Check whether the sum of prime elements of the array is prime or not
Given an array having N elements. The task is to check if the sum of prime elements of the array is prime or not.
Examples:
Input: arr[] = {1, 2, 3}
Output: Yes
As there are two primes in the array i.e. 2 and 3.
So, the sum of prime is 2 + 3 = 5 and 5 is also prime.
Input: arr[] = {2, 3, 2, 2}
Output: NoApproach: First find prime number up to 10^5 using Sieve. Then iterate over all elements of the array. If the number is prime then add it to sum. And finally, check whether the sum is prime or not. If prime then prints Yes otherwise No.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define ll long long int#define MAX 100000using namespace std;bool prime[MAX];// Sieve to find primevoid sieve(){ memset(prime, true, sizeof(prime)); prime[0] = prime[1] = false; for (int i = 2; i < MAX; i++) if (prime[i]) for (int j = 2 * i; j < MAX; j += i) prime[j] = false; }// Function to check if the sum of// prime is prime or notbool checkArray(int arr[], int n){ // find sum of all prime number ll sum = 0; for (int i = 0; i < n; i++) if (prime[arr[i]]) sum += arr[i]; // if sum is prime // then return yes if (prime[sum]) return true; return false;}// Driver codeint main(){ // array of elements int arr[] = { 1, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); sieve(); if (checkArray(arr, n)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the above approachimport java.io.*;class GFG { static int MAX =100000;static boolean prime[] = new boolean[MAX];// Sieve to find primestatic void sieve(){ for(int i=0;i<MAX;i++) { prime[i] =true; } prime[0] = prime[1] = false; for (int i = 2; i < MAX; i++) if (prime[i]) for (int j = 2 * i; j < MAX; j += i) prime[j] = false; }// Function to check if the sum of// prime is prime or notstatic boolean checkArray(int arr[], int n){ // find sum of all prime number int sum = 0; for (int i = 0; i < n; i++) if (prime[arr[i]]) sum += arr[i]; // if sum is prime // then return yes if (prime[sum]) return true; return false;}// Driver code public static void main (String[] args) { // array of elements int arr[] = { 1, 2, 3 }; int n = arr.length; sieve(); if (checkArray(arr, n)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed by shs.. |
Python3
# Python3 implementation of above approachfrom math import gcd, sqrtMAX = 100000prime = [True] * MAX# Sieve to find primedef sieve() : # 0 and 1 are not prime numbers prime[0] = False prime[1] = False for i in range(2, MAX) : if prime[i] : for j in range(2**i, MAX, i) : prime[j] = False # Function to check if the sum of# prime is prime or notdef checkArray(arr, n) : # find sum of all prime number sum = 0 for i in range(n) : if prime[arr[i]] : sum += arr[i] # if sum is prime # then return yes if prime[sum] : return True return False# Driver codeif __name__ == "__main__" : # list of elements arr = [1, 2, 3] n = len(arr) sieve() if checkArray(arr, n) : print("Yes") else : print("No") # This code is contributed by ANKITRAI1 |
C#
// C# implementation of the above approachusing System;class GFG{static int MAX = 100000;static bool[] prime = new bool[MAX];// Sieve to find primestatic void sieve(){ for(int i = 0; i < MAX; i++) { prime[i] = true; } prime[0] = prime[1] = false; for (int i = 2; i < MAX; i++) if (prime[i]) for (int j = 2 * i; j < MAX; j += i) prime[j] = false;}// Function to check if the sum of// prime is prime or notstatic bool checkArray(int[] arr, int n){ // find sum of all prime number int sum = 0; for (int i = 0; i < n; i++) if (prime[arr[i]]) sum += arr[i]; // if sum is prime // then return yes if (prime[sum]) return true; return false;}// Driver codepublic static void Main (){ // array of elements int[] arr = new int[] { 1, 2, 3 }; int n = arr.Length; sieve(); if (checkArray(arr, n)) Console.WriteLine("Yes"); else Console.WriteLine("No");}}// This code is contributed by mits |
PHP
<?php// PHP implementation of the// above approach// Sieve to find primefunction sieve(){ $MAX = 100000; $prime = array($MAX); for($i = 0; $i < $MAX; $i++) { $prime[$i] = true; } $prime[0] = $prime[1] = false; for ($i = 2; $i < $MAX; $i++) if ($prime[$i]) for ($j = 2 * $i; $j < $MAX; $j += $i) $prime[$j] = false;}// Function to check if the sum of// prime is prime or notfunction checkArray($arr, $n){ $prime = array(100000); // find sum of all prime number $sum = 0; for ($i = 0; $i < $n; $i++) if ($prime[$arr[$i]]) $sum += $arr[$i]; // if sum is prime // then return yes if ($prime[$sum]) return true; return false;}// Driver code$arr= array(1, 2, 3);$n = sizeof($arr);sieve();if (checkArray($arr, $n)) echo "Yes";else echo "No";// This code is contributed// by Akanksha Rai?> |
Javascript
<script>// JavaScript implementation of the// above approach// function check whether a number// is prime or notfunction isPrime(n){ // Corner case if (n <= 1) return 0; // Check from 2 to n-1 for (let i = 2; i < n; i++) if (n % i == 0) return 0; return 1;}var prime = new Array(5); // Sieve to find primefunction sieve(){ for(i = 0; i <=5; i++) { prime[i] = isPrime(i); }}// Function to check if the sum of// prime is prime or notfunction checkArray(arr, n){ // find sum of all prime number sum = 0; for (i = 0; i <= n; i++) if (prime[arr[i]]) sum += arr[i]; // if sum is prime // then return yes if (sum) return 1; return 0;}var arr= [1, 2, 3];n = 3;sieve();if (checkArray(arr, n)) document.write("Yes");else document.write("No");</script> |
Output:
Yes
Time Complexity: O(n * log(log n))
Auxiliary Space: O(MAX)
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