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Check if all rows of a matrix are circular rotations of each other
  • Difficulty Level : Medium
  • Last Updated : 14 Apr, 2021

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
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 <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;
}

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




<?php
// PHP 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.
function isPermutedMatrix( &$mat, $n)
{
    // Creating a string that contains
    // elements of first row.
    $str_cat = "";
    for ($i = 0 ; $i < $n ; $i++)
        $str_cat = $str_cat . "-" .
                   strval($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 = 1; $i < $n; $i++)
    {
        // Store the matrix into vector
        // in the form of strings
        $curr_str = "";
        for ($j = 0 ; $j < $n ; $j++)
            $curr_str = $curr_str . "-" .
                        strval($mat[$i][$j]);
 
        // Check if the current string is present
        // in the concatenated string or not
        if (strpos($str_cat, $curr_str))
            return true;
    }
 
    return false;
}
 
// Driver Code
$n = 4;
$mat = array(array(1, 2, 3, 4),
             array(4, 1, 2, 3),
             array(3, 4, 1, 2),
             array(2, 3, 4, 1));
if (isPermutedMatrix($mat, $n))
    echo "Yes";
else
    echo "No";
 
// This code is contributed by ita_c
?>

Javascript




<script>
 
// Javascript program to check if all rows of a matrix
// are rotations of each other
     
// Returns true if all rows of mat[0..n-1][0..n-1]
// are rotations of each other.
function isPermutedMatrix(mat, n)
{
     
    // Creating a string that contains
    // elements of first row.
    let str_cat = "";
    for (let 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(let i = 1; i < n; i++)
    {
         
        // Store the matrix into vector in the form
        // of strings
        let curr_str = "";
        for(let 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.includes(curr_str))
        {
            return true;
        }
    }
    return false;
}
 
// Drivers code
let n = 4;
let mat = [ [ 1, 2, 3, 4 ],
            [ 4, 1, 2, 3 ],
            [ 3, 4, 1, 2 ],
            [ 2, 3, 4, 1 ] ];
             
if (isPermutedMatrix(mat, n))
    document.write("Yes")
else
    document.write("No")
 
// This code is contributed by rag2127
 
</script>

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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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