Given an array arr[], the task is to find the longest subarray whose sum is a prime number.
Examples:
Input: arr[ ] = {1, 4, 2, 1}
Output: 3
Explanation: 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
Explanation: 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 notvoid 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 Codeint main(){ int n = 6; vector<int> arr = { 5, 2, 11, 4, 7, 19 }; cout << maxLenSubarrayWithSumPrime(arr, n);} |
Java
// Java program for above approachimport java.util.*;class GFG{// Utility function to find whether number is// prime or notstatic 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 Codepublic 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 approachfrom math import sqrt# Utility function to find whether number is# prime or notdef 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] = Falsedef 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 Codeif __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 approachusing 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(S*log(log(S)) + N2), where S and N are the total sum and length of the given array arr[] respectively.
Auxiliary Space: O(S)
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