# 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.

```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
Explanation : As 3, 2, 1 is not a rotated or
circular permutation of 1, 2, 3
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 C++ implementation of above steps.

```// C++ program to check if all rows of a matrix
// are rotations of each other
#include <bits/stdc++.h>
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[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 + "-" + to_string(mat[i][j]);

// Check if the current string is present in
// the concatenated string or not
if (str_cat.find(curr_str) == string::npos)
return false;
}

return true;
}

// Drivers code
int main()
{
int n = 4 ;
int mat[MAX][MAX] = {{1, 2, 3, 4},
{4, 1, 2, 3},
{3, 4, 1, 2},
{2, 3, 4, 1}
};
isPermutedMatrix(mat, n)? cout << "Yes" :
cout << "No";
return 0;
}
```

Output:

```Yes
```

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

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