Python3 Program for Minimum rotations required to get the same string
Given a string, we need to find the minimum number of rotations required to get the same string.
Examples:
Input : s = "Beginner" Output : 5 Input : s = "aaaa" Output : 1
Method 1: The idea is based on below post.
A Program to check if strings are rotations of each other or not
Step 1 : Initialize result = 0 (Here result is count of rotations)
Step 2 : Take a temporary string equals to original string concatenated with itself.
Step 3 : Now take the substring of temporary string of size same as original string starting from second character (or index 1).
Step 4 : Increase the count.
Step 5 : Check whether the substring becomes equal to original string. If yes, then break the loop. Else go to step 2 and repeat it from the next index.
Python3
# Python 3 program to determine minimum # number of rotations required to yield # same string. # Returns count of rotations to get the # same string back. def findRotations( str ): # tmp is the concatenated string. tmp = str + str n = len ( str ) for i in range ( 1 , n + 1 ): # substring from i index of # original string size. substring = tmp[i: i + n] # if substring matches with # original string then we will # come out of the loop. if ( str = = substring): return i return n # Driver code if __name__ = = '__main__' : str = "abc" print (findRotations( str )) # This code is contributed # by 29AjayKumar. |
3
Time Complexity: O(n2)
Auxiliary Space: O(n). We are using a temporary string of size n for the concatenated string.
Alternate Implementation in Python :
Python3
# Python 3 program to determine minimum # number of rotations required to yield # same string. # input string = 'aaaa' check = '' for r in range ( 1 , len (string) + 1 ): # checking the input after each rotation check = string[r:] + string[:r] # following if statement checks if input is # equals to check , if yes it will print r and # break out of the loop if check = = string: print (r) break # This code is contributed # by nagasowmyanarayanan. |
1
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(N)
Please refer complete article on Minimum rotations required to get the same string for more details!