Count of distinct sums that can be obtained by adding prime numbers from given arrays

Given two arrays arr1[] and arr2[]. The task is to count the distinct sums that can be obtained while choosing a prime element from arr1[] and another prime element from arr2[].

Examples:

Input: arr1[] = {2, 3}, arr2[] = {2, 2, 4, 7}
Output: 4
All possible prime pairs are (2, 2), (2, 2), (2, 7), (3, 2), (3, 2)
and (3, 7) with sums 4, 4, 9, 5, 5 and 10 respectively.



Input: arr1[] = {3, 1, 4, 2, 5}, arr2[] = {8, 7, 10, 6, 5}
Output: 5

Approach: Use Sieve of Eratosthenes to check whether a number is prime or not then for every prime pair, store it’s sum in a set in order to avoid duplicates. The size of the final set will be the required answer.

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
#define MAX 1000000
using namespace std;
  
bool prime[MAX];
void sieve()
{
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = false;
    for (int p = 2; p * p <= MAX; p++) {
        if (prime[p] == true) {
            for (int i = p * p; i <= MAX; i += p)
                prime[i] = false;
        }
    }
}
  
// Function to return the distinct sums
// that can be obtained by adding prime
// numbers from the given arrays
int distinctSum(int arr1[], int arr2[], int m, int n)
{
    sieve();
  
    // Set to store distinct sums
    set<int, greater<int> > sumSet;
  
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            if (prime[arr1[i]] && prime[arr2[j]])
                sumSet.insert(arr1[i] + arr2[j]);
  
    return sumSet.size();
}
  
// Driver code
int main()
{
    int arr1[] = { 2, 3 };
    int arr2[] = { 2, 2, 4, 7 };
    int m = sizeof(arr1) / sizeof(arr1[0]);
    int n = sizeof(arr2) / sizeof(arr2[0]);
    cout << distinctSum(arr1, arr2, m, n);
  
    return 0;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
MAX = 1000000
  
prime = [True for i in range(MAX + 1)]
  
def sieve():
  
    prime[0], prime[1] = False, False
  
    for p in range(2, MAX + 1):
        if p * p > MAX:
            break
        if (prime[p] == True):
            for i in range(2 * p, MAX + 1, p):
                prime[i] = False
  
# Function to return the distinct sums
# that can be obtained by adding prime
# numbers from the given arrays
def distinctSum(arr1, arr2, m, n):
    sieve()
  
    # Set to store distinct sums
    sumSet = dict()
  
    for i in range(m):
        for j in range(n):
            if (prime[arr1[i]] and 
                prime[arr2[j]]):
                sumSet[arr1[i] + arr2[j]] = 1
  
    return len(sumSet)
  
# Driver code
arr1 = [2, 3 ]
arr2 = [2, 2, 4, 7 ]
m = len(arr1)
n = len(arr2)
print(distinctSum(arr1, arr2, m, n))
  
# This code is contributed by mohit kumar
chevron_right

Output:
4



Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : mohit kumar 29



Article Tags :