Polígono regular que usa solo 1 en un círculo numerado binario

Dada una array de números enteros binarios, suponga que estos valores se mantienen en la circunferencia de un círculo a la misma distancia. Necesitamos saber si es posible dibujar un polígono regular usando solo 1 como vértices y, si es posible, imprimir el número máximo de lados que tiene ese polígono regular.

Ejemplo: 

Input : arr[] = [1, 1, 1, 0, 1, 0]
Output : Polygon possible with side length 3
We can draw a regular triangle having 1s as 
its vertices as shown in below diagram (a).

Input : arr[] = [1, 0, 1, 0, 1, 0, 1, 0, 1, 1]
Output : Polygon possible with side length 5
We can draw a regular pentagon having 1s as its 
vertices as shown in below diagram (b).

Podemos resolver este problema obteniendo una relación entre el número de vértices que puede tener un posible polígono y el número total de valores en la array. Deje que un posible polígono regular en el círculo tenga K vértices o K lados, entonces debería satisfacer dos cosas para ser la respuesta, 

Si el tamaño de array dado es N, entonces K debe dividir N; de lo contrario, K vértices no pueden dividir N vértices de la misma manera para ser un polígono regular. 
Lo siguiente es que debe haber uno en cada vértice del polígono elegido. 

Después de los puntos anteriores, podemos ver que para resolver este problema necesitamos iterar sobre los divisores de N y luego verificar si cada valor de la array a la distancia del divisor elegido es 1 o no. Si es 1, entonces encontramos nuestra solución. Podemos iterar sobre todos los divisores en tiempo O(sqrt(N)) simplemente iterando de 1 a sqrt(N). Puedes leer más sobre eso aquí . 

Implementación:

// C++ program to find whether a regular polygon
// is possible in circle with 1s as vertices
#include <bits/stdc++.h>
using namespace std;
 
// method returns true if polygon is possible with
// 'midpoints' number of midpoints
bool checkPolygonWithMidpoints(int arr[], int N,
                                  int midpoints)
{
    // loop for getting first vertex of polygon
    for (int j = 0; j < midpoints; j++)
    {
        int val = 1;
 
        // loop over array values at 'midpoints' distance
        for (int k = j; k < N; k += midpoints)
        {
            // and(&) all those values, if even one of
            // them is 0, val will be 0
            val &= arr[k];
        }
 
        /*  if val is still 1 and (N/midpoints) or (number
            of vertices) are more than two (for a polygon
            minimum) print result and return true */
        if (val && N/midpoints > 2)
        {
            cout << "Polygon possible with side length " <<(N/midpoints) << endl;
            return true;
        }
    }
    return false;
}
 
// method prints sides in the polygon or print not
// possible in case of no possible polygon
void isPolygonPossible(int arr[], int N)
{
    //  limit for iterating over divisors
    int limit = sqrt(N);
    for (int i = 1; i <= limit; i++)
    {
        // If i divides N then i and (N/i) will
        // be divisors
        if (N % i == 0)
        {
            //  check polygon for both divisors
            if (checkPolygonWithMidpoints(arr, N, i) ||
                checkPolygonWithMidpoints(arr, N, (N/i)))
                return;
        }
    }
 
    cout << "Not possiblen";
}
 
// Driver code to test above methods
int main()
{
    int arr[] = {1, 0, 1, 0, 1, 0, 1, 0, 1, 1};
    int N = sizeof(arr) / sizeof(arr[0]);
    isPolygonPossible(arr, N);
    return 0;
}

// Java program to find whether a regular polygon
// is possible in circle with 1s as vertices
 
class Test
{
    // method returns true if polygon is possible with
    // 'midpoints' number of midpoints
    static boolean checkPolygonWithMidpoints(int arr[], int N,
                                      int midpoints)
    {
        // loop for getting first vertex of polygon
        for (int j = 0; j < midpoints; j++)
        {
            int val = 1;
      
            // loop over array values at 'midpoints' distance
            for (int k = j; k < N; k += midpoints)
            {
                // and(&) all those values, if even one of
                // them is 0, val will be 0
                val &= arr[k];
            }
      
            /*  if val is still 1 and (N/midpoints) or (number
                of vertices) are more than two (for a polygon
                minimum) print result and return true */
            if (val != 0 && N/midpoints > 2)
            {
                System.out.println("Polygon possible with side length " +
                                               N/midpoints);
                return true;
            }
        }
        return false;
    }
      
    // method prints sides in the polygon or print not
    // possible in case of no possible polygon
    static void isPolygonPossible(int arr[], int N)
    {
        //  limit for iterating over divisors
        int limit = (int)Math.sqrt(N);
        for (int i = 1; i <= limit; i++)
        {
            // If i divides N then i and (N/i) will
            // be divisors
            if (N % i == 0)
            {
                //  check polygon for both divisors
                if (checkPolygonWithMidpoints(arr, N, i) ||
                    checkPolygonWithMidpoints(arr, N, (N/i)))
                    return;
            }
        }
      
        System.out.println("Not possible");
    }
     
    // Driver method
    public static void main(String args[])
    {
        int arr[] = {1, 0, 1, 0, 1, 0, 1, 0, 1, 1};
 
        isPolygonPossible(arr, arr.length);
    }
}

# Python3 program to find whether a
# regular polygon is possible in circle
# with 1s as vertices
from math import sqrt
 
# method returns true if polygon is
# possible with 'midpoints' number
# of midpoints
def checkPolygonWithMidpoints(arr, N, midpoints) :
 
    # loop for getting first vertex of polygon
    for j in range(midpoints) :
     
        val = 1
 
        # loop over array values at
        # 'midpoints' distance
        for k in range(j , N, midpoints) :
             
            # and(&) all those values, if even 
            # one of them is 0, val will be 0
            val &= arr[k]
         
        # if val is still 1 and (N/midpoints) or (number
        # of vertices) are more than two (for a polygon
        # minimum) print result and return true
        if (val and N // midpoints > 2) :
         
            print("Polygon possible with side length" ,
                                      (N // midpoints))
            return True
     
    return False
 
# method prints sides in the polygon or print
# not possible in case of no possible polygon
def isPolygonPossible(arr, N) :
 
    # limit for iterating over divisors
    limit = sqrt(N)
    for i in range(1, int(limit) + 1) :
         
        # If i divides N then i and (N/i)
        # will be divisors
        if (N % i == 0) :
         
            # check polygon for both divisors
            if (checkPolygonWithMidpoints(arr, N, i) or
                checkPolygonWithMidpoints(arr, N, (N // i))):
                return
         
    print("Not possiblen")
 
# Driver code
if __name__ == "__main__" :
 
    arr = [1, 0, 1, 0, 1, 0, 1, 0, 1, 1]
    N = len(arr)
    isPolygonPossible(arr, N)
 
# This code is contributed by Ryuga

// C# program to find whether
// a regular polygon is possible
// in circle with 1s as vertices
using System;
 
class GFG
{
     
// method returns true if
// polygon is possible
// with 'midpoints' number
// of midpoints
static bool checkPolygonWithMidpoints(int []arr, int N,
                                      int midpoints)
{
    // loop for getting first
    // vertex of polygon
    for (int j = 0; j < midpoints; j++)
    {
        int val = 1;
 
        // loop over array values
        // at 'midpoints' distance
        for (int k = j; k < N; k += midpoints)
        {
            // and(&) all those values,
            // if even one of them is 0,
            // val will be 0
            val &= arr[k];
        }
 
        /* if val is still 1 and
           (N/midpoints) or (number
           of vertices) are more than
           two (for a polygon minimum)
           print result and return true */
        if (val != 0 && N / midpoints > 2)
        {
            Console.WriteLine("Polygon possible with " +
                                        "side length " +
                                         N / midpoints);
            return true;
        }
    }
    return false;
}
 
// method prints sides in the
// polygon or print not possible
// in case of no possible polygon
static void isPolygonPossible(int []arr,
                              int N)
{
    // limit for iterating
    // over divisors
    int limit = (int)Math.Sqrt(N);
    for (int i = 1; i <= limit; i++)
    {
        // If i divides N then i
        // and (N/i) will be divisors
        if (N % i == 0)
        {
            // check polygon for
            // both divisors
            if (checkPolygonWithMidpoints(arr, N, i) ||
                checkPolygonWithMidpoints(arr, N, (N / i)))
                return;
        }
    }
 
    Console.WriteLine("Not possible");
}
 
// Driver Code
static public void Main ()
{
    int []arr = {1, 0, 1, 0, 1,
                 0, 1, 0, 1, 1};
 
    isPolygonPossible(arr, arr.Length);
}
}
 
// This code is contributed by jit_t

<?php
// PHP program to find whether
// a regular polygon is possible
// in circle with 1s as vertices
// method returns true if polygon
// is possible with 'midpoints'
// number of midpoints
 
function checkPolygonWithMidpoints($arr, $N,
                                   $midpoints)
{
    // loop for getting first
    // vertex of polygon
    for ($j = 0; $j < $midpoints; $j++)
    {
        $val = 1;
 
        // loop over array values
        // at 'midpoints' distance
        for ($k = $j; $k < $N; $k += $midpoints)
        {
            // and(&) all those values,
            // if even one of them is 0,
            // val will be 0
            $val &= $arr[$k];
        }
 
        /* if val is still 1 and
        (N/midpoints) or (number
        of vertices) are more than
        two (for a polygon minimum)
        print result and return true */
        if ($val && $N / $midpoints > 2)
        {
            echo "Polygon possible with side length " ,
                              ($N / $midpoints) , "\n";
            return true;
        }
    }
    return false;
}
 
// method prints sides in
// the polygon or print not
// possible in case of no
// possible polygon
function isPolygonPossible($arr, $N)
{
    // limit for iterating
    // over divisors
    $limit = sqrt($N);
    for ($i = 1; $i <= $limit; $i++)
    {
        // If i divides N then
        // i and (N/i) will be
        // divisors
        if ($N % $i == 0)
        {
            // check polygon for
            // both divisors
            if (checkPolygonWithMidpoints($arr, $N, $i) ||
                checkPolygonWithMidpoints($arr, $N, ($N / $i)))
                return;
        }
    }
 
echo "Not possiblen";
}
 
// Driver Code
$arr = array(1, 0, 1, 0, 1,
             0, 1, 0, 1, 1);
$N = sizeof($arr);
isPolygonPossible($arr, $N);
 
// This code is contributed by ajit
?>

<script>
// Javascript program to
// find whether a regular polygon
// is possible in circle with 1s as vertices
 
// method returns true if polygon is possible with
// 'midpoints' number of midpoints
function checkPolygonWithMidpoints(arr, N, midpoints)
{
    // loop for getting first vertex of polygon
    for (let j = 0; j < midpoints; j++)
    {
        let val = 1;
 
        // loop over array values at
        // 'midpoints' distance
        for (let k = j; k < N; k += midpoints)
        {
            // and(&) all those values,
            // if even one of
            // them is 0, val will be 0
            val &= arr[k];
        }
 
        /*  if val is still 1 and
            (N/midpoints) or (number
            of vertices) are more than
            two (for a polygon
            minimum) print result and return true */
        if (val && parseInt(N/midpoints) > 2)
        {
            document.write(
            "Polygon possible with side length " +
            parseInt(N/midpoints) + "<br>"
            );
            return true;
        }
    }
    return false;
}
 
// method prints sides in the polygon or print not
// possible in case of no possible polygon
function isPolygonPossible(arr, N)
{
    //  limit for iterating over divisors
    let limit = Math.sqrt(N);
    for (let i = 1; i <= limit; i++)
    {
        // If i divides N then i and (N/i) will
        // be divisors
        if (N % i == 0)
        {
            //  check polygon for both divisors
            if (checkPolygonWithMidpoints(arr, N, i) ||
                checkPolygonWithMidpoints(arr, N,
                parseInt(N/i)))
                return;
        }
    }
 
    document.write("Not possible");
}
 
// Driver code to test above methods
    let arr = [1, 0, 1, 0, 1, 0, 1, 0, 1, 1];
    let N = arr.length;
    isPolygonPossible(arr, N);
 
</script>

Polygon possible with side length 5

Complejidad de tiempo: O(sqrt(N) ² ) 
Espacio auxiliar: O(1)

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