# 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<``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); ` `    ``int` `n = ``sizeof``(arr2) / ``sizeof``(arr2); ` `    ``cout << distinctSum(arr1, arr2, m, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `static` `int` `MAX = ``1000000``; ` ` `  `static` `boolean` `[]prime = ``new` `boolean``[MAX + ``1``]; ` `static` `void` `sieve() ` `{ ` `    ``Arrays.fill(prime, ``true``); ` `    ``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 ` `static` `int` `distinctSum(``int` `arr1[], ` `                       ``int` `arr2[],  ` `                       ``int` `m, ``int` `n) ` `{ ` `    ``sieve(); ` ` `  `    ``// Set to store distinct sums ` `    ``Set sumSet = ``new` `HashSet(); ` ` `  `    ``for` `(``int` `i = ``0``; i < m; i++) ` `        ``for` `(``int` `j = ``0``; j < n; j++) ` `            ``if` `(prime[arr1[i]] && prime[arr2[j]]) ` `                ``sumSet.add(arr1[i] + arr2[j]); ` ` `  `    ``return` `sumSet.size(); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr1[] = { ``2``, ``3` `}; ` `    ``int` `arr2[] = { ``2``, ``2``, ``4``, ``7` `}; ` `    ``int` `m = arr1.length; ` `    ``int` `n = arr2.length; ` `    ``System.out.println(distinctSum(arr1, arr2, m, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

 `# 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 `

 `// C# implementation of the approach  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG ` `{ ` `static` `int` `MAX = 1000000; ` ` `  `static` `bool` `[]prime = ``new` `bool``[MAX + 1]; ` `static` `void` `sieve() ` `{ ` `    ``for` `(``int` `i = 0; i < MAX + 1; i++) ` `        ``prime[i] = ``true``; ` `    ``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 ` `static` `int` `distinctSum(``int` `[]arr1, ` `                       ``int` `[]arr2,  ` `                       ``int` `m, ``int` `n) ` `{ ` `    ``sieve(); ` ` `  `    ``// Set to store distinct sums ` `    ``HashSet<``int``> sumSet = ``new` `HashSet<``int``>(); ` ` `  `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``for` `(``int` `j = 0; j < n; j++) ` `            ``if` `(prime[arr1[i]] && prime[arr2[j]]) ` `                ``sumSet.Add(arr1[i] + arr2[j]); ` ` `  `    ``return` `sumSet.Count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr1 = { 2, 3 }; ` `    ``int` `[]arr2 = { 2, 2, 4, 7 }; ` `    ``int` `m = arr1.Length; ` `    ``int` `n = arr2.Length; ` `    ``Console.WriteLine(distinctSum(arr1, arr2, m, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:
```4
```

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Improved By : mohit kumar 29, Rajput-Ji

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