# Python – Check if two strings are Rotationally Equivalent

• Difficulty Level : Medium
• Last Updated : 05 Sep, 2022

Sometimes, while working with Python Strings, we can have problem in which we need to check if one string can be derived from other upon left or right rotation. This kind of problem can have application in many domains such as web development and competitive programming. Let’s discuss certain ways in which this task can be performed.

Input : test_str1 = ‘GFG’, test_str2 = ‘FGG’ Output : True Input : test_str1 = ‘geeks’, test_str2 = ‘ksege’ Output : False

Method #1 : Using loop + string slicing The combination of above functions can be used to solve this problem. In this, we perform the task of extracting strings for performing all possible rotations, to check if any rotation equals the other string.

## Python3

 `# Python3 code to demonstrate working of``# Check if two strings are Rotationally Equivalent``# Using loop + string slicing` `# initializing strings``test_str1 ``=` `'geeks'``test_str2 ``=` `'eksge'` `# printing original strings``print``(``"The original string 1 is : "` `+` `str``(test_str1))``print``(``"The original string 2 is : "` `+` `str``(test_str2))` `# Check if two strings are Rotationally Equivalent``# Using loop + string slicing``res ``=` `False``for` `idx ``in` `range``(``len``(test_str1)):``        ``if` `test_str1[idx: ] ``+` `test_str1[ :idx] ``=``=` `test_str2:``            ``res ``=` `True``            ``break` `# printing result``print``(``"Are two strings Rotationally equal ? : "` `+` `str``(res))`

Output :

```The original string 1 is : geeks
The original string 2 is : eksge
Are two strings Rotationally equal ? : True```

Method #2 : Using any() + join() + enumerate() This is one of the ways in which this task can be performed. In this, we perform the task of checking any rotational equivalent using any() extracted using nested generator expression and enumerate().

## Python3

 `# Python3 code to demonstrate working of``# Check if two strings are Rotationally Equivalent``# Using any() + join() + enumerate()` `# initializing strings``test_str1 ``=` `'geeks'``test_str2 ``=` `'eksge'` `# printing original strings``print``(``"The original string 1 is : "` `+` `str``(test_str1))``print``(``"The original string 2 is : "` `+` `str``(test_str2))` `# Check if two strings are Rotationally Equivalent``# Using any() + join() + enumerate()``res ``=` `any``(''.join([test_str2[idx2 ``-` `idx1]``        ``for` `idx2, val2 ``in` `enumerate``(test_str2)]) ``=``=` `test_str1``        ``for` `idx1, val1 ``in` `enumerate``(test_str1))` `# printing result``print``(``"Are two strings Rotationally equal ? : "` `+` `str``(res))`

Output :

```The original string 1 is : geeks
The original string 2 is : eksge
Are two strings Rotationally equal ? : True```

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)

Space Complexity: O(n)

My Personal Notes arrow_drop_up