# Count subarrays with Prime sum

Given an array A[] of integers. The task is to count total subarrays whose sum is prime with ( size > 1 ).

Examples:

```Input : A[] = { 1, 2, 3, 4, 5 }
Output : 3
Subarrays are -> {1, 2}, {2, 3}, {3, 4}

Input : A = { 22, 33, 4, 1, 10 };
Output : 4
```

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

Approach: Generate all possible subarrays and store their sum in a vector. Iterate the vector and check whether a sum is prime or not. It YES increment the count.

You can use sieve-of-eratosthenes to check whether a sum is prime in O(1).

Below is the implementation of above approach:

## C++

 `// C++ program to count subarrays ` `// with Prime sum ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count subarrays ` `// with Prime sum ` `int` `primeSubarrays(``int` `A[], ``int` `n) ` `{ ` `    ``int` `max_val = ``int``(``pow``(10, 7)); ` ` `  `    ``// 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<``bool``> 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``; ` `        ``} ` `    ``} ` ` `  `    ``int` `cnt = 0; ``// Initialize result ` ` `  `    ``// Traverse through the array ` `    ``for` `(``int` `i = 0; i < n - 1; ++i) { ` `        ``int` `val = A[i]; ` `        ``for` `(``int` `j = i + 1; j < n; ++j) { ` `            ``val += A[j]; ` ` `  `            ``if` `(prime[val]) ` `                ``++cnt; ` `        ``} ` `    ``} ` ` `  `    ``// return answer ` `    ``return` `cnt; ` `} ` ` `  `// Driver program ` `int` `main() ` `{ ` `    ``int` `A[] = { 1, 2, 3, 4, 5 }; ` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]); ` ` `  `    ``cout << primeSubarrays(A, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to count subarrays  ` `// with Prime sum  ` `import` `java.util.*; ` `class` `Solution ` `{ ` `   `  `// Function to count subarrays  ` `// with Prime sum  ` `static` `int` `primeSubarrays(``int` `A[], ``int` `n)  ` `{  ` `    ``int` `max_val = (``int``)(Math.pow(``10``, ``7``));  ` `   `  `    ``// 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=``new` `Vector(max_val + ``1``);  ` ` `  `     `  `    ``//initilize initial value ` `    ``for` `(``int` `p = ``0``; p

## Python3

 `# Python3 program to count subarrays  ` `# with Prime sum  ` ` `  `# Function to count subarrays  ` `# with Prime sum  ` `def` `primeSubarrays(A, n): ` ` `  `    ``max_val ``=` `10``*``*``7` ` `  `    ``# 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``] ``*` `(max_val ``+` `1``)  ` ` `  `    ``# Remaining part of SIEVE  ` `    ``prime[``0``] ``=` `False` `    ``prime[``1``] ``=` `False` `    ``for` `p ``in` `range``(``2``, ``int``(max_val``*``*``(``0.5``)) ``+` `1``):  ` ` `  `        ``# If prime[p] is not changed, then  ` `        ``# it is a prime  ` `        ``if` `prime[p] ``=``=` `True``:  ` ` `  `            ``# Update all multiples of p  ` `            ``for` `i ``in` `range``(``2` `*` `p, max_val ``+` `1``, p): ` `                ``prime[i] ``=` `False` `         `  `    ``cnt ``=` `0` `# Initialize result  ` ` `  `    ``# Traverse through the array  ` `    ``for` `i ``in` `range``(``0``, n ``-` `1``):  ` `        ``val ``=` `A[i]  ` `        ``for` `j ``in` `range``(i ``+` `1``, n):  ` `            ``val ``+``=` `A[j]  ` ` `  `            ``if` `prime[val] ``=``=` `True``:  ` `                ``cnt ``+``=` `1` ` `  `    ``# return answer  ` `    ``return` `cnt  ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``A ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]  ` `    ``n ``=` `len``(A)  ` ` `  `    ``print``(primeSubarrays(A, n)) ` ` `  `# This code is contributed by Rituraj Jain `

## C#

 `// C# program to count subarrays  ` `// with Prime sum  ` ` `  `class` `Solution ` `{ ` ` `  `// Function to count subarrays  ` `// with Prime sum  ` `static` `int` `primeSubarrays(``int``[] A, ``int` `n)  ` `{  ` `    ``int` `max_val = (``int``)(System.Math.Pow(10, 7));  ` ` `  `    ``// 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];  ` ` `  `     `  `    ``//initilize initial value ` `    ``for` `(``int` `p = 0; p

## PHP

 ` `

Output:

```3
```

Time Complexity: O(N2)

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