Dada una array de enteros arr[] , la tarea es verificar si esa array contiene todos los divisores de algún entero.
Ejemplos:
Entrada: arr[] = { 2, 3, 1, 6}
Salida: Sí
La array contiene todos los divisores de 6
Entrada: arr[] = { 12, 2, 5, 3, 6, 4, 1}
Salida: No
Enfoque: si la array contiene todos los divisores de un entero en particular, digamos X , entonces el elemento máximo en la array arr[] es el entero X. Ahora, encuentre el elemento máximo de la array arr[] y calcule todos sus divisores y guárdelo en un vector b . Si el arreglo arr[] y el vector b son iguales, entonces el arreglo contiene todos los divisores de un entero en particular; de lo contrario, no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if arr[]
// contains all the divisors of some integer
bool checkDivisors(int a[], int n)
{
// Maximum element from the array
int X = *max_element(a, a + n);
// Vector to store divisors
// of the maximum element i.e. X
vector<int> b;
// Store all the divisors of X
for (int i = 1; i * i <= X; i++) {
if (X % i == 0) {
b.push_back(i);
if (X / i != i)
b.push_back(X / i);
}
}
// If the lengths of a[]
// and b are different
// return false
if (b.size() != n)
return false;
// Sort a[] and b
sort(a, a + n);
sort(b.begin(), b.end());
for (int i = 0; i < n; i++) {
// If divisors are not
// equal return false
if (b[i] != a[i])
return false;
}
return true;
}
// Driver code
int main()
{
int arr[] = { 8, 1, 2, 12, 48,
6, 4, 24, 16, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
if (checkDivisors(arr, N))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// returns th maximum element of the array
static int max_element(int a[] )
{
int m = a[0];
for(int i = 0; i < a.length; i++)
m = Math.max(a[i], m);
return m;
}
// Function that returns true if arr[]
// contains all the divisors of some integer
static boolean checkDivisors(int a[], int n)
{
// Maximum element from the array
int X = max_element(a);
// Vector to store divisors
// of the maximum element i.e. X
Vector<Integer> b=new Vector<Integer>();
// Store all the divisors of X
for (int i = 1; i * i <= X; i++)
{
if (X % i == 0)
{
b.add(i);
if (X / i != i)
b.add(X / i);
}
}
// If the lengths of a[]
// and b are different
// return false
if (b.size() != n)
return false;
// Sort a[] and b
Arrays.sort(a);
Collections.sort(b);
for (int i = 0; i < n; i++)
{
// If divisors are not
// equal return false
if (b.get(i) != a[i])
return false;
}
return true;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 8, 1, 2, 12, 48,
6, 4, 24, 16, 3 };
int N = arr.length;
if (checkDivisors(arr, N))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Arnab Kundu
Python3
# Python 3 implementation of the approach
from math import sqrt
# Function that returns true if arr[]
# contains all the divisors of some integer
def checkDivisors(a,n):
# Maximum element from the array
X = max(a)
# Vector to store divisors
# of the maximum element i.e. X
b = []
# Store all the divisors of X
for i in range(1,int(sqrt(X))+1):
if (X % i == 0):
b.append(i)
if (X // i != i):
b.append(X // i)
# If the lengths of a[]
# and b are different
# return false
if (len(b) != n):
return False
# Sort a[] and b
a.sort(reverse = False)
b.sort(reverse = False)
for i in range(n):
# If divisors are not
# equal return false
if (b[i] != a[i]):
return False
return True
# Driver code
if __name__ == '__main__':
arr = [8, 1, 2, 12, 48,6, 4, 24, 16, 3]
N = len(arr)
if (checkDivisors(arr, N)):
print("Yes")
else:
print("No")
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// returns th maximum element of the array
static int max_element(int []a )
{
int m = a[0];
for(int i = 0; i < a.Length; i++)
m = Math.Max(a[i], m);
return m;
}
// Function that returns true if arr[]
// contains all the divisors of some integer
static bool checkDivisors(int []a, int n)
{
// Maximum element from the array
int X = max_element(a);
// Vector to store divisors
// of the maximum element i.e. X
List<int> b = new List<int>();
// Store all the divisors of X
for (int i = 1; i * i <= X; i++)
{
if (X % i == 0)
{
b.Add(i);
if (X / i != i)
b.Add(X / i);
}
}
// If the lengths of a[]
// and b are different
// return false
if (b.Count != n)
return false;
// Sort a[] and b
Array.Sort(a);
b.Sort();
for (int i = 0; i < n; i++)
{
// If divisors are not
// equal return false
if (b[i] != a[i])
return false;
}
return true;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 8, 1, 2, 12, 48,
6, 4, 24, 16, 3 };
int N = arr.Length;
if (checkDivisors(arr, N))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript implementation of the approach
// Function that returns true if arr[]
// contains all the divisors of some integer
function checkDivisors(a, n)
{
// Maximum element from the array
let X = Math.max(...a);
// Vector to store divisors
// of the maximum element i.e. X
let b = [];
// Store all the divisors of X
for (let i = 1; i * i <= X; i++) {
if (X % i == 0) {
b.push(i);
if (parseInt(X / i) != i)
b.push(parseInt(X / i));
}
}
// If the lengths of a[]
// and b are different
// return false
if (b.length != n)
return false;
// Sort a[] and b
a.sort((x,y) => x - y);
b.sort((x,y) => x - y);
for (let i = 0; i < n; i++) {
// If divisors are not
// equal return false
if (b[i] != a[i])
return false;
}
return true;
}
// Driver code
let arr = [ 8, 1, 2, 12, 48,
6, 4, 24, 16, 3 ];
let N = arr.length;
if (checkDivisors(arr, N))
document.write("Yes");
else
document.write("No");
</script>
Producción:
Yes
Complejidad de tiempo: O((n * log n) + max(arr)), donde max(arr) es el elemento más grande de la array arr.
Espacio auxiliar: O(max(arr))
Publicación traducida automáticamente
Artículo escrito por rupesh_rao y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA