Check if the sum of primes is divisible by any prime from the array

Given an array arr[], the task is to check whether the sum of primes in the array is divisible by any of the primes in the array. If it is then print YES, otherwise print NO.

Examples:

Input: arr[] = {2, 3} 
Output: NO 
Primes: 2, 3 
Sum = 2 + 3 = 5 which is neither divisible by 2 nor 3

Input: arr[] = {1, 2, 3, 4, 5} 
Output: YES 
2 + 3 + 5 = 10 is divisible by 2 as well as 5 

Approach: The idea is to generate all primes upto maximum element from the array using Sieve of Eratosthenes.  

• Traverse the array and check if the current element is prime or not. If it is prime then update sum = sum + arr[i].
• Traverse the array again and check if sum % arr[i] = 0 where arr[i] is prime. If it is then print YES. Otherwise print NO in the end.

Below is the implementation of the above approach: 

C++

// C++ program to check if sum of primes from an array
// is divisible by any of the primes from the same array
#include <bits/stdc++.h>
using namespace std;

// Function to print "YES" if sum of primes from an array
// is divisible by any of the primes from the same array
void SumDivPrime(int A[], int n)
{
    int max_val = *(std::max_element(A, A + n)) + 1;

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);

    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {

        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }

    int sum = 0;

    // Traverse through the array
    for (int i = 0; i < n; ++i) {
        if (prime[A[i]])
            sum += A[i];
    }

    for (int i = 0; i < n; ++i) {
        if (prime[A[i]] && sum % A[i] == 0) {
            cout << "YES";
            return;
        }
    }

    cout << "NO";
}

// Driver program
int main()
{
    int A[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(A) / sizeof(A[0]);

    SumDivPrime(A, n);

    return 0;
}

Java

// Java program to check if sum of primes from an array
// is divisible by any of the primes from the same array
class Solution
{
    //returns the maximum value
static int max_element(int A[])
{
    int max=Integer.MIN_VALUE;

    for(int i=0;i<A.length;i++)
        if(max<A[i])
            max=A[i];

    return max;
}


// Function to print "YES" if sum of primes from an array
// is divisible by any of the primes from the same array
static void SumDivPrime(int A[], int n)
{
    int max_val = (max_element(A)) + 1;

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    boolean prime[]=new boolean[max_val+1];

    //initialize the array
    for(int i=0;i<=max_val;i++)
    prime[i]=true;

    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {

        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {

            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }

    int sum = 0;

    // Traverse through the array
    for (int i = 0; i < n; ++i) {
        if (prime[A[i]])
            sum += A[i];
    }

    for (int i = 0; i < n; ++i) {
        if (prime[A[i]] && sum % A[i] == 0) {
            System.out.println( "YES");
            return;
        }
    }

    System.out.println("NO");
}

// Driver program
public static void main(String args[])
{
    int A[] = { 1, 2, 3, 4, 5 };
    int n = A.length;

    SumDivPrime(A, n);
}
}
//contributed by Arnab Kundu

Python3

# Python3 program to check if sum of
# primes from an array is divisible
# by any of the primes from the same array
import math

# Function to print "YES" if sum of primes
# from an array is divisible by any of the
# primes from the same array
def SumDivPrime(A, n):

    max_val = max(A) + 1

    # USE SIEVE TO FIND ALL PRIME NUMBERS
    # LESS THAN OR EQUAL TO max_val
    # Create a boolean array "prime[0..n]".
    # A value in prime[i] will finally be
    # false if i is Not a prime, else true.
    prime = [True] * (max_val + 1)

    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    for p in range(2, int(math.sqrt(max_val)) + 1):

        # If prime[p] is not changed,
        # then it is a prime
        if prime[p] == True :

            # Update all multiples of p
            for i in range(2 * p, max_val + 1, p):
                prime[i] = False

    sum = 0

    # Traverse through the array
    for i in range(0, n):
        if prime[A[i]]:
            sum += A[i]

    for i in range(0, n):
        if prime[A[i]] and sum % A[i] == 0:
            print("YES")
            return

    print("NO")

# Driver Code
A = [ 1, 2, 3, 4, 5 ]
n = len(A)

SumDivPrime(A, n)

# This code is contributed
# by saurabh_shukla

C#

// C# program to check if sum of primes
// from an array is divisible by any of
// the primes from the same array
class GFG
{

//returns the maximum value
static int max_element(int[] A)
{
    int max = System.Int32.MinValue;

    for(int i = 0; i < A.Length; i++)
        if(max < A[i])
            max = A[i];

    return max;
}


// Function to print "YES" if sum of
// primes from an array  is divisible
// by any of the primes from the same array
static void SumDivPrime(int[] A, int n)
{
    int max_val = (max_element(A)) + 1;

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    bool[] prime=new bool[max_val+1];

    //initialize the array
    for(int i = 0; i <= max_val; i++)
        prime[i] = true;

    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++)
    {

        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
    int sum = 0;

    // Traverse through the array
    for (int i = 0; i < n; ++i)
    {
        if (prime[A[i]])
            sum += A[i];
    }

    for (int i = 0; i < n; ++i)
    {
        if (prime[A[i]] && sum % A[i] == 0)
        {
            System.Console.WriteLine( "YES");
            return;
        }
    }
    System.Console.WriteLine("NO");
}

// Driver code
public static void Main()
{
    int []A = { 1, 2, 3, 4, 5 };
    int n = A.Length;
    SumDivPrime(A, n);
}
}

// This code is contributed by mits

PHP

<?php
// PHP program to check if sum of primes
// from an array is divisible by any of
// the primes from the same array

// Function to print "YES" if sum of primes
// from an array is divisible by any of the
// primes from the same array
function SumDivPrime($A, $n)
{
    $max_val = max($A);

    // USE SIEVE TO FIND ALL PRIME NUMBERS
    // LESS THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    $prime = array_fill(1, $max_val + 1, true);

    // Remaining part of SIEVE
    $prime[0] = false;
    $prime[1] = false;
    for ($p = 2; $p * $p <= $max_val; $p++)
    {

        // If prime[p] is not changed, then
        // it is a prime
        if ($prime[$p] == true)
        {

            // Update all multiples of p
            for ($i = $p * 2;
                 $i <= $max_val; $i += $p)
                $prime[$i] = false;
        }
    }

    $sum = 0;

    // Traverse through the array
    for ($i = 0; $i < $n; ++$i)
    {
        if ($prime[$A[$i]])
            $sum += $A[$i];
    }

    for ($i = 0; $i < $n; ++$i)
    {
        if ($prime[$A[$i]] &&
            $sum % $A[$i] == 0)
        {
            echo "YES";
            return;
        }
    }

    echo "NO";
}

// Driver Code
$A = array( 1, 2, 3, 4, 5 );
$n = sizeof($A) ;

SumDivPrime($A, $n);

// This code is contributed by Ryuga
?>

Javascript

<script>
// Javascript program to check if sum
// of primes from an array is divisible
// by any of the primes from the same array

// Returns the maximum value
function max_element(A)
{
    var max = Number.MIN_VALUE;

    for(var i = 0; i < A.length; i++)
        if (max < A[i])
            max = A[i];

    return max;
}

// Function to print "YES" if sum of primes
// from an array is divisible by any of the
// primes from the same array
function SumDivPrime(A, n)
{
    var max_val = (max_element(A)) + 1;

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    var prime = new Array(max_val + 1);

    // Initialize the array
    for(var i = 0; i <= max_val; i++)
        prime[i] = true;

    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;

    for(var p = 2; p * p <= max_val; p++)
    {

        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true)
        {

            // Update all multiples of p
            for(var i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }

    var sum = 0;

    // Traverse through the array
    for(var i = 0; i < n; ++i)
    {
        if (prime[A[i]])
            sum += A[i];
    }

    for(var i = 0; i < n; ++i)
    {
        if (prime[A[i]] && sum % A[i] == 0)
        {
            document.write("YES");
            return;
        }
    }
    document.write("NO");
}

// Driver code
var A = [ 1, 2, 3, 4, 5 ];
var n = A.length;
SumDivPrime(A, n);

// This code is contributed by SoumikMondal

</script>

Output

YES

Complexity Analysis:

• Time Complexity: O(max_val*log(log(max_val))), where max_val is the largest element of the array.
• Auxiliary Space: O(max_val)