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

Last Updated : 12 Dec, 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[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 :

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 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 
import java.io.*;
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# 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

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

Largest Cross Bordered Square

Sahil Chhabra (akku)

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

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