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

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

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:

 // C++ implementation of the approach #include #define MAX 1000000 using namespace std;    bool prime[MAX]; void sieve() {     memset(prime, true, sizeof(prime));     prime = prime = 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 > 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);     int n = sizeof(arr2) / sizeof(arr2);     cout << distinctSum(arr1, arr2, m, n);        return 0; }

 # Python3 implementation of the approach MAX = 1000000    prime = [True for i in range(MAX + 1)]    def sieve():        prime, prime = 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

Output:
4

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