Count the number of primes in the prefix sum array of the given array

Last Updated : 31 May, 2022

Given an array arr[] of N integers, the task is to count the number of primes in the prefix sum array of the given array.
Examples:

Input: arr[] = {1, 4, 8, 4}
Output: 3
The prefix sum array is {1, 5, 13, 17}
and the three primes are 5, 13 and 17.
Input: arr[] = {1, 5, 2, 3, 7, 9}
Output: 1

Approach: Create the prefix sum array and then use Sieve of Eratosthenes to count the number of primes in the prefix sum array.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of primes
// in the given array
int primeCount(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = *max_element(arr, arr + n);

    // 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;
        }
    }

    // Find all primes in arr[]
    int count = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            count++;

    return count;
}

// Function to generate the prefix array
void getPrefixArray(int arr[], int n, int pre[])
{

    // Fill the prefix array
    pre[0] = arr[0];
    for (int i = 1; i < n; i++) {
        pre[i] = pre[i - 1] + arr[i];
    }
}

// Driver code
int main()
{

    int arr[] = { 1, 4, 8, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Prefix array of arr[]
    int pre[n];
    getPrefixArray(arr, n, pre);

    // Count of primes in the prefix array
    cout << primeCount(pre, n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{ 
    
//returns the max element
static int max_element(int a[])
{
    int m = a[0];
    for(int i = 0; i < a.length; i++)
        m = Math.max(a[i], m);
    
    return m;
}

// Function to return the count of primes
// in the given array
static int primeCount(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = max_element(arr);

    // 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.
    boolean prime[] = new boolean[max_val + 1];
    for (int p = 0; p <= max_val; 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;
        }
    }

    // Find all primes in arr[]
    int count = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            count++;

    return count;
}

// Function to generate the prefix array
static int[] getPrefixArray(int arr[], int n, int pre[])
{

    // Fill the prefix array
    pre[0] = arr[0];
    for (int i = 1; i < n; i++)
    {
        pre[i] = pre[i - 1] + arr[i];
    }
    return pre;
}

// Driver code
public static void main(String args[])
{

    int arr[] = { 1, 4, 8, 4 };
    int n = arr.length;

    // Prefix array of arr[]
    int pre[]=new int[n];
    pre=getPrefixArray(arr, n, pre);

    // Count of primes in the prefix array
    System.out.println(primeCount(pre, n));

}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach 

# Function to return the count 
# of primes in the given array 
def primeCount(arr, n): 
 
    # Find maximum value in the array 
    max_val = max(arr) 

    # 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] = prime[1] = False
    p = 2
    while p * p <= max_val:  

        # If prime[p] is not changed, 
        # then it is a prime 
        if prime[p] == True:  

            # Update all multiples of p 
            for i in range(p * 2, max_val+1, p): 
                prime[i] = False
                
        p += 1
         
    # Find all primes in arr[] 
    count = 0 
    for i in range(0, n): 
        if prime[arr[i]]: 
            count += 1 

    return count 
 
# Function to generate the prefix array 
def getPrefixArray(arr, n, pre): 
 
    # Fill the prefix array 
    pre[0] = arr[0] 
    for i in range(1, n):  
        pre[i] = pre[i - 1] + arr[i] 

# Driver code 
if __name__ == "__main__":
 
    arr = [1, 4, 8, 4]  
    n = len(arr) 

    # Prefix array of arr[] 
    pre = [None] * n 
    getPrefixArray(arr, n, pre) 

    # Count of primes in the prefix array 
    print(primeCount(pre, n)) 

# This code is contributed by Rituraj Jain

// C# implementation of the approach
using System;

class GFG
{ 
    
// returns the max element
static int max_element(int[] a)
{
    int m = a[0];
    for(int i = 0; i < a.Length; i++)
        m = Math.Max(a[i], m);
    
    return m;
}

// Function to return the count of primes
// in the given array
static int primeCount(int[] arr, int n)
{
    // Find maximum value in the array
    int max_val = max_element(arr);

    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a bool 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];
    for (int p = 0; p <= max_val; 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;
        }
    }

    // Find all primes in arr[]
    int count = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            count++;

    return count;
}

// Function to generate the prefix array
static int[] getPrefixArray(int[] arr, int n, int[] pre)
{

    // Fill the prefix array
    pre[0] = arr[0];
    for (int i = 1; i < n; i++)
    {
        pre[i] = pre[i - 1] + arr[i];
    }
    return pre;
}

// Driver code
public static void Main()
{

    int[] arr = { 1, 4, 8, 4 };
    int n = arr.Length;

    // Prefix array of arr[]
    int[] pre = new int[n];
    pre = getPrefixArray(arr, n, pre);

    // Count of primes in the prefix array
    Console.Write(primeCount(pre, n));

}
}

// This code is contributed by ChitraNayal

PHP

<?php
// PHP implementation of the approach 

// Function to return the count of primes 
// in the given array 
function primeCount($arr, $n) 
{ 
    // Find maximum value in the array 
    $max_val = max($arr); 

    // 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; 
        } 
    } 

    // Find all primes in arr[] 
    $count = 0; 
    for ($i = 0; $i < $n; $i++) 
        if ($prime[$arr[$i]]) 
            $count++; 

    return $count; 
} 

// Function to generate the prefix array 
function getPrefixArray($arr, $n, $pre) 
{ 

    // Fill the prefix array 
    $pre[0] = $arr[0]; 
    for ($i = 1; $i < $n; $i++) 
    { 
        $pre[$i] = $pre[$i - 1] + $arr[$i]; 
    } 
    return $pre ;
} 

// Driver code 
$arr = array( 1, 4, 8, 4 ); 
$n = count($arr) ;

// Prefix array of arr[] 
$pre = array(); 
$pre = getPrefixArray($arr, $n, $pre); 

// Count of primes in the prefix array 
echo primeCount($pre, $n); 

// This code is contributed by AnkitRai01
?>

JavaScript

<script>

// Javascript implementation of the approach

// Function to return the count of primes
// in the given array
function primeCount(arr, n)
{
    // Find maximum value in the array
    let max_val = Math.max(...arr);

    // 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.
    let prime = new Array(max_val + 1).fill(true);

    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (let 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 (let i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }

    // Find all primes in arr[]
    let count = 0;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]])
            count++;

    return count;
}

// Function to generate the prefix array
function getPrefixArray(arr, n, pre)
{

    // Fill the prefix array
    pre[0] = arr[0];
    for (let i = 1; i < n; i++) {
        pre[i] = pre[i - 1] + arr[i];
    }
}

// Driver code

    let arr = [ 1, 4, 8, 4 ];
    let n = arr.length;

    // Prefix array of arr[]
    let pre = new Array(n);
    getPrefixArray(arr, n, pre);

    // Count of primes in the prefix array
    document.write(primeCount(pre, n));

</script>

Output:

Time Complexity: O(n + sqrt(max(arr))), where max(arr) is the largest element of the array arr.

Auxiliary Space: O(n + max(arr))

Find the sum of prime numbers in the Kth array

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Count the number of primes in the prefix sum array of the given array

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