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



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++

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// C++ program to count subarrays
// with Prime sum
  
#include <bits/stdc++.h>
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;
}

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Java 














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// 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<Boolean> prime=new Vector<Boolean>(max_val + 1);  


  


      


    //initilize initial value 


    for (int p = 0; p <max_val + 1; p++) 


    prime.add(p,true); 


    


    // Remaining part of SIEVE  


    prime.set(0, false);  


    prime.set(1, false);  


    for (int p = 2; p * p <= max_val; p++) {  


    


        // If prime[p] is not changed, then  


        // it is a prime  


        if (prime.get(p) == true) {  


    


            // Update all multiples of p  


            for (int i = p * 2; i <= max_val; i += p)  


                prime.set(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.get(val))  


                ++cnt;  


        }  


    }  


    


    // return answer  


    return cnt;  


}  


    


// Driver program  


public static void main(String args[]) 


{  


    int A[] = { 1, 2, 3, 4, 5 };  


    int n = A.length;  


    


    System.out.print( primeSubarrays(A, n));  


    


}  


} 


//contributed by Arnab Kundu 



















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Python3














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# 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 



















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C#














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// 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 <max_val + 1; p++) 


    prime[p]=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  


static void Main() 


{  


    int[] A = { 1, 2, 3, 4, 5 };  


    int n = A.Length;  


  


    System.Console.WriteLine( primeSubarrays(A, n));  


  


}  


} 


//contributed by mits 



















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PHP














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<?php 


// PHP program to count subarrays  


// with Prime sum  


  


  


// Function to count subarrays  


// with Prime sum  


function primeSubarrays($A, $n)  


{  


    $max_val = pow(10, 5);  


  


    // 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.  


    $prime=array_fill(0,$max_val + 1,true);  


  


    // Remaining part of SIEVE  


    $prime[0] = false;  


    $prime[1] = false;  


    for ($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 ($i = $p * 2; $i <= $max_val; $i += $p)  


                $prime[$i] = false;  


        }  


    }  


  


    $cnt = 0; // Initialize result  


  


    // Traverse through the array  


    for ($i = 0; $i < $n - 1; ++$i) {  


        $val = $A[$i];  


        for ($j = $i + 1; $j < $n; ++$j) {  


            $val += $A[$j];  


  


            if ($prime[$val])  


                ++$cnt;  


        }  


    }  


  


    // return answer  


    return $cnt;  


}  


  


// Driver program  


   


    $A = array( 1, 2, 3, 4, 5 );  


    $n = count($A);  


  


    echo primeSubarrays($A, $n);  


  


// This code is contributed by mits  


?> 



















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Output:


3





Time Complexity: O(N2)



          
          
          

          

    

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