# 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:

## C++

`// 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*

## Python3

# 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

**Output:**

4

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