Check if the given array contains all the divisors of some integer

Given an integer array arr[], the task is to check if that array contains all the divisor of some integer.

Examples:

Input: arr[] = { 2, 3, 1, 6}
Output: Yes
The array contains all the divisors of 6



Input: arr[] = { 12, 2, 5, 3, 6, 4, 1}
Output: No

Approach: If the array contains all the divisors of a particular integer say X then the maximum element in the array arr[] is the integer X. Now, find the maximum element of the array arr[] and calculate all of its divisors and store it in a vector b. If array arr[] and vector b are equal then the array contains all the divisors of a particular integer, otherwise no.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

Yes


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