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Count subarrays with Prime sum
• Difficulty Level : Medium
• Last Updated : 09 May, 2020

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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