Absolute Difference between the Sum of Non-Prime numbers and Prime numbers of an Array?

Here we will see how we can find the absolute difference between the sum of all prime numbers and all non-prime numbers of an array. To solve this problem, we have to check whether a number is prime or not. One possible way for primality testing is by checking a number is not divisible by any number between 2 to square root of that number. So this process will take 𝑂(√𝑛) amount of time. Then get the sum and try to find the absolute difference.



   sum_p := sum of all prime numbers in arr
   sum_np := sum of all non-prime numbers in arr
   return |sum_p – sum_np|


 Live Demo

#include <iostream>
#include <cmath>
using namespace std;
bool isPrime(int n){
   for(int i = 2; i<=sqrt(n); i++){
      if(n % i == 0){
         return false; //not prime
   return true; //prime
int diffPrimeNonPrimeSum(int arr[], int n) {
   int sum_p = 0, sum_np = 0;
   for(int i = 0; i<n; i++){
         sum_p += arr[i];
      } else {
         sum_np += arr[i];
   return abs(sum_p - sum_np);
main() {
   int arr[] = { 5, 8, 9, 6, 21, 27, 3, 13};
   int n = sizeof(arr) / sizeof(arr[0]);
   cout << "Difference: " << diffPrimeNonPrimeSum(arr, n);


Difference: 50