# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 = ``false``; ` `    ``prime = ``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); ` ` `  `    ``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

## 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];  ` `     `  `    ``//initilize the array  ` `    ``for``(``int` `i = 0; i <= max_val; i++)  ` `        ``prime[i] = ``true``;  ` ` `  `    ``// Remaining part of SIEVE  ` `    ``prime = ``false``;  ` `    ``prime = ``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

 ` `

Output:

```YES
```

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