Sub array sum is prime or not
Last Updated :
16 Nov, 2022
Given an array and limits (lower and upper limits), check the sum of the subarray in the given limit is prime or not
Examples :
Input : a[] = {1, 2, 3, 5, 5, 4, 7, 8, 9};
lower = 3, upper = 6
Output : Yes
Explanation:- subarray is {3, 5, 5, 4} and
sum of subarray 3+5+5+4 = 17 which is prime, so
the output is yes
Input : a[] = {1, 6, 4, 5, 5, 4, 7, 8, 9};
lower = 2, upper = 5
Output : No
Explanation:- subarray is {6, 4, 5, 5} and sum
of subarray 6+4+5+5 = 20 which is Not prime so the
output is No
- First calculate the sum of sub-array using upper limit and lower limit
- Then check the sum is prime or not.
- If it is prime then return true otherwise return false.
Let’s understand this approach using code below.
C++
#include <iostream>
using namespace std;
bool isPrime( int a[], int lower,
int upper)
{
int n = 0;
for ( int i = lower - 1;
i <= upper - 1;
i++)
n += a[i];
if (n <= 1)
return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
int main()
{
int a[] = { 1, 2, 3, 5, 5,
4, 7, 8, 9 };
int lower = 3, upper = 6;
if (isPrime(a, lower, upper))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
|
Java
import java.io.*;
public class GFG {
static boolean isPrime( int a[],
int lower,
int upper)
{
int n = 0 ;
for ( int i = lower - 1 ;
i <= upper - 1 ; i++)
n += a[i];
if (n <= 1 )
return false ;
for ( int i = 2 ; i < n; i++)
if (n % i == 0 )
return false ;
return true ;
}
public static void main(String[] args)
{
int a[] = { 1 , 2 , 3 , 5 , 5 , 4 , 7 , 8 , 9 };
int lower = 3 , upper = 6 ;
if (isPrime(a, lower, upper))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def isPrime(a, lower, upper) :
n = 0
for i in range (lower - 1 , upper) :
n = n + a[i]
if (n < = 1 ) :
return False
for i in range ( 2 , n) :
if (n % i = = 0 ) :
return False
return True
a = [ 1 , 2 , 3 , 5 , 5 , 4 , 7 , 8 , 9 ]
lower = 3
upper = 6
if (isPrime(a, lower, upper)) :
print ( "Yes" )
else :
print ( "No" )
|
C#
using System;
class GFG {
static bool isPrime( int [] a,
int lower,
int upper)
{
int n = 0;
for ( int i = lower - 1;
i <= upper - 1;
i++)
n += a[i];
if (n <= 1)
return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
public static void Main()
{
int [] a = { 1, 2, 3, 5, 5,
4, 7, 8, 9 };
int lower = 3, upper = 6;
if (isPrime(a, lower, upper))
Console.Write( "Yes" );
else
Console.Write( "No" );
}
}
|
PHP
<?php
function isPrime( $a , $lower , $upper )
{
$n = 0;
for ( $i = $lower - 1;
$i <= $upper - 1; $i ++)
$n += $a [ $i ];
if ( $n <= 1)
return false;
for ( $i = 2; $i < $n ; $i ++)
if ( $n % $i == 0)
return false;
return true;
}
$a = array (1, 2, 3, 5, 5,
4, 7, 8, 9);
$lower = 3; $upper = 6;
if (isPrime( $a , $lower , $upper ))
echo "Yes" , " \n" ;
else
echo "No" , " \n" ;
?>
|
Javascript
<script>
function isPrime(a, lower, upper)
{
let n = 0;
for (let i = lower - 1;
i <= upper - 1; i++)
n += a[i];
if (n <= 1)
return false ;
for (let i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
let a = [ 1, 2, 3, 5, 5, 4, 7, 8, 9 ];
let lower = 3, upper = 6;
if (isPrime(a, lower, upper))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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