Comprobar si todas las filas de una array son rotaciones circulares entre sí

Dada una array de tamaño n*n, la tarea es encontrar si todas las filas son rotaciones circulares entre sí o no. 

Ejemplos: 

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

La idea se basa en el siguiente artículo. 
Un programa para verificar si las strings son rotaciones entre sí o no.

Pasos :  

1. Cree una string de elementos de la primera fila y concatene la string consigo misma para que las operaciones de búsqueda de strings se puedan realizar de manera eficiente. Deje que esta string sea str_cat.
2. Atraviese todas las filas restantes. Para cada fila que se recorre, cree una string str_curr de los elementos de la fila actual. Si str_curr no es una substring de str_cat, devuelve falso.
3. Devolver verdadero.

A continuación se muestra la implementación de los pasos anteriores. 

// 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 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 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 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
?>

<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>

Yes

Complejidad temporal : O(n ³ ) 
Espacio auxiliar : O(n)

Este artículo es una contribución de Sahil Chhabra (akku) . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.