# Check if all rows of a matrix are circular rotations of each other

• Difficulty Level : Medium
• Last Updated : 19 Aug, 2022

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 = 1, 2, 3
3, 2, 1
1, 3, 2
Output:  No
Explanation : As 3, 2, 1 is not a rotated or
circular permutation of 1, 2, 3```

The idea is based on 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 above steps.

## C++

 `// C++ program to check if all rows of a matrix` `// are rotations of each other` `#include ` `using` `namespace` `std;` `const` `int` `MAX = 1000;`   `// Returns true if all rows of mat[0..n-1][0..n-1]` `// are rotations of each other.` `bool` `isPermutedMatrix( ``int` `mat[MAX][MAX], ``int` `n)` `{` `    ``// Creating a string that contains elements of first` `    ``// row.` `    ``string str_cat = ``""``;` `    ``for` `(``int` `i = 0 ; i < n ; i++)` `        ``str_cat = str_cat + ``"-"` `+ to_string(mat[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` `(``int` `i=1; i

## Java

 `// Java program to check if all rows of a matrix ` `// are rotations of each other ` `class` `GFG ` `{`   `    ``static` `int` `MAX = ``1000``;`   `    ``// Returns true if all rows of mat[0..n-1][0..n-1] ` `    ``// are rotations of each other. ` `    ``static` `boolean` `isPermutedMatrix(``int` `mat[][], ``int` `n) ` `    ``{` `        ``// Creating a string that contains` `        ``// elements of first row. ` `        ``String str_cat = ``""``;` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``{` `            ``str_cat = str_cat + ``"-"` `+ String.valueOf(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` `(``int` `i = ``1``; i < n; i++)` `        ``{` `            ``// Store the matrix into vector in the form ` `            ``// of strings ` `            ``String curr_str = ``""``;` `            ``for` `(``int` `j = ``0``; j < n; j++) ` `            ``{` `                ``curr_str = curr_str + ``"-"` `+ String.valueOf(mat[i][j]);` `            ``}`   `            ``// Check if the current string is present in ` `            ``// the concatenated string or not ` `            ``if` `(str_cat.contentEquals(curr_str)) ` `            ``{` `                ``return` `false``;` `            ``}` `        ``}`   `        ``return` `true``;` `    ``}`   `    ``// Drivers code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``int` `n = ``4``;` `        ``int` `mat[][] = {{``1``, ``2``, ``3``, ``4``},` `        ``{``4``, ``1``, ``2``, ``3``},` `        ``{``3``, ``4``, ``1``, ``2``},` `        ``{``2``, ``3``, ``4``, ``1``}` `        ``};` `        ``if` `(isPermutedMatrix(mat, n)) ` `        ``{` `            ``System.out.println(``"Yes"``);` `        ``}` `        ``else` `        ``{` `            ``System.out.println(``"No"``);` `        ``}` `    ``}` `}`   `/* This code contributed by PrinciRaj1992 */`

## 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`

## C#

 `// C# program to check if all rows of a matrix ` `// are rotations of each other ` `using` `System;`   `class` `GFG ` `{`   `    ``//static int MAX = 1000;`   `    ``// Returns true if all rows of mat[0..n-1,0..n-1] ` `    ``// are rotations of each other. ` `    ``static` `bool` `isPermutedMatrix(``int` `[,]mat, ``int` `n) ` `    ``{` `        ``// Creating a string that contains` `        ``// elements of first row. ` `        ``string` `str_cat = ``""``;` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{` `            ``str_cat = str_cat + ``"-"` `+ mat[0,i].ToString();` `        ``}`   `        ``// 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` `(``int` `i = 1; i < n; i++)` `        ``{` `            ``// Store the matrix into vector in the form ` `            ``// of strings ` `            ``string` `curr_str = ``""``;` `            ``for` `(``int` `j = 0; j < n; j++) ` `            ``{` `                ``curr_str = curr_str + ``"-"` `+ mat[i,j].ToString();` `            ``}`   `            ``// Check if the current string is present in ` `            ``// the concatenated string or not ` `            ``if` `(str_cat.Equals(curr_str)) ` `            ``{` `                ``return` `false``;` `            ``}` `        ``}`   `        ``return` `true``;` `    ``}`   `    ``// Driver code ` `    ``static` `void` `Main() ` `    ``{` `        ``int` `n = 4;` `        ``int` `[,]mat = {{1, 2, 3, 4},` `        ``{4, 1, 2, 3},` `        ``{3, 4, 1, 2},` `        ``{2, 3, 4, 1}` `        ``};` `        `  `        ``if` `(isPermutedMatrix(mat, n)) ` `        ``{` `            ``Console.WriteLine(``"Yes"``);` `        ``}` `        ``else` `        ``{` `            ``Console.WriteLine(``"No"``);` `        ``}` `    ``}` `}`   `/* This code contributed by mits */`

## PHP

 ``

## Javascript

 ``

Output

`Yes`

Time complexity : O(n3
Auxiliary Space : O(n), since n extra space has been taken.

This article is contributed by Sahil Chhabra (akku). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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