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Python3 Program to Check if all rows of a matrix are circular rotations of each other

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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, 3

                   3, 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 :  

  1. 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.
  2. 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.
  3. 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)

Please refer complete article on Check if all rows of a matrix are circular rotations of each other for more details!



Last Updated : 12 Jul, 2022
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