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

• Difficulty Level : Easy
• Last Updated : 05 May, 2021

Given an array of positive numbers, the task is to calculate the absolute difference between sum of non-prime numbers and prime numbers.
Note: 1 is neither prime nor non-prime.
Examples:

Input : 1 3 5 10 15 7
Output : 10
Explanation: Sum of non-primes = 25
Sum of primes = 15

Input : 3 4 6 7
Output : 0

Naive Approach: A simple solution is to traverse the array and keep checking for every element if it is prime or not. If number is prime, then add it to sum S2 which represents the sum of primes else check if its not 1 then add it to sum of non-primes let’s say S1. After traversing the whole array, take the absolute difference between the two(S1-S2).
Time complexity: O(Nsqrt(N))
Efficient Approach: Generate all primes up to the maximum element of the array using the sieve of Eratosthenes and store them in a hash. Now, traverse the array and check if the number is present in the hash map. Then, add these numbers to sum S2 else check if it’s not 1, then add it to sum S1.
After traversing the whole array, display the absolute difference between the two.
Time Complexity: O(Nlog(log(N))

## C++

 // C++ program to find the Absolute Difference// between the Sum of Non-Prime numbers// and Prime numbers of an Array #include using namespace std; // Function to find the difference between// the sum of non-primes and the// sum of primes of an array.int CalculateDifference(int arr[], int n){    // Find maximum value in the array    int max_val = *max_element(arr, arr + n);     // 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 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;        }    }     // Store the sum of primes in S1 and    // the sum of non primes in S2    int S1 = 0, S2 = 0;    for (int i = 0; i < n; i++) {         if (prime[arr[i]]) {             // the number is prime            S1 += arr[i];        }        else if (arr[i] != 1) {             // the number is non-prime            S2 += arr[i];        }    }     // Return the absolute difference    return abs(S2 - S1);} int main(){     // Get the array    int arr[] = { 1, 3, 5, 10, 15, 7 };    int n = sizeof(arr) / sizeof(arr[0]);     // Find the absolute difference    cout << CalculateDifference(arr, n);     return 0;}

## Java

 // Java program to find the Absolute// Difference between the Sum of// Non-Prime numbers and Prime numbers// of an Arrayimport java.util.*; class GFG{ // Function to find the difference// between the sum of non-primes// and the sum of primes of an array.static int CalculateDifference(int arr[],                               int n){    // Find maximum value in the array    int max_val = Integer.MIN_VALUE;    for(int i = 0; i < n; i++)    {        if(arr[i] > max_val)            max_val = arr[i];    }     // 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];         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;        }    }     // Store the sum of primes in    // S1 and the sum of non    // primes in S2    int S1 = 0, S2 = 0;    for (int i = 0; i < n; i++)    {         if (prime[arr[i]])        {             // the number is prime            S1 += arr[i];        }        else if (arr[i] != 1)        {             // the number is non-prime            S2 += arr[i];        }    }     // Return the absolute difference    return Math.abs(S2 - S1);} // Driver Codepublic static void main(String []args){     // Get the array    int arr[] = { 1, 3, 5, 10, 15, 7 };    int n = arr.length;     // Find the absolute difference    System.out.println(CalculateDifference(arr, n));}} // This code is contributed// by ihritik

## Python3

 # Python3 program to find the Absolute# Difference between the Sum of Non-Prime# numbers and Prime numbers of an Arrayimport sys # Function to find the difference# between the sum of non-primes# and the sum of primes of an array.def CalculateDifference(arr, n):     # Find maximum value in the array    max_val = -1    for i in range(0, n):        if(arr[i] > max_val):            max_val = arr[i]         # 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 for i in range(max_val + 1)]     # Remaining part of SIEVE    prime[0] = False    prime[1] = False    p = 2    while (p * p <= max_val):                 # If prime[p] is not changed,        # then it is a prime        if prime[p] == True:                         # Update all multiples of p            for i in range(p * 2,                           max_val + 1, p):                prime[i] = False        p += 1     # Store the sum of primes in    # S1 and the sum of non primes    # in S2    S1 = 0    S2 = 0    for i in range (0, n):         if prime[arr[i]]:             # the number is prime            S1 += arr[i]                 elif arr[i] != 1:             # the number is non-prime            S2 += arr[i]     # Return the absolute difference    return abs(S2 - S1) # Driver code     # Get the arrayarr = [ 1, 3, 5, 10, 15, 7 ]n = len(arr) # Find the absolute differenceprint(CalculateDifference(arr, n)) # This code is contributed# by ihritik

## C#

 // C# program to find the Absolute// Difference between the Sum of// Non-Prime numbers and Prime// numbers of an Arrayusing System; class GFG{ // Function to find the difference// between the sum of non-primes// and the sum of primes of an array.static int CalculateDifference(int []arr,                               int n){    // Find maximum value in the array    int max_val = int.MinValue;    for(int i = 0; i < n; i++)    {        if(arr[i] > max_val)            max_val = arr[i];    }     // 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];         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;        }    }     // Store the sum of primes in    // S1 and the sum of non primes    // in S2    int S1 = 0, S2 = 0;    for (int i = 0; i < n; i++)    {        if (prime[arr[i]])        {             // the number is prime            S1 += arr[i];        }        else if (arr[i] != 1)        {             // the number is non-prime            S2 += arr[i];        }    }     // Return the absolute difference    return Math.Abs(S2 - S1);} // Driver Codepublic static void Main(string []args){     // Get the array    int []arr = { 1, 3, 5, 10, 15, 7 };    int n = arr.Length;     // Find the absolute difference    Console.WriteLine(CalculateDifference(arr, n));}} // This code is contributed// by ihritik

## PHP

 \$max_val)            \$max_val = \$arr[\$i];    }     // 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(0, \$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;        }    }     // Store the sum of primes in    // S1 and the sum of non    // primes in S2    \$S1 = 0;    \$S2 = 0;    for (\$i = 0; \$i < \$n; \$i++)    {         if (\$prime[\$arr[\$i]])        {             // the number is prime            \$S1 += \$arr[\$i];        }        else if (\$arr[\$i] != 1)        {             // the number is non-prime            \$S2 += \$arr[\$i];        }    }     // Return the absolute difference    return abs(\$S2 - \$S1);} // Driver code     // Get the array\$arr = array( 1, 3, 5, 10, 15, 7 );\$n = sizeof(\$arr); // Find the absolute differenceecho CalculateDifference(\$arr, \$n); // This code is contributed// by ihritik?>

## Javascript



Output:

10

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