Python3 Program to Check if all rows of a matrix are circular rotations of each other
Given a matrix of n*n size, the task is to find whether all rows are circular rotations of each other or not.
Examples:
Input: mat[][] = 1, 2, 3
3, 1, 2
2, 3, 1
Output: Yes,All rows are rotated permutation of each other.
Input: mat[3][3] = 1, 2, 33, 2, 1
1, 3, 2
Output: No,As 3, 2, 1 is not a rotated or circular permutation of 1, 2, 3
The idea is based on the below article.
A Program to check if strings are rotations of each other or not
Steps :
- Create a string of first row elements and concatenate the string with itself so that string search operations can be efficiently performed. Let this string be str_cat.
- Traverse all remaining rows. For every row being traversed, create a string str_curr of current row elements. If str_curr is not a substring of str_cat, return false.
- Return true.
Below is the implementation of the above steps.
Python3
# Python3 program to check if all rows # of a matrix are rotations of each other MAX = 1000 # Returns true if all rows of mat[0..n-1][0..n-1] # are rotations of each other. def isPermutedMatrix(mat, n) : # Creating a string that contains # elements of first row. str_cat = "" for i in range (n) : str_cat = str_cat + "-" + str (mat[ 0 ][i]) # Concatenating the string with itself # so that substring search operations # can be performed on this str_cat = str_cat + str_cat # Start traversing remaining rows for i in range ( 1 , n) : # Store the matrix into vector # in the form of strings curr_str = "" for j in range (n) : curr_str = curr_str + "-" + str (mat[i][j]) # Check if the current string is present # in the concatenated string or not if (str_cat.find(curr_str)) : return True return False # Driver code if __name__ = = "__main__" : n = 4 mat = [[ 1 , 2 , 3 , 4 ], [ 4 , 1 , 2 , 3 ], [ 3 , 4 , 1 , 2 ], [ 2 , 3 , 4 , 1 ]] if (isPermutedMatrix(mat, n)): print ( "Yes" ) else : print ( "No" ) # This code is contributed by Ryuga |
Output
Yes
Time complexity: O(n3)
Auxiliary Space: O(n)
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