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Find longest subarray with Prime sum in given Array

  • Difficulty Level : Easy
  • Last Updated : 19 Sep, 2021

Given an array arr[], the task is to find the longest subarray whose sum is a prime number.

Examples:

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Input:   arr[ ] = {1, 4, 2, 1}
Output:  3
Explaination:  4+2+1=7 and 7 is a prime number thus the subarray we get is {4, 2, 1} .



Input: arr[ ] = {5, 2, 11, 4, 7, 19}
Output:  5
Explaination:  5+2+11+4+7=29 and 29 is a prime number thus the subarray we get is {5, 2, 11, 4, 7, 19} .

Approach: This problem can be solved by generating all subarrays and check for each subarray whether the sum is prime or not. For checking prime Sieve of Eratosthenes can be used. Any subarray can have a maximum sum equal to the sum of all the elements of the original array. Follow the steps to solve the given problem.

  • Create a variable of total_sum which stores the sum of all the elements in the arr[].
  • Do Sieve of Eratosthenes till total_sum because any subarray can’t have a sum greater than that.
  • Create a max_sum variable to keep track of the maximum subarray we get while generating all the subarrays.
  • Use two loops to find the sum of all the subarrays of arr[] and for each subarray, check whether the sum is prime or not.
  • Take maximum among those subarrays whose sum is prime.
  • Print max_sum as the required answer.

Below is the implementation of the above approach.

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to find whether number is
// prime or not
void SieveOfEratosthenes(
    vector<bool>& prime, int total_sum)
{
    // Create a boolean array "prime[0..n]" and
    // initialize all entries it as true.
    // A value in prime[i] will finally be false
    // if i is Not a prime, else true.
    for (int i = 0; i <= total_sum; i++) {
        prime[i] = true;
    }
 
    for (int p = 2; p * p <= total_sum; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p
            // greater than or equal to the square
            // of it numbers which are multiple of p
            // and are less than p^2 are already
            // been marked.
            for (int i = p * p; i <= total_sum; i += p)
                prime[i] = false;
        }
    }
}
 
int maxLenSubarrayWithSumPrime(
    vector<int>& arr, int n)
{
    // to store total_sum of original array
    int total_sum = 0;
 
    // calculate total_sum
    for (int i = 0; i < n; i++) {
        total_sum += arr[i];
    }
 
    // to store whether the number is
    // prime or not
    vector<bool> prime(total_sum + 1);
 
    // calling sieve to get prime values
    SieveOfEratosthenes(prime, total_sum);
 
    // to keep track of current and
    // maximum sum till now
    int max_sum = 0, cur_sum = 0;
    for (int i = 0; i < n; i++) {
        cur_sum = 0;
        for (int j = i; j < n; j++) {
            cur_sum += arr[j];
 
            // is current sum is prime
            if (prime[cur_sum]) {
                max_sum = max(max_sum, j - i + 1);
            }
        }
    }
 
    // return maximum sum founded.
    return max_sum;
}
 
// Driver Code
int main()
{
    int n = 6;
    vector<int> arr = { 5, 2, 11, 4, 7, 19 };
 
    cout << maxLenSubarrayWithSumPrime(arr, n);
}

Java




// Java program for above approach
import java.util.*;
 
class GFG{
 
// Utility function to find whether number is
// prime or not
static void SieveOfEratosthenes(
    boolean []prime, int total_sum)
{
   
    // Create a boolean array "prime[0..n]" and
    // initialize all entries it as true.
    // A value in prime[i] will finally be false
    // if i is Not a prime, else true.
    for (int i = 0; i <= total_sum; i++) {
        prime[i] = true;
    }
 
    for (int p = 2; p * p <= total_sum; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p
            // greater than or equal to the square
            // of it numbers which are multiple of p
            // and are less than p^2 are already
            // been marked.
            for (int i = p * p; i <= total_sum; i += p)
                prime[i] = false;
        }
    }
}
 
static int maxLenSubarrayWithSumPrime(
    int[] arr, int n)
{
    // to store total_sum of original array
    int total_sum = 0;
 
    // calculate total_sum
    for (int i = 0; i < n; i++) {
        total_sum += arr[i];
    }
 
    // to store whether the number is
    // prime or not
    boolean []prime = new boolean[total_sum + 1];
 
    // calling sieve to get prime values
    SieveOfEratosthenes(prime, total_sum);
 
    // to keep track of current and
    // maximum sum till now
    int max_sum = 0, cur_sum = 0;
    for (int i = 0; i < n; i++) {
        cur_sum = 0;
        for (int j = i; j < n; j++) {
            cur_sum += arr[j];
 
            // is current sum is prime
            if (prime[cur_sum]) {
                max_sum = Math.max(max_sum, j - i + 1);
            }
        }
    }
 
    // return maximum sum founded.
    return max_sum;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 6;
    int []arr = { 5, 2, 11, 4, 7, 19 };
 
    System.out.print(maxLenSubarrayWithSumPrime(arr, n));
}
}
 
// This code is contributed by shikhasingrajput.

Python3




# Python 3 program for above approach
 
from math import sqrt
# Utility function to find whether number is
# prime or not
def SieveOfEratosthenes(prime, total_sum):
    # Create a boolean array "prime[0..n]" and
    # initialize all entries it as true.
    # A value in prime[i] will finally be false
    # if i is Not a prime, else true.
    for i in range(total_sum+1):
        prime[i] = True
 
    for p in range(2,int(sqrt(total_sum)),1):
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == True):
 
            # Update all multiples of p
            # greater than or equal to the square
            # of it numbers which are multiple of p
            # and are less than p^2 are already
            # been marked.
            for i in range(p * p,total_sum+1,p):
                prime[i] = False
 
def maxLenSubarrayWithSumPrime(arr, n):
    # to store total_sum of original array
    total_sum = 0
 
    # calculate total_sum
    for i in range(n):
        total_sum += arr[i]
 
    # to store whether the number is
    # prime or not
    prime = [ False for i in range(total_sum + 1)]
 
    # calling sieve to get prime values
    SieveOfEratosthenes(prime, total_sum)
 
    # to keep track of current and
    # maximum sum till now
    max_sum = 0
    cur_sum = 0
    for i in range(n):
        cur_sum = 0
        for j in range(i,n,1):
            cur_sum += arr[j]
 
            # is current sum is prime
            if (prime[cur_sum]):
                max_sum = max(max_sum, j - i + 1)
 
    # return maximum sum founded.
    return max_sum
 
# Driver Code
if __name__ == '__main__':
    n = 6
    arr = [5, 2, 11, 4, 7, 19]
    print(maxLenSubarrayWithSumPrime(arr, n))
     
    # This code is contributed by SURENDRA_GANGWAR.

C#




// C# program for above approach
using System;
class GFG {
 
    // Utility function to find whether number is
    // prime or not
    static void SieveOfEratosthenes(bool[] prime,
                                    int total_sum)
    {
 
        // Create a boolean array "prime[0..n]" and
        // initialize all entries it as true.
        // A value in prime[i] will finally be false
        // if i is Not a prime, else true.
        for (int i = 0; i <= total_sum; i++) {
            prime[i] = true;
        }
 
        for (int p = 2; p * p <= total_sum; p++) {
 
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
 
                // Update all multiples of p
                // greater than or equal to the square
                // of it numbers which are multiple of p
                // and are less than p^2 are already
                // been marked.
                for (int i = p * p; i <= total_sum; i += p)
                    prime[i] = false;
            }
        }
    }
 
    static int maxLenSubarrayWithSumPrime(int[] arr, int n)
    {
        // to store total_sum of original array
        int total_sum = 0;
 
        // calculate total_sum
        for (int i = 0; i < n; i++) {
            total_sum += arr[i];
        }
 
        // to store whether the number is
        // prime or not
        bool[] prime = new bool[total_sum + 1];
 
        // calling sieve to get prime values
        SieveOfEratosthenes(prime, total_sum);
 
        // to keep track of current and
        // maximum sum till now
        int max_sum = 0, cur_sum = 0;
        for (int i = 0; i < n; i++) {
            cur_sum = 0;
            for (int j = i; j < n; j++) {
                cur_sum += arr[j];
 
                // is current sum is prime
                if (prime[cur_sum]) {
                    max_sum = Math.Max(max_sum, j - i + 1);
                }
            }
        }
 
        // return maximum sum founded.
        return max_sum;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 6;
        int[] arr = { 5, 2, 11, 4, 7, 19 };
 
        Console.WriteLine(
            maxLenSubarrayWithSumPrime(arr, n));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Utility function to find whether number is
        // prime or not
        function SieveOfEratosthenes(
            prime, total_sum)
        {
         
            // Create a boolean array "prime[0..n]" and
            // initialize all entries it as true.
            // A value in prime[i] will finally be false
            // if i is Not a prime, else true.
            for (let i = 0; i <= total_sum; i++) {
                prime[i] = true;
            }
 
            for (let p = 2; p * p <= total_sum; p++) {
 
                // If prime[p] is not changed,
                // then it is a prime
                if (prime[p] == true) {
 
                    // Update all multiples of p
                    // greater than or equal to the square
                    // of it numbers which are multiple of p
                    // and are less than p^2 are already
                    // been marked.
                    for (let i = p * p; i <= total_sum; i += p)
                        prime[i] = false;
                }
            }
            return prime;
        }
 
        function maxLenSubarrayWithSumPrime(
            arr, n)
        {
         
            // to store total_sum of original array
            let total_sum = 0;
 
            // calculate total_sum
            for (let i = 0; i < n; i++) {
                total_sum += arr[i];
            }
 
            // to store whether the number is
            // prime or not
            let prime = new Array(total_sum + 1);
            // calling sieve to get prime values
            prime = SieveOfEratosthenes(prime, total_sum);
 
            // to keep track of current and
            // maximum sum till now
            let max_sum = 0, cur_sum = 0;
            for (let i = 0; i < n; i++) {
                cur_sum = 0;
                for (let j = i; j < n; j++) {
                    cur_sum += arr[j];
 
                    // is current sum is prime
                    if (prime[cur_sum]) {
                        max_sum = Math.max(max_sum, j - i + 1);
                    }
                }
            }
 
            // return maximum sum founded.
            return max_sum;
        }
 
        // Driver Code
        let n = 6;
        let arr = [5, 2, 11, 4, 7, 19];
 
        document.write(maxLenSubarrayWithSumPrime(arr, n));
 
     // This code is contributed by Potta Lokesh
 
    </script>

 
 

Output
5

Time Complexity: O(N^2) 
Auxiliary Complexity: O(N)

 




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