Find the sum of non-prime elements in the given array
Last Updated :
29 Nov, 2022
Given an array arr[] and the task is to print the sum of the non-prime elements from the array.
Examples:
Input: arr[] = {1, 3, 7, 4, 9, 8}
Output: 22
Non-prime elements are {1, 4, 9, 8} and 1 + 4 + 9 + 8 = 22
Input: arr[] = {11, 4, 10, 7}
Output: 14
Approach: Initialize sum = 0 and start traversing the array element by element, if current element is not a prime then update sum = sum + arr[i]. Print the sum in the end. Primality can be optimally tested using Sieve of Eratosthenes.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int nonPrimeSum( int arr[], int n)
{
int max_val = *max_element(arr, arr + n);
vector< bool > prime(max_val + 1, true );
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= max_val; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= max_val; i += p)
prime[i] = false ;
}
}
int sum = 0;
for ( int i = 0; i < n; i++)
if (!prime[arr[i]])
sum += arr[i];
return sum;
}
int main()
{
int arr[] = { 1, 3, 7, 4, 9, 8 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << nonPrimeSum(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int max_element( int arr[])
{
int max_e = Integer.MIN_VALUE;
for ( int i = 0 ; i < arr.length; i++)
{
max_e = Math.max(max_e, arr[i]);
}
return max_e;
}
static int nonPrimeSum( int arr[], int n)
{
int max_val = max_element(arr);
boolean prime[] = new boolean [max_val + 1 ];
for ( int i = 0 ; i < prime.length; i++)
prime[i] = true ;
prime[ 0 ] = false ;
prime[ 1 ] = false ;
for ( int p = 2 ; p * p <= max_val; p++)
{
if (prime[p] == true )
{
for ( int i = p * 2 ; i <= max_val; i += p)
prime[i] = false ;
}
}
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
if (!prime[arr[i]])
sum += arr[i];
return sum;
}
public static void main(String args[])
{
int arr[] = { 1 , 3 , 7 , 4 , 9 , 8 };
int n = arr.length;
System.out.println( nonPrimeSum(arr, n));
}
}
|
Python3
from math import sqrt
def nonPrimeSum(arr, n) :
max_val = max (arr)
prime = [ True ] * (max_val + 1 )
prime[ 0 ] = False
prime[ 1 ] = False
for p in range ( 2 , int (sqrt(max_val)) + 1 ) :
if (prime[p] = = True ) :
for i in range (p * 2 , max_val + 1 , p) :
prime[i] = False
sum = 0
for i in range ( 0 , n) :
if ( not prime[arr[i]]) :
sum + = arr[i]
return sum
if __name__ = = "__main__" :
arr = [ 1 , 3 , 7 , 4 , 9 , 8 ]
n = len (arr)
print (nonPrimeSum(arr, n))
|
C#
using System;
class GFG
{
static int max_element( int [] arr)
{
int max_e = int .MinValue;
for ( int i = 0; i < arr.Length; i++)
{
max_e = Math.Max(max_e, arr[i]);
}
return max_e;
}
static int nonPrimeSum( int [] arr, int n)
{
int max_val = max_element(arr);
bool [] prime = new bool [max_val + 1];
for ( int i = 0; i < prime.Length; i++)
prime[i] = true ;
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= max_val; p++)
{
if (prime[p] == true )
{
for ( int i = p * 2;
i <= max_val; i += p)
prime[i] = false ;
}
}
int sum = 0;
for ( int i = 0; i < n; i++)
if (!prime[arr[i]])
sum += arr[i];
return sum;
}
public static void Main()
{
int [] arr = { 1, 3, 7, 4, 9, 8 };
int n = arr.Length;
Console.WriteLine(nonPrimeSum(arr, n));
}
}
|
PHP
<?php
function nonPrimeSum( $arr , $n )
{
$max_val = max( $arr );
$prime = array_fill (0, $max_val + 1, true);
$prime [0] = false;
$prime [1] = false;
for ( $p = 2; $p * $p <= $max_val ; $p ++)
{
if ( $prime [ $p ] == true)
{
for ( $i = $p * 2;
$i <= $max_val ; $i += $p )
$prime [ $i ] = false;
}
}
$sum = 0;
for ( $i = 0; $i < $n ; $i ++)
if (! $prime [ $arr [ $i ]])
$sum += $arr [ $i ];
return $sum ;
}
$arr = array ( 1, 3, 7, 4, 9, 8 );
$n = count ( $arr );
echo nonPrimeSum( $arr , $n );
?>
|
Javascript
<script>
function max_element(arr)
{
let max_e = Number.MIN_VALUE;
for (let i = 0; i < arr.length; i++)
{
max_e = Math.max(max_e, arr[i]);
}
return max_e;
}
function nonPrimeSum(arr,n)
{
let max_val = max_element(arr);
let prime = new Array(max_val + 1);
for (let i = 0; i < prime.length; i++)
prime[i] = true ;
prime[0] = false ;
prime[1] = false ;
for (let p = 2; p * p <= max_val; p++)
{
if (prime[p] == true )
{
for (let i = p * 2; i <= max_val; i += p)
prime[i] = false ;
}
}
let sum = 0;
for (let i = 0; i < n; i++)
if (!prime[arr[i]])
sum += arr[i];
return sum;
}
let arr=[1, 3, 7, 4, 9, 8 ];
let n = arr.length;
document.write( nonPrimeSum(arr, n));
</script>
|
Time Complexity: O(max(n, m * log(log(m)))), where n is the size of the array and m is the maximum element in the array.
Auxiliary Space: O(m), where m is the maximum element in the array.
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