 1
 print('Numbers\t\tSquares')
 print('1 \t\t '+str(1*1))
@@ -6,20 +7,7 @@
 print('4 \t\t '+str(4*4))
 
 
-# method 2
-def number_squares(n):
-    print(str(n)+' \t\t '+str(n*n))
-
-start_num,end_num = input('enter the starting and ending number:').split(' ')
-if start_num<=end_num:
-    for n in range(int(start_num),int(end_num)+1):
-        number_squares(n)
-else:
-    print('enter the proper range')
-
-
-# method 3 : optimized 
-
+# method 2:
 def number_squares(start_num,end_num):
     if start_num>end_num:
         print('Invalid Range: Starting number must be less than or equal to the ending number.')
@@ -32,4 +20,34 @@ def number_squares(start_num,end_num):
     print('Numbers\t\tSquares')
     number_squares(start_num,end_num)
 except ValueError:
-    print("Invalid input! Please enter two integers separated by a space.")
+    print("Invalid input! Please enter two integers separated by a space.")
+
+
+# Assignment
+# In Python to find the range of numbers that are both squares and cubes within a specified limit 
+def number_square_cubes(start_num,end_num):
+    if start_num>end_num:
+        print('Invalid Range: Starting number must be less than or equal to the ending number.')
+        return
+    for n in range(start_num,end_num+1):
+        print(f"{n}\t\t{n*n}\t\t{n*n*n}")
+
+try:
+    start_num,end_num=map(int,input('Enter the starring and ending range with single space:').split(' '))
+    print('Numbers\t\tSquares\t\tCubes')
+    number_square_cubes(start_num,end_num)
+except ValueError:
+    print("Invalid input! Please enter two integers separated by a space.")
+
+
+# Time Complexity
+
+# The main time complexity comes from the loop, which iterates over the numbers in the range [start_num, end_num]. 
+# As mentioned above, the loop runs N=end_num−start_num+1
+# N=end_num−start_num+1 times, and each iteration involves constant-time operations.
+
+# Thus, the time complexity of the code is:
+# O(N)whereN=end_num−start_num+1
+
+# If the size of the input range increases, the number of iterations increases linearly, 
+# so the time complexity is O(N).